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97817_11_ch11_p732-741.qk_97817_11_ch11_p732-741 11/3/10 5:29 PM Page 739

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS

739

So the sum of the areas of the rectangles is

ϱ

1

1

1

1

1

1

ϩ

ϩ

ϩ

ϩ

ϩ

и

и

и

͚

2

2

2

2

2

2

1

2

3

4

5

n1 n

If we exclude the first rectangle, the total area of the remaining rectangles is smaller than

the area under the curve y 1͞x 2 for x ജ 1, which is the value of the integral x1ϱ ͑1͞x 2 ͒ dx.

In Section 7.8 we discovered that this improper integral is convergent and has value 1. So

the picture shows that all the partial sums are less than

1

ϱ 1

ϩ y 2 dx 2

2

1

1

x

Thus the partial sums are bounded. We also know that the partial sums are increasing

(because all the terms are positive). Therefore the partial sums converge (by the Monotonic

Sequence Theorem) and so the series is convergent. The sum of the series (the limit of the

partial sums) is also less than 2:

ϱ

͚

n1

1

1

1

1

1

2 ϩ 2 ϩ 2 ϩ 2 ϩ иии Ͻ 2

2

n

1

2

3

4

[The exact sum of this series was found by the Swiss mathematician Leonhard Euler

(1707–1783) to be 2͞6, but the proof of this fact is quite difficult. (See Problem 6 in the

Problems Plus following Chapter 15.)]

Now let’s look at the series

ϱ

n

n

sn

͚

i1

5

10

50

100

500

1000

5000

͚

1

si

n1

1

1

1

1

1

1

ϩ

ϩ

ϩ

ϩ

ϩ иии

sn

s1

s2

s3

s4

s5

The table of values of sn suggests that the partial sums aren’t approaching a finite number,

so we suspect that the given series may be divergent. Again we use a picture for confirmation. Figure 2 shows the curve y 1͞sx , but this time we use rectangles whose tops lie

above the curve.

3.2317

5.0210

12.7524

18.5896

43.2834

61.8010

139.9681

FIGURE 2

y

y= 1

x

œ„

0

1

2

area= 1

1

œ„

3

area= 1

2

œ„

4

area= 1

3

œ„

5

x

area= 1

4

œ„

The base of each rectangle is an interval of length 1. The height is equal to the value of

the function y 1͞sx at the left endpoint of the interval. So the sum of the areas of all the

rectangles is

ϱ

1

1

1

1

1

1

ϩ

ϩ

ϩ

ϩ

ϩ иии ͚

s1

s2

s3

s4

s5

n1 sn

This total area is greater than the area under the curve y 1͞sx for x ജ 1, which is equal

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p732-741.qk_97817_11_ch11_p732-741 11/3/10 5:29 PM Page 740

740

CHAPTER 11

INFINITE SEQUENCES AND SERIES

to the integral x1ϱ (1͞sx ) dx. But we know from Section 7.8 that this improper integral is

divergent. In other words, the area under the curve is infinite. So the sum of the series must

be infinite; that is, the series is divergent.

The same sort of geometric reasoning that we used for these two series can be used to

prove the following test. (The proof is given at the end of this section.)

The Integral Test Suppose f is a continuous, positive, decreasing function on ͓1, ϱ͒

and let a n f ͑n͒. Then the series ϱn1 a n is convergent if and only if the improper

integral x1ϱ f ͑x͒ dx is convergent. In other words:

ϱ

ϱ

͚a

(i) If y f ͑x͒ dx is convergent, then

1

n

is convergent.

n1

ϱ

ϱ

͚a

(ii) If y f ͑x͒ dx is divergent, then

1

n

is divergent.

n1

NOTE When we use the Integral Test, it is not necessary to start the series or the integral

at n 1. For instance, in testing the series

ϱ

͚

n4

1

͑n Ϫ 3͒2

y

we use

ϱ

4

1

dx

͑x Ϫ 3͒2

Also, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. Then ϱnN a n is

convergent, so ϱn1 a n is convergent by Note 4 of Section 11.2.

