12INFINITE SEQUENCES AND SERIESINFINITE SEQUENCES AND SERIES

Infinite sequences and series were

introduced briefly in A Preview of Calculus

in connection with Zenorsquos paradoxes and

the decimal representation of numbers

INFINITE SEQUENCES AND SERIES

Their importance in calculus stems from

Newtonrsquos idea of representing functions as

sums of infinite series

For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series

INFINITE SEQUENCES AND SERIES

We will pursue his idea in Section 1210

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in

mathematical physics and chemistry

such as Bessel functions are defined

as sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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Infinite sequences and series were

introduced briefly in A Preview of Calculus

in connection with Zenorsquos paradoxes and

the decimal representation of numbers

INFINITE SEQUENCES AND SERIES

Their importance in calculus stems from

Newtonrsquos idea of representing functions as

sums of infinite series

For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series

INFINITE SEQUENCES AND SERIES

We will pursue his idea in Section 1210

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in

mathematical physics and chemistry

such as Bessel functions are defined

as sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

Their importance in calculus stems from

Newtonrsquos idea of representing functions as

sums of infinite series

For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series

INFINITE SEQUENCES AND SERIES

We will pursue his idea in Section 1210

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in

mathematical physics and chemistry

such as Bessel functions are defined

as sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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We will pursue his idea in Section 1210

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in

mathematical physics and chemistry

such as Bessel functions are defined

as sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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Many of the functions that arise in

mathematical physics and chemistry

such as Bessel functions are defined

as sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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Physicists also use series in another

way as we will see in Section 1211

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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121Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

SEQUENCE

A sequence can be thought of as a list

of numbers written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term

SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

We will deal exclusively with infinite

sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

Notice that for every positive integer n

there is a corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

However we usually write an instead

of the function notation f(n) for the value

of the function at the number n

SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

The sequence a1 a2 a3 is also

denoted by

1orn n n

a a

Notation

SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

Some sequences can be

defined by giving a formula

for the nth term

Example 1

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

• Slide 1
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SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

In this and the subsequent examples notice that n doesnrsquot have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

• Slide 1
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• Slide 126

SEQUENCES

Find a formula for the general term an

of the sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

Notice that the numerators of these fractions

start with 3 and increase by 1 whenever we

go to the next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

The denominators are the powers

of 5

Thus an has denominator 5n

Example 2

SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

The signs of the terms are alternately

positive and negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

• Slide 1
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• Slide 126

SEQUENCES

In Example 1 b the factor (ndash1)n meant

we started with a negative term

Here we want to start with a positive term

Example 2

SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

Thus we use (ndash1)nndash1 or (ndash1)n+1

Therefore

1 2( 1)

5n

n n

na

Example 2

SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

We now look at some sequences

that donrsquot have a simple defining

equation

Example 3

SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

The sequence pn where pn

is the population of the world as

of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

• Slide 1
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SEQUENCES

If we let an be the digit in the nth decimal place

of the number e then an is a well-defined

sequence whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined

recursively by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n ge 3

Each is the sum of the two preceding term terms

The first few terms are 1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 124
• Slide 125
• Slide 126

FIBONACCI SEQUENCE

This sequence arose when the 13th-century

Italian mathematician Fibonacci solved a

problem concerning the breeding of rabbits

See Exercise 71

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can

be pictured either

by Plotting its terms on

a number line Plotting its graph

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

• Slide 1
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• Slide 125
• Slide 126

SEQUENCES

Note that since

a sequence is

a function whose

domain is the set

of positive integers

its graph consists of isolated points with

coordinates

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

Fig 1212 p 712

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

Fig 1211 p 712

Fig 1212 p 712

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

SEQUENCES

In general the notation

means that the terms of the sequence an

approach L as n becomes large

lim nna L

SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SEQUENCES

Notice that the following definition of

the limit of a sequence is very similar to

the definition of a limit of a function at infinity

given in Section 26

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L

as we like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nna

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing

the graphs of two sequences that have

the limit L

Fig 1213 p 713

LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

A more precise version of

Definition 1 is as follows

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if for every ε gt 0 there is a corresponding

integer N such that

if n gt N then |an ndash L| lt ε

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure

in which the terms a1 a2 a3 are plotted

on a number line

Fig 1214 p 713

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε)

is chosen there exists an N such that all

terms of the sequence from aN+1 onward must

lie in that interval

Fig 1214 p 713

LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

Another illustration of Definition 2

is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

Fig 1215 p 713

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 125
• Slide 126

LIMIT OF A SEQUENCE

This picture must be valid no matter how

small ε is chosen

Usually however a smaller ε requires a larger N

Fig 1215 p 713

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7

in Section 26 you will see that the only

difference between and

is that n is required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If and f(n) = an when n

is an integer then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Theorem 3 is illustrated here

