Math 104Calculus I

SEQUENCES and

INFINITE SERIES

Sequences

The lists of numbers you generate using a numerical method like Newton's method to get better and better approximations to the root of an equation are examples of (mathematical) sequences .

Sequences are infinite lists of numbers, Sometimes it is useful to think of them as functions from the positive integers into the reals, in other words,

,...,, 321 aaa

forth. so and ,a(2) ,a(1) 21 aa

The feeling we have about numerical methods like the bisection method, is that if we kept doing it more and more times, we would get numbers that are closer and closer to the actual root of the equation. In other words

where r is the root.

Sequences for which exists

and is finite are called convergent, other sequences are called divergent

Convergent and Divergent

nn

a lim

rann

lim

For example...

The sequence

1, 1/2, 1/4, 1/8, 1/16, .... , 1/2 , ... is convergent (and converges to zero, since

), whereas:

the sequence 1, 4, 9, 16, .…n , ... is divergent.

2

n

0 lim21

n

n

Practice

The sequence

...,2

1,......,

5

4,

4

3,

3

2

n

n

A. Converges to 0

B. Converges to 1

C. Converges to n

D. Converges to ln 2

E. Diverges

Another...

The sequence

...,2

1)1(,......,

5

4,

4

3,

3

2

n

nn

A. Converges to 0

B. Converges to 1

C. Converges to -1

D. Converges to ln 2

E. Diverges

A powerful existence theoremIt is sometimes possible to assert that a sequence is convergent even if we can't find the limit right away. We do this by using the least upper bound property of the real numbers:

If a sequence has the property that a <a <a < .... is called a "monotonically increasing" sequence. For such a sequence, either the sequence is bounded (all the terms are less than some fixed number) or else it increases without bound to infinity. The latter case is divergent, and the former must converge to the least upper bound of the set of numbers {a , a , ... } . So if we find some upper bound, we are guaranteed convergence, even if we can't find the least upper bound.

1 2 3

1 2

Consider the sequence...

222,22 ,2

To get each term from the previous one, you add 2 and then take the square root.

It is clear that this is a monotonically increasing sequence. It is convergent because all the terms are less than 2. To see this, note that if x>2, then

So our terms can't be greater than 2, since adding 2 and taking the square root makes our terms bigger, not smaller.

Therefore, the sequence has a limit, by the theorem.

etc.

.2 so and ,222 xxxxx

QUESTION:

What is the limit?

...,222,22 ,2

Series of Constants

We’ve looked at limits and sequences. Now, we look at a specific kind of sequential limit, namely the limit (or sum) of a series.

Zeno’s Paradox

How can an infinite number of things happen in a finite amount of time?

(Zeno's paradox concerned Achilles and a tortoise)

Discussion Questions

1. Is Meg Ryan’s reasoning correct? If it isn't, what is wrong with it?

2. If the ball bounces an infinite number of times, how come it stops? How do you figure out the total distance traveled by the ball?

Resolution

The resolution of these problems is accomplished by the use of limits.

In particular, each is resolved by understanding why it is possible to "add together" an infinite number of numbers and get a finite sum.

An example

Meg Ryan worried about adding together

...16

1

8

1

4

1

2

1

Picture ThisThe picture suggests that

the "infinite sum"

should be 1. This is in fact true, but requires some proof.

We'll provide the proof, but in a more general context.

...16

1

8

1

4

1

2

1

The idea of a seriesA "series" is any "infinite sum" of numbers. Usually there is some pattern to the numbers, so we can give an idea of the pattern either by giving the first few numbers, or by giving an actual formula for the nth number in the list. For example, we could write

n

nn

1n1 2

1 asor

2

1 as ...

16

1

8

1

4

1

2

1

The things being added together are called “terms” of the series.

Other series we will consider...

1). equal todefined is 0! (since !

1or ...,

!4

1

!3

1

!2

1

1

11

1

1 as recognize couldyou which ...,

30

1

20

1

12

1

6

1

2

1

series" harmonic galternatin" thecalled is this1

or ...4

1

3

1

2

11

series" harmonic" thecalled sometimes is this1

or ...4

1

3

1

2

11

0

1

1

1

1

n

n

n

n

n

n

nn

n

n

Two obvious questions

1. Does the series have a sum? (Officially: "Does the series converge?")

2. What is the sum? (Officially: "What does the series converge to?")

A less obvious question is...