ϱ

EXAMPLE 1 Test the series

͚

n1

1

for convergence or divergence.

n ϩ1

2

SOLUTION The function f ͑x͒ 1͑͞x 2 ϩ 1͒ is continuous, positive, and decreasing on

͓1, ϱ͒ so we use the Integral Test:

y

1

t

1

dx lim y 2

dx lim tanϪ1x

tlϱ 1 x ϩ 1

tlϱ

x2 ϩ 1

ϱ

]

1

ͩ

lim tanϪ1t Ϫ

tlϱ

4

ͪ

t

1

Ϫ

2

4

4

Thus x1ϱ 1͑͞x 2 ϩ 1͒ dx is a convergent integral and so, by the Integral Test, the series

1͑͞n 2 ϩ 1͒ is convergent.

ϱ

v

EXAMPLE 2 For what values of p is the series

͚

n1

1

convergent?

np

SOLUTION If p Ͻ 0, then lim n l ϱ ͑1͞n ͒ ϱ. If p 0, then lim n l ϱ ͑1͞n p ͒ 1. In

p

In order to use the Integral Test we need to be

able to evaluate x1ϱ f ͑x͒ dx and therefore we

have to be able to find an antiderivative of f .

Frequently this is difficult or impossible, so we

need other tests for convergence too.

either case lim n l ϱ ͑1͞n p ͒ 0, so the given series diverges by the Test for Divergence

(11.2.7).

If p Ͼ 0, then the function f ͑x͒ 1͞x p is clearly continuous, positive, and decreasing

on ͓1, ϱ͒. We found in Chapter 7 [see (7.8.2)] that

y

ϱ

1

1

dx converges if p Ͼ 1 and diverges if p ഛ 1

xp

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS

741

It follows from the Integral Test that the series 1͞n p converges if p Ͼ 1 and diverges

if 0 Ͻ p ഛ 1. (For p 1, this series is the harmonic series discussed in Example 8 in

Section 11.2.)

The series in Example 2 is called the p-series. It is important in the rest of this chapter,

so we summarize the results of Example 2 for future reference as follows.

ϱ

1

The p-series

͚

n1

1

is convergent if p Ͼ 1 and divergent if p ഛ 1.

np

EXAMPLE 3

(a) The series

ϱ

͚

n1

1

1

1

1

1

3 ϩ 3 ϩ 3 ϩ 3 ϩ иии

n3

1

2

3

4

is convergent because it is a p-series with p 3 Ͼ 1.

(b) The series

ϱ

ϱ

1

1

1

1

1

͚ 3 1 ϩ 3 ϩ 3 ϩ 3 ϩ иии

͚

1͞3

s2

s3

s4

n1 n

n1 sn

1

is divergent because it is a p-series with p 3 Ͻ 1.

NOTE We should not infer from the Integral Test that the sum of the series is equal to

the value of the integral. In fact,

ϱ

͚

n1

1

2

n2

6

Therefore, in general,

ϱ

͚a

n1

n

y

ϱ

1

EXAMPLE 4 Determine whether the series

ϱ

1

1

dx 1

x2

f ͑x͒ dx

ϱ

v

y

whereas

͚

n1

ln n

converges or diverges.

n

SOLUTION The function f ͑x͒ ͑ln x͒͞x is positive and continuous for x Ͼ 1 because the

logarithm function is continuous. But it is not obvious whether or not f is decreasing, so

we compute its derivative:

f Ј͑x͒

͑1͞x͒x Ϫ ln x

1 Ϫ ln x

x2

x2

Thus f Ј͑x͒ Ͻ 0 when ln x Ͼ 1, that is, x Ͼ e. It follows that f is decreasing when x Ͼ e

and so we can apply the Integral Test:

y

ϱ

1

ln x

t ln x

͑ln x͒2

dx lim y

dx lim

tlϱ 1

tlϱ

x

x

2

ͬ

t

1

͑ln t͒

ϱ

2

2

lim

tlϱ

Since this improper integral is divergent, the series ͑ln n͒͞n is also divergent by the

