ALTERNATING SERIES The convergence tests that we have looked at so far apply only to series with positive

11.5 ALTERNATING SERIES The convergence tests that we have looked at so far apply only to series with positive

terms. In this section and the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance are alternating series, whose terms alternate in sign.

An alternating series is a series whose terms are alternately positive and negative. Here are two examples:

1 1 1 1 1 ⬁ 共⫺1兲 n⫺ 1

2 3 4 5 6 n苷 兺 1 n

共⫺1兲 n

2 3 4 5 6 7 n苷 兺 1 n⫹ 1

We see from these examples that the th term of an alternating series is of the form n

a n 苷 共⫺1兲 n⫺ 1 b n

or

a n 苷 共⫺1兲 n b n

where b n is a positive number. ( In fact, b n 苷 ⱍ a n ⱍ . )

The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.

THE ALTERNATING SERIES TEST If the alternating series

共⫺1兲 n⫺ 1 b n 苷 b 1 ⫺b 2 ⫹b 3 ⫺b 4 ⫹b 5 ⫺b 6 ⫹⭈⭈⭈

then the series is convergent.

Before giving the proof let’s look at Figure 1, which gives a picture of the idea behind the proof. We first plot s 1 苷 b 1 on a number line. To find s 2 we subtract , so b 2 s 2 is to the left of . Then to find s 1 s 3 we add , so b 3 s 3 is to the right of . But, since s 2 b 3 ⬍b 2 , s 3 is to the left of . Continuing in this manner, we see that the partial sums oscillate back and s 1 forth. Since b n l 0 , the successive steps are becoming smaller and smaller. The even par- tial sums s 2 , s 4 , s 6 , . . . are increasing and the odd partial sums , s 1 s 3 , s 5 , . . . are decreasing. Thus it seems plausible that both are converging to some number , which is the sum of s the series. Therefore we consider the even and odd partial sums separately in the follow- ing proof.

b¡ -b™

+b£ -b¢ +b∞ -bß

FIGURE 1

0 s™

s¢

sß

s∞

s£

s¡

SECTION 11.5 ALTERNATING SERIES

PROOF OF THE ALTERNATING SERIES TEST We first consider the even partial sums: s 2 苷 b 1 ⫺b 2 艌 0 since b 2 艋b 1

s 4 苷 s 2 ⫹ 共b 3 ⫺b 4 兲艌s 2 since b 4 艋b 3

In general

s 2n 苷 s 2n⫺2 ⫹ 共b 2n⫺1 ⫺b 2n 兲艌s 2n⫺2

since b 2n 艋b 2n⫺1

Thus

0艋s 2 艋s 4 艋s 6 艋⭈⭈⭈艋s 2n 艋⭈⭈⭈

But we can also write

s 2n 苷 b 1 ⫺ 共b 2 ⫺b 3 兲 ⫺ 共b 4 ⫺b 5 兲 ⫺ ⭈ ⭈ ⭈ ⫺ 共b 2n⫺2 ⫺b 2n⫺1 兲⫺b 2n

Every term in brackets is positive, so s 2n 艋b 1 for all . Therefore the sequence n 兵s 2n 其 of even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Let’s call its limit , that is, s

nl⬁ lim s 2n 苷 s Now we compute the limit of the odd partial sums: nl⬁ lim s 2n⫹1 苷 nl⬁ lim 共s 2n ⫹b 2n⫹1 兲

苷 lim nl⬁ s 2n ⫹ nl⬁ lim b 2n⫹1

苷s⫹0

[by condition (ii)]

苷s

Since both the even and odd partial sums converge to , we have s lim nl⬁ s n 苷 s [see Exercise 80(a) in Section 11.1] and so the series is convergent.

N Figure 2 illustrates Example 1 by showing the

graphs of the terms a n 苷 共⫺1兲 n⫺ 1 兾n and the

partial sums . Notice how the values of s n s n V EXAMPLE 1 The alternating harmonic series

zigzag across the limiting value, which appears to be about 0.7 . In fact, it can be proved that

1 1 1 ⬁ 共⫺1兲 n⫺ 1

the exact sum of the series is ln 2 ⬇ 0.693

2 3 4 n苷 兺 1 n

(see Exercise 36).

(ii) lim b n 苷 lim nl⬁ 1 nl⬁ 苷 0 n

so the series is convergent by the Alternating Series Test.

