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6: Absolute Convergence and the Ratio and Root Tests

6: Absolute Convergence and the Ratio and Root Tests

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97817_11_ch11_p752-761.qk_97817_11_ch11_p752-761 11/3/10 5:30 PM Page 757



ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS



SECTION 11.6



757



2 Definition A series ͸ a n is called conditionally convergent if it is convergent

but not absolutely convergent.



Example 2 shows that the alternating harmonic series is conditionally convergent. Thus

it is possible for a series to be convergent but not absolutely convergent. However, the next

theorem shows that absolute convergence implies convergence.



Theorem If a series



3



͸ a n is absolutely convergent, then it is convergent.



PROOF Observe that the inequality



Խ Խ



Խ Խ



0 ഛ an ϩ an ഛ 2 an



is true because a n is either a n or Ϫa n . If ͸ a n is absolutely convergent, then ͸ a n is

convergent, so ͸ 2 a n is convergent. Therefore, by the Comparison Test, ͸ (a n ϩ a n ) is

convergent. Then



Խ Խ

Խ Խ



Խ Խ

Խ Խ



͚a



n



Խ Խ) Ϫ ͚ Խ a Խ



෇ ͚ (a n ϩ a n



n



is the difference of two convergent series and is therefore convergent.



v



EXAMPLE 3 Determine whether the series

ϱ



͚



n෇1



cos n

cos 1

cos 2

cos 3



ϩ

ϩ

ϩ иии

n2

12

22

32



is convergent or divergent.

Figure 1 shows the graphs of the terms a n and

partial sums sn of the series in Example 3.

Notice that the series is not alternating but

has positive and negative terms.



SOLUTION This series has both positive and negative terms, but it is not alternating.



(The first term is positive, the next three are negative, and the following three are positive: The signs change irregularly.) We can apply the Comparison Test to the series of

absolute values

ϱ



͚



0.5



n෇1



͕sn ͖



Խ



Ϳ Ϳ



Խ



ϱ

cos n

cos n

෇ ͚

2

n

n2

n෇1



Խ



Խ



Since cos n ഛ 1 for all n, we have



Խ cos n Խ ഛ



͕a n ͖

0



FIGURE 1



n



n2



1

n2



We know that ͸ 1͞n 2 is convergent ( p-series with p ෇ 2) and therefore ͸ cos n ͞n 2 is

convergent by the Comparison Test. Thus the given series ͸ ͑cos n͒͞n 2 is absolutely

convergent and therefore convergent by Theorem 3.



Խ



Խ



The following test is very useful in determining whether a given series is absolutely

convergent.



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_p752-761.qk_97817_11_ch11_p752-761 11/3/10 5:30 PM Page 758



758



CHAPTER 11



INFINITE SEQUENCES AND SERIES



The Ratio Test



Ϳ Ϳ



ϱ

a nϩ1

෇ L Ͻ 1, then the series ͚ a n is absolutely convergent

nlϱ

an

n෇1

(and therefore convergent).



(i) If lim



Ϳ Ϳ



Ϳ Ϳ



ϱ

a nϩ1

a nϩ1

෇ L Ͼ 1 or lim

෇ ϱ, then the series ͚ a n

nlϱ

nlϱ

an

an

n෇1

is divergent.



(ii) If lim



Ϳ Ϳ



a nϩ1

෇ 1, the Ratio Test is inconclusive; that is, no conclusion can

an

be drawn about the convergence or divergence of ͸ a n .



(iii) If lim



nlϱ



PROOF



(i) The idea is to compare the given series with a convergent geometric series. Since

L Ͻ 1, we can choose a number r such that L Ͻ r Ͻ 1. Since

lim



nlϱ



Խ



Խ



Ϳ Ϳ



a nϩ1

෇L

an



LϽr



and



the ratio a nϩ1͞a n will eventually be less than r ; that is, there exists an integer N

such that

a nϩ1

whenever n ജ N

Ͻr

an



Ϳ Ϳ



or, equivalently,



Խa Խ Ͻ Խa Խr



4



nϩ1



whenever n ജ N



n



Putting n successively equal to N , N ϩ 1, N ϩ 2, . . . in 4 , we obtain



Խa Խ Ͻ Խa Խr

Խa Խ Ͻ Խa Խr Ͻ Խa Խr

Խa Խ Ͻ Խa Խr Ͻ Խa Խr

Nϩ1



N



Nϩ2



Nϩ1



N



Nϩ3



Nϩ2



N



2



3



and, in general,



Խa Խ Ͻ Խa Խr



5



Nϩk



N



k



for all k ജ 1



Now the series

ϱ



͚ Խa Խr

N



k



k෇1



Խ Խ



Խ Խ



Խ Խ



෇ aN r ϩ aN r 2 ϩ aN r 3 ϩ и и и



is convergent because it is a geometric series with 0 Ͻ r Ͻ 1. So the inequality 5

together with the Comparison Test, shows that the series

ϱ



ϱ



͚ Խa Խ ෇ ͚ Խa Խ ෇ Խa Խ ϩ Խa Խ ϩ Խa Խ ϩ и и и

n



n෇Nϩ1



Nϩk



Nϩ1



Nϩ2



Nϩ3



k෇1



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_p752-761.qk_97817_11_ch11_p752-761 11/3/10 5:30 PM Page 759



ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS



SECTION 11.6



759



is also convergent. It follows that the series ͸ϱn෇1 a n is convergent. (Recall that a finite

number of terms doesn’t affect convergence.) Therefore ͸ a n is absolutely convergent.

(ii) If a nϩ1͞a n l L Ͼ 1 or a nϩ1͞a n l ϱ, then the ratio a nϩ1͞a n will eventually be

greater than 1; that is, there exists an integer N such that



Խ Խ



Խ



Խ



Խ



Խ



Խ



Ϳ Ϳ



a nϩ1

Ͼ1

an



Խ



Խ



whenever n ജ N



Խ Խ Խ



This means that a nϩ1 Ͼ a n whenever n ജ N and so

lim a n



0



nlϱ



Therefore ͸ a n diverges by the Test for Divergence.



Խ



Խ



NOTE Part (iii) of the Ratio Test says that if lim n l ϱ a nϩ1͞a n ෇ 1, the test gives no

information. For instance, for the convergent series ͸ 1͞n 2 we have



Ϳ Ϳ

anϩ1

an



1

͑n ϩ 1͒2

n2







1

͑n ϩ 1͒2

n2



1



ͩ ͪ

1



n



2



as n l ϱ



l1



whereas for the divergent series ͸ 1͞n we have

1

a nϩ1

nϩ1

n

1







l1

an

1

nϩ1

1



n

n



Ϳ Ϳ

The Ratio Test is usually conclusive if the nth

term of the series contains an exponential or a

factorial, as we will see in Examples 4 and 5.



as n l ϱ



Therefore, if lim n l ϱ a nϩ1͞a n ෇ 1, the series ͸ a n might converge or it might diverge. In

this case the Ratio Test fails and we must use some other test.



Խ



Խ



ϱ



EXAMPLE 4 Test the series



͚ ͑Ϫ1͒



n



n෇1



n3

for absolute convergence.

3n



SOLUTION We use the Ratio Test with a n ෇ ͑Ϫ1͒nn 3͞3 n:

Estimating Sums

In the last three sections we used various methods for estimating the sum of a series—the

method depended on which test was used to

prove convergence. What about series for

which the Ratio Test works? There are two

possibilities: If the series happens to be an alternating series, as in Example 4, then it is best to

use the methods of Section 11.5. If the terms are

all positive, then use the special methods

explained in Exercise 38.



Ϳ Ϳ

a nϩ1

an



͑Ϫ1͒nϩ1͑n ϩ 1͒3

3 nϩ1

͑n ϩ 1͒3 3 n





ؒ 3

n 3

͑Ϫ1͒ n

3 nϩ1

n

n

3



|







1

3



|



ͩ ͪ ͩ ͪ

nϩ1

n



3







1

3







1

n



3



l



1

Ͻ1

3



Thus, by the Ratio Test, the given series is absolutely convergent and therefore

convergent.



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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760



CHAPTER 11



INFINITE SEQUENCES AND SERIES

ϱ



v



͚



EXAMPLE 5 Test the convergence of the series



n෇1



nn

.

n!



SOLUTION Since the terms a n ෇ n ͞n! are positive, we don’t need the absolute value

n



signs.

a nϩ1

͑n ϩ 1͒nϩ1 n!



ؒ n

an

͑n ϩ 1͒!

n







͑n ϩ 1͒͑n ϩ 1͒n n!

ؒ n

͑n ϩ 1͒n!

n



ͩ ͪ ͩ ͪ

nϩ1

n



n



n



1

n



෇ 1ϩ



le



as n l ϱ



(see Equation 6.4.9 or 6.4*.9). Since e Ͼ 1, the given series is divergent by the Ratio

Test.

NOTE Although the Ratio Test works in Example 5, an easier method is to use the Test

for Divergence. Since

nn

n ؒ n ؒ n ؒ иии ؒ n

an ෇



ജn

n!

1 ؒ 2 ؒ 3 ؒ иии ؒ n



it follows that a n does not approach 0 as n l ϱ. Therefore the given series is divergent by

the Test for Divergence.