Integral Test.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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742

CHAPTER 11

INFINITE SEQUENCES AND SERIES

Estimating the Sum of a Series

Suppose we have been able to use the Integral Test to show that a series a n is convergent

and we now want to find an approximation to the sum s of the series. Of course, any partial

sum sn is an approximation to s because lim n l ϱ sn s. But how good is such an approximation? To find out, we need to estimate the size of the remainder

Rn s Ϫ sn a nϩ1 ϩ a nϩ2 ϩ a nϩ3 ϩ и и и

The remainder Rn is the error made when sn , the sum of the first n terms, is used as an approximation to the total sum.

We use the same notation and ideas as in the Integral Test, assuming that f is decreasing on ͓n, ϱ͒. Comparing the areas of the rectangles with the area under y f ͑x͒ for x Ͼ n

in Figure 3, we see that

y

y=ƒ

an+1 an+2

0

ϱ

Rn a nϩ1 ϩ a nϩ2 ϩ и и и ഛ y f ͑x͒ dx

...

n

x

n

Similarly, we see from Figure 4 that

FIGURE 3

Rn a nϩ1 ϩ a nϩ2 ϩ и и и ജ y

ϱ

nϩ1

f ͑x͒ dx

y

y=ƒ

an+1 an+2

0

n+1

So we have proved the following error estimate.

2 Remainder Estimate for the Integral Test Suppose f ͑k͒ a k , where f is a

continuous, positive, decreasing function for x ജ n and a n is convergent. If

Rn s Ϫ sn , then

...

x

y

ϱ

ϱ

nϩ1

f ͑x͒ dx ഛ Rn ഛ y f ͑x͒ dx

n

FIGURE 4

v

EXAMPLE 5

(a) Approximate the sum of the series 1͞n 3 by using the sum of the first 10 terms.

Estimate the error involved in this approximation.

(b) How many terms are required to ensure that the sum is accurate to within 0.0005?

SOLUTION In both parts (a) and (b) we need to know

xnϱ f ͑x͒ dx. With

f ͑x͒ 1͞x 3, which

satisfies the conditions of the Integral Test, we have

y

ϱ

n

ͫ ͬ

1

1

dx lim Ϫ 2

tlϱ

x3

2x

t

n

ͩ

lim Ϫ

tlϱ

1

1

ϩ 2

2t 2

2n

ͪ

1

2n2

(a) Approximating the sum of the series by the 10th partial sum, we have

ϱ

͚

n1

1

1

1

1

1

Ϸ s10 3 ϩ 3 ϩ 3 ϩ и и и ϩ 3 Ϸ 1.1975

n3

1

2

3

10

According to the remainder estimate in 2 , we have

R10 ഛ y

ϱ

10

1

1

1

dx

3

2

x

2͑10͒

200

So the size of the error is at most 0.005.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 11.3

THE INTEGRAL TEST AND ESTIMATES OF SUMS

743

(b) Accuracy to within 0.0005 means that we have to find a value of n such that

Rn ഛ 0.0005. Since

ϱ 1

1

Rn ഛ y 3 dx

n x

2n 2

1

Ͻ 0.0005

2n 2

we want

Solving this inequality, we get

n2 Ͼ

1

1000

0.001

or

n Ͼ s1000 Ϸ 31.6

We need 32 terms to ensure accuracy to within 0.0005.

If we add sn to each side of the inequalities in 2 , we get

3

sn ϩ y

ϱ

ϱ

nϩ1

f ͑x͒ dx ഛ s ഛ sn ϩ y f ͑x͒ dx

n

because sn ϩ Rn s. The inequalities in 3 give a lower bound and an upper bound for s.

They provide a more accurate approximation to the sum of the series than the partial sum

sn does.