Fig 1216 p 714

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 (Theorem 5

in Section 26) we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If an becomes large as n becomes large

we use the notation

The following precise definition is similar to Definition 9 in Section 26

lim nna

LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT OF A SEQUENCE

means that for every positive

number M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nna

LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If then the sequence an

is divergent but in a special way

We say that an diverges to infin

lim nna

LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The Limit Laws given in Section 23

also hold for the limits of sequences

and their proofs are similar

LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT LAWS FOR SEQUENCES

Suppose an and bn are convergent

sequences and

c is a constant

LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT LAWS FOR SEQUENCES

Then

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

LIMITS OF SEQUENCES

The Squeeze Theorem

can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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SQUEEZE THEOREM FOR SEQUENCES

If an le bn le cn for n ge n0

and

thenlim nnb L

lim limn nn na c L

Fig 1217 p 715

LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Another useful fact about limits of

sequences is given by the following

theorem

The proof is left as Exercise 75

LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Theorem 6

lim 0nn

a

lim 0nna

If

then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 124
• Slide 125
• Slide 126

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

LIMITS OF SEQUENCES

Here we canrsquot apply lrsquoHospitalrsquos Rule

directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule

to the related function f(x) = (ln x)x and

obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Therefore by Theorem 3

we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Determine whether

the sequence an = (ndash1)n

is convergent or divergent

Example 6

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

LIMITS OF SEQUENCES

If we write out the terms of the sequence

we obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

Fig 1218 p 715

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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• Slide 126

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)nn

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

Fig 1219 p 716

LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The following theorem says that if we

apply a continuous function to the terms

of a convergent sequence the result is also

convergent

LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If and the function f is

continuous at L then

The proof is left as Exercise 76

Theorem 7

lim nna L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Discuss the convergence

of the sequence an = nnn

where n = 1 2 3 hellip n

Example 9

LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Both the numerator and denominator

approach infinity as n rarr infin

However here we have no corresponding

function for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Letrsquos write out a few terms to get a feeling

for what happens to an as n gets large

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Therefore

1 2 3n

na

n n n n

E g 9mdashEquation 8

LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

From these expressions and the graph here

it appears that the terms are decreasing and

perhaps approach 0

Example 9

Fig 12110 p 716

LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

To confirm this observe from Equation 8

that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Thus 0 lt an le

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

1

n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From Section 26 and the graphs of the exponential functions in Section 15 we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3

we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

If r le ndash1 then r n diverges as in

Example 6

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The figure shows the graphs for various

values of r

Example 10

Fig 12111 p 717

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

Fig 1218 p 715

LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The results of Example 10

are summarized for future use

as follows

LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

The sequence r n is convergent

if ndash1 lt r le 1 and divergent for all other

values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n ge 1 that is a1 lt a2 lt a3 lt

Decreasing if an gt an+1 for all n ge 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

The sequence is decreasing

because

and so an gt an+1 for all n ge 1

Example 11

3

5n

3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

We must show that an+1 lt an

that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

This inequality is equivalent to the one

we get by cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

Since n ge 1 we know that the inequality

n2 + n gt 1 is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

Consider the function2

( ) 1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f(n) gt f(n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an le M for all n ge 1

Below if there is a number m such that m le an for all n ge 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

We know that not every bounded

sequence is convergent

For instance the sequence an = (ndash1)n satisfies ndash1 le an le 1 but is divergent from Example 6

BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

Similarly not every

monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

However if a sequence is both bounded

and monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

If an is increasing and an le M for all n

then the terms are forced to crowd together

and approach some number L

Fig 12112 p 718

BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

The proof of Theorem 12 is based on

the Completeness Axiom for the set

of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x le M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b le M

BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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BOUNDED SEQUENCES

The Completeness Axiom is

an expression of the fact that there is

no gap or hole in the real number line

Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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Every bounded monotonic

sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Suppose an is an increasing

sequence

Since an is bounded the set S = an|n ge 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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However the sequence is increasing

So an ge aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nna L

MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

A similar proof (using

the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that

a sequence that is increasing and bounded

above is convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

This fact is used many times in

dealing with infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Investigate the sequence an defined by

the recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

We begin by computing the first several

terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing

we use mathematical induction to show

that an+1 gt an for all n ge 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

That is true for n = 1 because

a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true

for n = k + 1

Therefore the inequality is true for all n

by induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Next we verify that an is bounded by

showing that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an ge a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

This shows by mathematical

induction that an lt 6 for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Since the sequence an is increasing

and bounded Theorem 12 guarantees

that it has a limit

However the theorem doesnrsquot tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Nevertheless now that we know

exists we can use the recurrence relation

to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr infin n + 1 rarr infin too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

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