3. How fast does the series converge?

Convergence

The word convergence suggests a limiting process. Fortunately, we don't have to invent a new kind of limit for series.

Think of series as a process of adding together the terms starting from the beginning. Then the nth "partial sum" of the series is simply the sum of the first n terms of the series.

For example...

the partial sums of the IQ series are:

1st partial sum = 1/2

2nd partial sum = 1/2 + 1/4 = 3/4

3rd partial sum = 1/2 + 1/4 + 1/8 = 7/8

and so forth.

It looks line the nth partial sum of the IQ series is

n

n

2

12

It is only natural

It is natural to define (and this is even the official definition!) the sum or limit of the series to be equal to the limit of the sequence of its partial sums, if the latter limit exists.

For the IQ series, we really do have:

12

12 lim lim n

ns

nn

n

This bears out our earlier suspicion.

This presents a problem...

The problem is that it is often difficult or impossible to get an explicit expression for the partial sums of a series.

So, as with integrals, we'll learn a few basic examples, and then do the best we can -- sometimes only answering question 1, other times managing 1 and 2, and still other times 1, 2, and 3.

Geometric series

The IQ series is a specific example of a geometric series .

A geometric series has terms that are (possibly a constant times) the successive powers of a number.

The IQ series has successive powers of 1/2.

Other examples

5

0

0

1

1

2

3...

256

3

128

3

64

3

32

3

7

15...

343

5

49

5

7

55

)4(3...19248123

10

13...

10000

3

1000

3

100

3

10

3...3333333.0

1...1111

nn

n

n

n

n

n

n

n

n

Convergence of geometric series

Start (how else?) with partial sums

Finite geometric sum:

Therefore

and so

)1(1()1( nn raSr

nn

nn

rSararararar

Sarararara

)1(432

32

...

...

r

raS

n

n

1

)1( )1(

We conclude that...

otherwise. diverges and,1r

whenprecisely converges series geometric theTherefore

otherwise.exist not does and ,1r if 1 toequal is lim

r

aSn

n

connect

Which of the geometric series on the previous slide (reproduced on the next slide) converge?

What do they converge to?

Some questions

Other examples

5

0

0

1

1

2

3...

256

3

128

3

64

3

32

3

7

15...

343

5

49

5

7

55

)4(3...19248123

10

13...

10000

3

1000

3

100

3

10

3...3333333.0

1...1111

nn

n

n

n

n

n

n

n

n

Telescoping series

Another kind of series that we can sum: telescoping series

This seems silly at first, but it's not!

A series is said to telescope if all the terms in the partial sums cancel except perhaps for the first and the last.

Example

1. toconverges series theso

,1n

1-1 is series thisof sum partialnth eClearly th

...)1

11...()

4

1

3

1()

3

1

2

1()

2

11(

nn

What’s the big deal?

Well, you could rewrite the series as

which is not so obvious (in fact, it was one of the examples given near the beginning of today’s class).

...)1(

1...

12

1

6

1

2

1

nn

Now you try one...

A) 1

B) 3/4

C) 1/2

D) 1/4

E) 1/8

?1

1 series theof sum theisWhat

2k2

k

Occasionally it helps to recognize a series as a

telescoping series. One important example of such a

series is provided by improper integrals.

Suppose F '(x) = f(x). Then we can think of the

improper integral

as being the sum of the series

Improper integrals

1)(f dxx

...F(3))-(F(4)F(2))-(F(3)F(1))-(F(2)

...)(f)(f)(f4

3

3

2

2

1

dxxdxxdxx

Continued...Since the nth partial sum of this series is F(n+1) - F(1), it's

clear that the series converges to

just as the integral would

be equal to

(Note the subtle difference between the two limits -- the limit of the series might exist even when the improper integral does not).

F(1)))(Flim(

nn

F(1)))(Flim(

xx

The convergence question

For a while, we’ll concentrate on the question

1: Does the series converge?

One obvious property that convergent series must have is that their terms must get smaller and smaller in order for the limit of the partial sums to exist.