Although Euler was able to calculate the exact

sum of the p-series for p 2, nobody has been

able to find the exact sum for p 3. In Example

6, however, we show how to estimate this sum.

EXAMPLE 6 Use 3 with n 10 to estimate the sum of the series

ϱ

͚

n1

1

.

n3

SOLUTION The inequalities in 3 become

s10 ϩ y

ϱ

11

1

ϱ 1

dx ഛ s ഛ s10 ϩ y 3 dx

10 x

x3

From Example 5 we know that

y

ϱ

n

s10 ϩ

so

1

1

dx

3

x

2n 2

1

1

ഛ s ഛ s10 ϩ

2͑11͒2

2͑10͒2

Using s10 Ϸ 1.197532, we get

1.201664 ഛ s ഛ 1.202532

If we approximate s by the midpoint of this interval, then the error is at most half the

length of the interval. So

ϱ

͚

n1

1

Ϸ 1.2021

n3

with error Ͻ 0.0005

If we compare Example 6 with Example 5, we see that the improved estimate in 3 can

be much better than the estimate s Ϸ sn . To make the error smaller than 0.0005 we had to

use 32 terms in Example 5 but only 10 terms in Example 6.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p742-751.qk_97817_11_ch11_p742-751 11/3/10 5:29 PM Page 744

744

CHAPTER 11

INFINITE SEQUENCES AND SERIES

Proof of the Integral Test

y

We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and

2 for the series 1͞n 2 and 1͞sn . For the general series a n, look at Figures 5 and 6. The

area of the first shaded rectangle in Figure 5 is the value of f at the right endpoint of ͓1, 2͔,

that is, f ͑2͒ a 2 . So, comparing the areas of the shaded rectangles with the area under

y f ͑x͒ from 1 to n, we see that

y=ƒ

a™ a£ a¢ a∞

0

1

2

3

an

5 ...

4

n x

FIGURE 5

y

n

a 2 ϩ a 3 ϩ и и и ϩ a n ഛ y f ͑x͒ dx

4

1

(Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6

shows that

y=ƒ

y

5

an-1

1

2

3

4

f ͑x͒ dx ഛ a 1 ϩ a 2 ϩ и и и ϩ a nϪ1

ϱ

(i) If y f ͑x͒ dx is convergent, then 4 gives

1

a¡ a™ a£ a¢

0

n

1

5 ...

n

͚a

n x

1

i2

FIGURE 6

ϱ

n

ഛ y f ͑x͒ dx ഛ y f ͑x͒ dx

i

1

since f ͑x͒ ജ 0. Therefore

n

͚a

sn a 1 ϩ

ϱ

i

ഛ a 1 ϩ y f ͑x͒ dx M, say

1

i2

Since sn ഛ M for all n, the sequence ͕sn ͖ is bounded above. Also

snϩ1 sn ϩ a nϩ1 ജ sn

since a nϩ1 f ͑n ϩ 1͒ ജ 0. Thus ͕sn ͖ is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (11.1.12). This means that a n is convergent.

(ii) If x1ϱ f ͑x͒ dx is divergent, then x1n f ͑x͒ dx l ϱ as n l ϱ because f ͑x͒ ജ 0. But 5

gives

y

n

1

nϪ1

f ͑x͒ dx ഛ

͚a

i

snϪ1

i1

and so snϪ1 l ϱ. This implies that sn l ϱ and so a n diverges.

11.3

Exercises

1. Draw a picture to show that

ϱ

͚

n2

3–8 Use the Integral Test to determine whether the series is

convergent or divergent.

1

ϱ 1

Ͻ y 1.3 dx

1 x

n 1.3

ϱ

3.

͚

n1

What can you conclude about the series?

ϱ

2. Suppose f is a continuous positive decreasing function for

x ജ 1 and an f ͑n͒. By drawing a picture, rank the following

three quantities in increasing order:

y

6

1

5

f ͑x͒ dx

͚a

i1

CAS Computer algebra system required

6

i

͚a

5.

