12INFINITE SEQUENCES AND SERIESINFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES

The convergence tests that we have

looked at so far apply only to series

with positive terms.

12.5Alternating Series

In this section, we will learn:

How to deal with series

whose terms alternate in sign.

INFINITE SEQUENCES AND SERIES

ALTERNATING SERIES

An alternating series is a series whose

terms are alternately positive and negative.

Here are two examples:

1

1

1

1 1 1 1 1 ( 1)1 ...2 3 4 5 6

1 2 3 4 5 6... ( 1)

2 3 4 5 6 7 1

n

n

n

n

n

n

n

ALTERNATING SERIES

From these examples, we see that

the nth term of an alternating series is

of the form

an = (–1)n – 1bn or an = (–1)nbn

where bn is a positive number.

In fact, bn = |an|

ALTERNATING SERIES

The following test states that, if the terms

of an alternating series decrease toward 0

in absolute value, the series converges.

ALTERNATING SERIES TEST

If the alternating series

satisfies

i. bn+1 ≤ bn for all n

ii.

then the series is convergent.

11 2 3 4 5 6

1

( 1) ...

0

nn

n

n

b b b b b b b

b

lim 0nnb

ALTERNATING SERIES

Before giving the proof, let’s look at this figure

—which gives a picture of the idea behind the

proof.

Fig. 12.5.1, p. 746

ALTERNATING SERIES

First, we plot s1 = b1 on a number line.

To find s2,we subtract b2.

So, s2 is to the left of s1.

Fig. 12.5.1, p. 746

ALTERNATING SERIES

Then, to find s3, we add b3.

So, s3 is to the right of s2.

However, since b3 < b2, s3 is to the left of s1.

Fig. 12.5.1, p. 746

ALTERNATING SERIES

Continuing in this manner, we see that

the partial sums oscillate back and forth.

Since bn → 0, the successive steps are becoming smaller and smaller.

Fig. 12.5.1, p. 746

ALTERNATING SERIES

The even partial sums s2, s4, s6, . . .

are increasing.

The odd partial sums s1, s3, s5, . . .

are decreasing.

Fig. 12.5.1, p. 746

ALTERNATING SERIES

Thus, it seems plausible that both

are converging to some number s,

which is the sum of the series.

So, we consider the even and odd partial sums separately in the following proof.

ALTERNATING SERIES TEST—PROOF

First, we consider the even partial sums:

s2 = b1 – b2 ≥ 0 since b2 ≤ b1

s4 = s2 + (b3 – b4) ≥ s2 since b4 ≤ b3

ALTERNATING SERIES TEST—PROOF

In general,

s2n = s2n – 2 + (b2n – 1 – b2n) ≥ s2n – 2

since b2n ≤ b2n – 1

Thus, 0 ≤ s2 ≤ s4 ≤ s6 ≤ … ≤ s2n ≤ …

ALTERNATING SERIES TEST—PROOF

However, we can also write:

s2n = b1 – (b2 – b3) – (b4 – b5) – …

– (b2n – 2 – b2n – 1) – b2n

Every term in brackets is positive. So, s2n ≤ b1 for all n.

ALTERNATING SERIES TEST—PROOF

Thus, the sequence {s2n} of even

partial sums is increasing and bounded

above.

Therefore, it is convergent by the Monotonic Sequence Theorem.

ALTERNATING SERIES TEST—PROOF

Let’s call its limit s, that is,

Now, we compute the limit of the odd partial sums:

2lim nns s

2 1 2 2 1

2 2 1

lim lim( )

lim lim

0 (condition ii)

n n nn n

n nn n

s s b

s b

s

s

ALTERNATING SERIES TEST—PROOF

As both the even and odd partial sums

converge to s, we have

See Exercise 80(a) in Section 11.1

Thus, the series is convergent.

lim nns s

ALTERNATING SERIES

The alternating harmonic series

satisfies

i. bn+1 < bn because

ii.

It is convergent by the Alternating Series Test.

Example 1

1

1

1 1 1 ( 1)1 ...2 3 4

n

n n

1 1

1n n

1lim lim 0nn nb

n

ALTERNATING SERIES

The figure illustrates Example 1 by showing

the graphs of the terms an = (–1)n – 1/n

and the partial sums sn.

Fig. 12.5.2, p. 747

ALTERNATING SERIES

Notice how the values of sn zigzag across

the limiting value, which appears to be

about 0.7

In fact, it can be proved that the exact sum of the series is ln 2 ≈ 0.693(Exercise 36).