兵a n 其

n苷 兺 1 4n ⫺ 1

V 共⫺1兲 EXAMPLE 2 3n The series is alternating but

3n

3 3 n lim l ⬁ 苷 b n 苷 n lim l ⬁ 苷 n 4n ⫺ 1 lim l ⬁ 1 4

FIGURE 2

CHAPTER 11 INFINITE SEQUENCES AND SERIES

so condition (ii) is not satisfied. Instead, we look at the limit of the nth term of the series:

共⫺1兲 n 3n n lim l ⬁ a n 苷 n lim l ⬁ 4n ⫺ 1

This limit does not exist, so the series diverges by the Test for Divergence.

n⫹ 1 n EXAMPLE 3 2 Test the series

n苷 兺 1 n 3 ⫹ 1

共⫺1兲 for convergence or divergence.

SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test.

Unlike the situation in Example 1, it is not obvious that the sequence given by

b n 苷 n 2 兾共n 3 ⫹ 1 兲 is decreasing. However, if we consider the related function

f 共x兲 苷 x 2 兾共x 3 ⫹ 1 兲 , we find that x 共2 ⫺ x 3 兲

f⬘ 共x兲 苷 共x 3 ⫹ 1 兲 2

Since we are considering only positive , we see that x f⬘ 共x兲 ⬍ 0 if 2⫺x 3 ⬍0 , that is,

x⬎ 3 s 2 . Thus is decreasing on the interval f ( 3 s 2 ,⬁ ) . This means that f 共n ⫹ 1兲 ⬍ f 共n兲

N Instead of verifying condition (i) of the Alter-

and therefore b n⫹ 1 ⬍b n when n艌 2 . (The inequality b 2 ⬍b 1 can be verified directly but

nating Series Test by computing a derivative,

all that really matters is that the sequence 兵b n 其 is eventually decreasing.)

we could verify that b n⫹ 1 ⬍b n directly by using

Condition (ii) is readily verified:

the technique of Solution 1 of Example 12 in Section 11.1.

lim n l b n 苷 lim 3 n 苷 l n lim l 苷 0

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

ESTIMATING SUMS

A partial sum of any convergent series can be used as an approximation to the total sum s n s , but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using s ⬇s n is the remainder R n 苷 s⫺s n . The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is

smaller than b n⫹ 1 , which is the absolute value of the first neglected term.

ALTERNATING SERIES ESTIMATION THEOREM If s苷 冘 共⫺1兲 n⫺ 1 b n is the sum of an

N You can see geometrically why the Alternating Series Estimation Theorem is true

alternating series that satisfies

by looking at Figure 1 (on page 710). Notice that

(ii) nl⬁ lim b n 苷 ⱍ 0 ⱍ

s⫺s 4 ⬍b 5 , s⫺s 5 ⬍b 6 , and so on. Notice

(i) 0艋b n⫹ 1 艋b n

and

also that lies between any two consecutive s partial sums.

then

ⱍ R n 苷 ⱍ ⱍ s⫺s n ⱍ 艋b n⫹ 1

PROOF We know from the proof of the Alternating Series Test that s lies between any two

consecutive partial sums s n and s n⫹ 1 . It follows that

ⱍ s⫺s n ⱍ 艋 ⱍ s n⫹ 1 ⫺s n 苷 ⱍ b n⫹ 1 M

SECTION 11.5 ALTERNATING SERIES

V 共⫺1兲 EXAMPLE 4 n Find the sum of the series 兺 correct to three decimal places.

n苷 0 n ! (By definition, 0! 苷 1 .)

SOLUTION We first observe that the series is convergent by the Alternating Series Test because

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:

Notice that

b 7 苷 1 ⬍ 5040 1 5000 苷 0.0002

s 苷 1⫺1⫹ 1 ⫺ 1 ⫹ 1 ⫺ 1 ⫹ and 1 6 2 6 24 120 720 ⬇ 0.368056

By the Alternating Series Estimation Theorem we know that

ⱍ s⫺s 6 ⱍ 艋b 7 ⬍ 0.0002

N In Section 11.10 we will prove that

This error of less than 0.0002 does not affect the third decimal place, so we have

e x 苷 冘 n苷 ⬁ 0 x n 兾n! for all , so what we have x

s ⬇ 0.368 correct to three decimal places.

obtained in Example 4 is actually an approxi- mation to the number e ⫺ 1 .

NOTE The rule that the error (in using s n to approximate ) is smaller than the first s 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.