͚

n1

ϱ

7.

͚

n1

1

5

sn

ϱ

4.

n1

1

n5

ϱ

1

͑2n ϩ 1͒ 3

6.

n

n ϩ1

8.

2

͚

͚

n1

ϱ

1

sn ϩ 4

͚ne

2 Ϫn 3

n1

i

i2

1. Homework Hints available at stewartcalculus.com

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p742-751.qk_97817_11_ch11_p742-751 11/3/10 5:29 PM Page 745

CHAPTER 11.3

9–26 Determine whether the series is convergent or divergent.

ϱ

9.

͚

n1

ϱ

1

n

10.

s2

͚n

p-series with p 2:

Ϫ0.9999

͑2͒

n3

2 s2

ϩ

1

3 s3

ϩ

1

4 s4

(a)

ϩ иии

5 s5

͚

n2

ϱ

(c)

1

1

1

1

13. 1 ϩ ϩ ϩ ϩ ϩ и и и

3

5

7

9

͚

n1

ϱ

1

n2

(b)

n3

35. Euler also found the sum of the p-series with p 4:

͑4͒

ϱ

͚

n1

15.

͚

n1

ϱ

sn ϩ 4

n2

16.

͚

n1

n2

3

n ϩ1

17.

͚

n1

ϱ

19.

͚

n1

ϱ

21.

͚

n2

ϱ

23.

͚

n1

ϱ

1

n2 ϩ 4

18.

͚

n3

ϱ

ln n

n3

20.

1

n ln n

22.

e 1͞n

n2

24.

͚

n1

ϱ

͚

n2

ϱ

͚

n3

1

4

4

n

90

Use Euler’s result to find the sum of the series.

ϱ

ϱ

1

͑n ϩ 1͒2

͚

1

͑2n͒2

1

1

1

1

1

14.

ϩ ϩ

ϩ

ϩ

ϩ иии

5

8

11

14

17

ϱ

1

2

2

n

6

(See page 739.) Use this fact to find the sum of each series.

1

ϩ

ϱ

͚

n1

ϱ

1

745

34. Leonhard Euler was able to calculate the exact sum of the

1

1

1

1

11. 1 ϩ ϩ

ϩ

ϩ

ϩ иии

8

27

64

125

12. 1 ϩ

THE INTEGRAL TEST AND ESTIMATES OF SUMS

3n Ϫ 4

n 2 Ϫ 2n

(a)

͚

n1

ͩͪ

3

n

4

ϱ

(b)

͚

k5

1

͑k Ϫ 2͒4

36. (a) Find the partial sum s10 of the series ϱn1 1͞n 4. Estimate the

1

n 2 ϩ 6n ϩ 13

error in using s10 as an approximation to the sum of the

series.

(b) Use 3 with n 10 to give an improved estimate of the

sum.

(c) Compare your estimate in part (b) with the exact value

given in Exercise 35.

(d) Find a value of n so that s n is within 0.00001 of the sum.

1

n͑ln n͒ 2

n2

en

37. (a) Use the sum of the first 10 terms to estimate the sum of the

ϱ

25.

͚

n1

ϱ

1

n2 ϩ n3

26.

͚

n1

series ϱn1 1͞n 2. How good is this estimate?

(b) Improve this estimate using 3 with n 10.

(c) Compare your estimate in part (b) with the exact value

given in Exercise 34.

(d) Find a value of n that will ensure that the error in the

approximation s Ϸ sn is less than 0.001.

n

n4 ϩ 1

27–28 Explain why the Integral Test can’t be used to determine

whether the series is convergent.

ϱ

27.

͚

n1

cos n

sn

ϱ

28.

͚

n1

cos 2 n

1 ϩ n2

38. Find the sum of the series ϱn1 1͞n 5 correct to three decimal

places.