Fig. 12.5.2, p. 747

ALTERNATING SERIES

The series is alternating.

However,

So, condition ii is not satisfied.

Example 2

1

( 1) 3

4 1

n

n

n

n

3 3 3lim lim lim

14 1 44n

n n n

nb

nn

ALTERNATING SERIES

Instead, we look at the limit of the nth term

of the series:

This limit does not exist.

So, the series diverges by the Test for Divergence.

Example 2

( 1) 3lim lim

4 1

n

nn n

na

n

ALTERNATING SERIES

Test the series

for convergence or divergence.

The given series is alternating.

So, we try to verify conditions i and ii of the Alternating Series Test.

Example 3

21

31

( 1)1

n

n

n

n

ALTERNATING SERIES

Unlike the situation in Example 1, it is

not obvious that the sequence given by

bn = n2/(n3 + 1) is decreasing.

However, if we consider the related function

f(x) = x2/(x3 + 1), we find that:

Example 3

3

3 2

(2 )'( )

( 1)

x xf x

x

ALTERNATING SERIES

Since we are considering only positive x,

we see that f’(x) < 0 if 2 – x3 < 0, that is,

x > .

Thus, f is decreasing on the interval ( , ∞).

Example 3

3 2

3 2

ALTERNATING SERIES

This means that f(n + 1) < f(n) and,

therefore, bn+1 < bn when n ≥ 2.

The inequality b2 < b1 can be verified directly.

However, all that really matters is that the sequence {bn} is eventually decreasing.

Example 3

ALTERNATING SERIES

Condition ii is readily verified:

Thus, the given series is convergent by the Alternating Series Test.

2

3

3

1

lim lim lim 011 1

nn n n

n nbn

n

Example 3

ESTIMATING SUMS

A partial sum sn of any convergent series

can be used as an approximation to the total

sum s.

However, this is not of much use unless we can estimate the accuracy of the approximation.

The error involved in using s ≈ sn is the remainder Rn = s – sn.

ESTIMATING SUMS

The next theorem says that, for series

that satisfy the conditions of the Alternating

Series Test, the size of the error is smaller

than bn+1.

This is the absolute value of the first neglected term.

ALTERNATING SERIES ESTIMATION THEOREM

If s = Σ (–1)n-1bn is the sum of an alternating

series that satisfies

i. 0 ≤ bn+1 ≤ bn

ii.

then |Rn| = |s – sn| ≤ bn+1

lim 0nnb

ALTERNATING SERIES ESTIMATION THM.—PROOF

From the proof of the Alternating Series Test,

we know that s lies between any two

consecutive partial sums sn and sn+1.

It follows that:

|s – sn| ≤ |sn+1 – sn| = bn+1

ALTERNATING SERIES ESTIMATION THEOREM

You can see geometrically why the theorem

is true by looking at this figure.

Notice that s – s4 < b5, |s – s5| < b6, and so on.

Notice also that s lies between any two consecutive partial sums.

ESTIMATING SUMS

Find the sum of the series

correct to three decimal places.

By definition, 0! = 1.

Example 4

0

( 1)

!

n

n n

ESTIMATING SUMS

First, we observe that the series is convergent

by the Alternating Series Test because:

i.

ii.

1 1 1

( 1)! !( 1) !n n n n

1 1 10 0 so 0 as

! !n

n n n

Example 4

ESTIMATING SUMS

To get a feel for how many terms we need

to use in our approximation, let’s write out

the first few terms of the series:

1 1 1 1 1 12 6 24 120 720 5040

1 1 1 1 1 1 1 1...

0! 1! 2! 3! 4! 5! 6! 7!1 1 ...

s

Example 4

ESTIMATING SUMS

Notice that

and

1 17 5040 5000 0.0002b

1 1 1 1 16 2 6 24 120 7201 1

0.368056

s

1 1 1 1 1 12 6 24 120 720 5040

1 1 1 1 1 1 1 1...

0! 1! 2! 3! 4! 5! 6! 7!1 1 ...

s

Example 4

ESTIMATING SUMS

By the Alternating Series Estimation Theorem,

we know that:

| s – s6 | ≤ b7 < 0.0002

This error of less than 0.0002 does not affect the third decimal place.

So, we have s ≈ 0.368 correct to three decimal places.

Example 4

NOTE

The rule that the error (in using sn to

approximate s) is smaller than the first

neglected term is, in general, valid only for

alternating series that satisfy the conditions of

the Alternating Series Estimation Theorem.

The rule does not apply to other types of series.