Fundamental necessary condition for convergence:

0lim unless convergecannot seriesA 1

n

nn

n aa

This is only a test you can use to prove that a series does

NOT converge

111 ))arctan( does as diverges, (e.g.,

nnnn n

HarmonicJust because the nth term goes to zero doesn't mean that

the series converges. An important example is the harmonic series

1

1

nn

We can show that the harmonic series diverges by the following argument using the partial sums:

25

81

81

81

81

24

81

71

61

51

41

31

21

8

24

41

41

23

41

31

21

4

23

21

2

1

1

1

1

1

S

S

S

SFor the harmonic series,

Harmonic (cont.)

and so on -- every time we double the number of terms, we add at least one more half. This indicates (and by induction we could prove) that

diverges. series harmonic theso ,lim

thatcase y thenecessaril isit so ,23

2

nn

n

S

S n

Cantilever tower:The divergence of the harmonic series makes the following trick possible. It is possible to stack books (or cards, or any other kind of stackable, identical objects) near the edge of a table so that the top object is completely off the table (and as far off as one wishes, provided you have enough objects to stack).

Series of positive termsConvergence questions for series of positive terms are easiest

to understand conceptually.

Since all the terms a are assumed to be positive, the sequence of partial sums {S } must be an increasing sequence.

So the least upper bound property discussed earlier comes into play -- either the sequence of partial sums has an upper bound or it doesn't.

If the sequence of partial sums is bounded above, then it must converge and so will the series. If not, then the series diverges. That's it.

nn

Tests for convergence of series of positive terms:

The upper bound observations give rise to several "tests" for convergence of series of positive terms. They all are based pretty much on common sense ways to show that the partial sums of the series being tested is bounded are all less than those of a series that is known to converge (or greater than those of a series that is known to diverge). The names of the tests we will discuss are...

Tests...

1. The integral test

2. The comparison test

3. The ratio test

4. The limit comparison test (sometimes called the ratio comparison test)

5. The root test

TODAY

TODAY

The integral test

Since improper integrals of the form

provide us with many examples of telescoping series whose convergence is readily determined, we can use integrals to determine convergence of series:

1dx)x(f

For example, consider the series

From the following picture, it is evident that the nth partial sum of this series is less than

Integral test cont.

1

12

nn

n

x1 dx

1 21

What is the sum?The sum of the terms is equal to the sum of the areas

of the shaded rectangles, and if we start integrating at 1 instead of 0, the

improper integral converges

(question: what is the integral? so what bound to you conclude for the series?).

Since the value of the improper integral (plus 1) provides us with an upper bound for all of the partial sums, the series must converge.

It is an interesting question as to exactly what the sum is. We will answer it next week.

1 2 dxx1

The integral test...

11

1

1

integral the

of divergence knew thealready we-because-series

harmonic theof divergence theof proofeasier

new,a us gives This integral). for the checked be

toneedsinfinity at econvergenc(only diverge

bothor converge botheither f integral

improper theand f series the,decreasing

and positive is f(x) function theif that says

.dx

dx)x(

)x(

x

x

Discussion and Connect

-- for which exponents p does the series

converge?

(These are sometimes called p-series, for obvious reasons -- these together with the geometric series give us lots of useful examples of series whose convergence or divergence we know).

1

1

nn p

Question…

Error estimates:Using the picture that proves the integral test for

convergent series, we can get an estimate on how far off we are from the limit of the series if we stop adding after N terms for any finite value of N.

If we approximate the convergent series

by the partial sum

then the error we commit is less than the value of the integral

1

)(fn

n

N

nN ns

1

)(f

Ndxx)(f

Take a closer look...

1.6n bigger tha little a is sum actual

the--offfar t isn' estimate thislaer, see shall we

(As . sum infitie thefrom 0.2,or ,

thanlessby differs This 1.46.ely approximat is

which,1 sum theexample,For

1

151

5

1

36005269

251

161

91

41

22

nnx

dx

Question

A) Converge

B) Diverge

diverge?or converge series theDoes1

1 2

nnn

Question

A) Converge

B) Diverge

diverge?or converge series theDoes1

1

)arctan(2

nn

n

Connect

For this latter series, find a bound on the error if we use the sum of the first 100 terms to approximate the limit. (answer: it is less than about .015657444)

Exercise

The comparison testThis convergence test is even more common-

sensical than the integral test. It says that if

all the terms of the series are less than

the corresponding terms of the series

and if converges, then

converges also.

1nna

1nnb

1nna

1nnb

Reverse

This test can also be used in reversed -- if

the b series diverges and the a’s are bigger

than the corresponding b’s, then

diverges also.

1nna

Examples:

diverges.

converges.

1)sin(

12

1

nnn

n

nnn

Question

diverge?or converge series theDoes5

21

kk

A) Converge

B) Diverge

Question

diverge?or converge series theDoes1

21

2

nnn

A) Converge

B) Diverge