39. Estimate ϱn1 ͑2n ϩ 1͒Ϫ6 correct to five decimal places.

29–32 Find the values of p for which the series is convergent.

ϱ

29.

͚

n2

ϱ

31.

ϱ

1

n͑ln n͒ p

͚ n͑1 ϩ n

30.

͚

n3

ϱ

͒

2 p

32.

n1

͚

n1

40. How many terms of the series ϱn2 1͓͞n͑ln n͒ 2 ͔ would you

1

n ln n ͓ln͑ln n͔͒ p

need to add to find its sum to within 0.01?

ln n

np

41. Show that if we want to approximate the sum of the series

ϱn1 nϪ1.001 so that the error is less than 5 in the ninth decimal

place, then we need to add more than 10 11,301 terms!

33. The Riemann zeta-function is defined by

͑x͒

ϱ

͚

n1

1

nx

and is used in number theory to study the distribution of prime

numbers. What is the domain of ?

CAS

42. (a) Show that the series ϱn1 ͑ln n͒2͞n 2 is convergent.

(b) Find an upper bound for the error in the approximation

s Ϸ sn .

(c) What is the smallest value of n such that this upper bound

is less than 0.05?

(d) Find sn for this value of n.

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746

CHAPTER 11

INFINITE SEQUENCES AND SERIES

43. (a) Use 4 to show that if s n is the nth partial sum of the har-

(b) Interpret

monic series, then

tn Ϫ tnϩ1 ͓ln͑n ϩ 1͒ Ϫ ln n͔ Ϫ

sn ഛ 1 ϩ ln n

(b) The harmonic series diverges, but very slowly. Use part (a)

to show that the sum of the first million terms is less than

15 and the sum of the first billion terms is less than 22.

44. Use the following steps to show that the sequence

tn 1 ϩ

1

1

1

ϩ ϩ и и и ϩ Ϫ ln n

2

3

n

has a limit. (The value of the limit is denoted by ␥ and is called

Euler’s constant.)

(a) Draw a picture like Figure 6 with f ͑x͒ 1͞x and interpret

tn as an area [or use 5 ] to show that tn Ͼ 0 for all n.

11.4

1

nϩ1

as a difference of areas to show that tn Ϫ tnϩ1 Ͼ 0. Therefore ͕tn ͖ is a decreasing sequence.

(c) Use the Monotonic Sequence Theorem to show that ͕tn ͖ is

convergent.

45. Find all positive values of b for which the series ϱn1 b ln n

converges.

46. Find all values of c for which the following series converges.

ϱ

͚

ͩ

n1

1

c

Ϫ

n

nϩ1

ͪ

The Comparison Tests

In the comparison tests the idea is to compare a given series with a series that is known to

be convergent or divergent. For instance, the series

ϱ

͚

1

n1

1

2n ϩ 1

reminds us of the series ϱn1 1͞2 n, which is a geometric series with a 12 and r 12 and is

therefore convergent. Because the series 1 is so similar to a convergent series, we have the

feeling that it too must be convergent. Indeed, it is. The inequality

1

1

Ͻ n

2n ϩ 1

2

shows that our given series 1 has smaller terms than those of the geometric series and

therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This

means that its partial sums form a bounded increasing sequence, which is convergent. It

also follows that the sum of the series is less than the sum of the geometric series:

ϱ

͚

n1

1

Ͻ1

2 ϩ1

n

Similar reasoning can be used to prove the following test, which applies only to series

whose terms are positive. The first part says that if we have a series whose terms are

smaller than those of a known convergent series, then our series is also convergent. The

second part says that if we start with a series whose terms are larger than those of a known

divergent series, then it too is divergent.

a n and bn are series with positive terms.

If bn is convergent and a n ഛ bn for all n, then a n is also convergent.

If bn is divergent and a n ജ bn for all n, then a n is also divergent.

The Comparison Test Suppose that

(i)

(ii)

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p742-751.qk_97817_11_ch11_p742-751 11/3/10 5:30 PM Page 747

SECTION 11.4

It is important to keep in mind the distinction

between a sequence and a series. A sequence

is a list of numbers, whereas a series is a sum.

With every series a n there are associated two

sequences: the sequence ͕a n ͖ of terms and the

sequence ͕sn ͖ of partial sums.

Standard Series for Use

with the Comparison Test

THE COMPARISON TESTS

747

PROOF

n

(i) Let

sn

͚a

n

tn

i

i1

͚b

t

i

i1

ϱ

͚b

n

n1

Since both series have positive terms, the sequences ͕sn ͖ and ͕tn ͖ are increasing

͑snϩ1 sn ϩ a nϩ1 ജ sn ͒. Also tn l t, so tn ഛ t for all n. Since a i ഛ bi , we have sn ഛ tn .

Thus sn ഛ t for all n. This means that ͕sn ͖ is increasing and bounded above and therefore

converges by the Monotonic Sequence Theorem. Thus a n converges.

(ii) If bn is divergent, then tn l ϱ (since ͕tn ͖ is increasing). But a i ജ bi so sn ജ tn .

Thus sn l ϱ. Therefore a n diverges.

In using the Comparison Test we must, of course, have some known series bn for the

purpose of comparison. Most of the time we use one of these series:

■

■

A p -series [ 1͞n p converges if p Ͼ 1 and diverges if p ഛ 1; see (11.3.1)]

A geometric series [ ar nϪ1 converges if r Ͻ 1 and diverges if r ജ 1;

see (11.2.4)]

Խ Խ

ϱ

v

EXAMPLE 1 Determine whether the series

Խ Խ

5

converges or diverges.

2n ϩ 4n ϩ 3

͚

2

n1

SOLUTION For large n the dominant term in the denominator is 2n 2, so we compare the

given series with the series 5͑͞2n 2 ͒. Observe that

5

5

Ͻ 2

2n ϩ 4n ϩ 3

2n

2

because the left side has a bigger denominator. (In the notation of the Comparison Test,

a n is the left side and bn is the right side.) We know that

ϱ

͚

n1

5

5

2n 2

2

ϱ

͚

n1

1

n2

is convergent because it’s a constant times a p-series with p 2 Ͼ 1. Therefore

ϱ

͚

n1

5

2n ϩ 4n ϩ 3

2

is convergent by part (i) of the Comparison Test.

NOTE 1 Although the condition a n ഛ bn or a n ജ bn in the Comparison Test is given for

all n, we need verify only that it holds for n ജ N, where N is some fixed integer, because

the convergence of a series is not affected by a finite number of terms. This is illustrated in

the next example.

ϱ

v

EXAMPLE 2 Test the series

͚

k1

ln k

for convergence or divergence.

k

SOLUTION We used the Integral Test to test this series in Example 4 of Section 11.3, but

we can also test it by comparing it with the harmonic series. Observe that ln k Ͼ 1 for

k ജ 3 and so

ln k

1

Ͼ

kജ3

k

k

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p742-751.qk_97817_11_ch11_p742-751 11/3/10 5:30 PM Page 748

748

CHAPTER 11

INFINITE SEQUENCES AND SERIES

We know that 1͞k is divergent ( p-series with p 1). Thus the given series is divergent by the Comparison Test.

NOTE 2 The terms of the series being tested must be smaller than those of a convergent

series or larger than those of a divergent series. If the terms are larger than the terms of a

convergent series or smaller than those of a divergent series, then the Comparison Test

doesn’t apply. Consider, for instance, the series

ϱ

͚

n1

1

2n Ϫ 1

The inequality

1

1

Ͼ n

2n Ϫ 1

2

is useless as far as the Comparison Test is concerned because bn ( 12 ) is convergent

and a n Ͼ bn. Nonetheless, we have the feeling that 1͑͞2 n Ϫ 1͒ ought to be convergent

n

because it is very similar to the convergent geometric series ( 12 ) . In such cases the following test can be used.

n

The Limit Comparison Test Suppose that

a n and bn are series with positive

terms. If

Exercises 40 and 41 deal with the

cases c 0 and c ϱ.

lim

nlϱ

an

c

bn

where c is a finite number and c Ͼ 0, then either both series converge or both

diverge.

PROOF Let m and M be positive numbers such that m Ͻ c Ͻ M . Because a n ͞bn is close

to c for large n, there is an integer N such that

mϽ

an

ϽM

bn

mbn Ͻ a n Ͻ Mbn

and so

when n Ͼ N

when n Ͼ N

If bn converges, so does Mbn . Thus a n converges by part (i) of the Comparison

Test. If bn diverges, so does mbn and part (ii) of the Comparison Test shows that a n

diverges.

ϱ

EXAMPLE 3 Test the series

͚

n1

1

for convergence or divergence.

2n Ϫ 1

SOLUTION We use the Limit Comparison Test with

an

1

2n Ϫ 1

bn

1

2n

and obtain

lim

nlϱ

an

1͑͞2 n Ϫ 1͒

2n

1

lim

lim n

lim

1Ͼ0

n

nlϱ

nlϱ 2 Ϫ 1

n l ϱ 1 Ϫ 1͞2 n

bn

1͞2

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_11_ch11_p742-751.qk_97817_11_ch11_p742-751 11/3/10 5:30 PM Page 749

SECTION 11.4

THE COMPARISON TESTS

749

Since this limit exists and 1͞2 n is a convergent geometric series, the given series converges by the Limit Comparison Test.

ϱ

͚

EXAMPLE 4 Determine whether the series

n1

2n 2 ϩ 3n

converges or diverges.

s5 ϩ n 5

SOLUTION The dominant part of the numerator is 2n 2 and the dominant part of the denom-

inator is sn 5 n 5͞2. This suggests taking

an

lim

nlϱ

2n 2 ϩ 3n

s5 ϩ n 5

bn

2n 2

2

1͞2

5͞2

n

n

an

2n 2 ϩ 3n n 1͞2

2n 5͞2 ϩ 3n 3͞2

lim

ؒ

lim

n l ϱ s5 ϩ n 5

n l ϱ 2s5 ϩ n 5

bn

2

3

n

2ϩ

ͱ

lim

nlϱ

2

5

ϩ1

n5

2ϩ0

1

2s0 ϩ 1

1

Since bn 2 1͞n 1͞2 is divergent ( p-series with p 2 Ͻ 1), the given series diverges

by the Limit Comparison Test.

Notice that in testing many series we find a suitable comparison series bn by keeping

only the highest powers in the numerator and denominator.

Estimating Sums

If we have used the Comparison Test to show that a series a n converges by comparison

with a series bn, then we may be able to estimate the sum a n by comparing remainders.

As in Section 11.3, we consider the remainder

Rn s Ϫ sn a nϩ1 ϩ a nϩ2 ϩ и и и

For the comparison series bn we consider the corresponding remainder

Tn t Ϫ tn bnϩ1 ϩ bnϩ2 ϩ и и и

Since a n ഛ bn for all n, we have Rn ഛ Tn . If bn is a p-series, we can estimate its remainder Tn as in Section 11.3. If bn is a geometric series, then Tn is the sum of a geometric

series and we can sum it exactly (see Exercises 35 and 36). In either case we know that Rn

is smaller than Tn .

v

EXAMPLE 5 Use the sum of the first 100 terms to approximate the sum of the series

1͑͞n 3 ϩ 1͒. Estimate the error involved in this approximation.

SOLUTION Since

1

1

Ͻ 3

n3 ϩ 1

n

the given series is convergent by the Comparison Test. The remainder Tn for the comparison series 1͞n 3 was estimated in Example 5 in Section 11.3 using the Remainder Estimate for the Integral Test. There we found that

Tn ഛ y

ϱ

n

1

1

dx

x3

2n 2

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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