The following test is convenient to apply when n th powers occur. Its proof is similar to

the proof of the Ratio Test and is left as Exercise 41.

The Root Test



Խ Խ



n

(i) If lim s

a n ෇ L Ͻ 1, then the series



nlϱ



ϱ



͚a



n



is absolutely convergent



n෇1



(and therefore convergent).



Խ Խ



Խ Խ



n

n

(ii) If lim s

a n ෇ L Ͼ 1 or lim s

a n ෇ ϱ, then the series



nlϱ



nlϱ



ϱ



͚a



n



is divergent.



n෇1



Խ Խ



n

(iii) If lim s

a n ෇ 1, the Root Test is inconclusive.



nlϱ



Խ Խ



n

If lim n l ϱ s

a n ෇ 1, then part (iii) of the Root Test says that the test gives no information. The series ͸ a n could converge or diverge. (If L ෇ 1 in the Ratio Test, don’t try the

Root Test because L will again be 1. And if L ෇ 1 in the Root Test, don’t try the Ratio Test

because it will fail too.)



ϱ



v



EXAMPLE 6 Test the convergence of the series



͚



n෇1



SOLUTION



an ෇



Խ Խ



n

an

s



ͩ



2n ϩ 3

3n ϩ 2



ͪ



ͩ



2n ϩ 3

3n ϩ 2



ͪ



n



.



n



3

2n ϩ 3

n

2





l Ͻ1

3n ϩ 2

2

3



n





Thus the given series converges by the Root 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.



97817_11_ch11_p752-761.qk_97817_11_ch11_p752-761 11/3/10 5:30 PM Page 761



SECTION 11.6



ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS



761



Rearrangements

The question of whether a given convergent series is absolutely convergent or conditionally

convergent has a bearing on the question of whether infinite sums behave like finite sums.

If we rearrange the order of the terms in a finite sum, then of course the value of the sum

remains unchanged. But this is not always the case for an infinite series. By a rearrangement of an infinite series ͸ a n we mean a series obtained by simply changing the order of

the terms. For instance, a rearrangement of ͸ a n could start as follows:

a 1 ϩ a 2 ϩ a 5 ϩ a 3 ϩ a 4 ϩ a 15 ϩ a 6 ϩ a 7 ϩ a 20 ϩ и и и

It turns out that

if ͸ a n is an absolutely convergent series with sum s,

then any rearrangement of ͸ a n has the same sum s.

However, any conditionally convergent series can be rearranged to give a different sum. To

illustrate this fact let’s consider the alternating harmonic series

1 Ϫ 12 ϩ 13 Ϫ 14 ϩ 15 Ϫ 16 ϩ 17 Ϫ 18 ϩ и и и ෇ ln 2



6



(See Exercise 36 in Section 11.5.) If we multiply this series by 12 , we get

1

2



Ϫ 14 ϩ 16 Ϫ 18 ϩ и и и ෇ 12 ln 2



Inserting zeros between the terms of this series, we have



Adding these zeros does not affect the sum of

the series; each term in the sequence of partial

sums is repeated, but the limit is the same.



0 ϩ 12 ϩ 0 Ϫ 14 ϩ 0 ϩ 16 ϩ 0 Ϫ 18 ϩ и и и ෇ 12 ln 2



7



Now we add the series in Equations 6 and 7 using Theorem 11.2.8:

1 ϩ 13 Ϫ 12 ϩ 15 ϩ 17 Ϫ 14 ϩ и и и ෇ 32 ln 2



8



Notice that the series in 8 contains the same terms as in 6 , but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are

different. In fact, Riemann proved that

if ͸ a n is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of ͸ a n that has a sum equal to r.

A proof of this fact is outlined in Exercise 44.



11.6



Exercises



1. What can you say about the series ͸ a n in each of the following



cases?



Ϳ Ϳ

Ϳ Ϳ



a nϩ1

෇8

(a) lim

nlϱ

an

(c) lim



nlϱ



Ϳ Ϳ



ϱ



5.



n෇0



a nϩ1

෇ 0.8

(b) lim

nlϱ

an



ϱ



7.



2.



͚



n෇1

ϱ



3.



͚



a nϩ1

෇1

an



n෇1



ϱ



6.



2 k

3



͚ ͑Ϫ1͒



n෇1

ϱ



11.



͚



n෇1

ϱ



4.



͚ ͑Ϫ1͒



nϪ1



n෇1



n

n ϩ4

2



ϱ



13.



͚



n෇1



͚



n෇0

ϱ



͚ k( )



8.



͚



n෇1



ϱ



9.



͑Ϫ2͒ n

n2

n

5n



͑Ϫ1͒ n

5n ϩ 1



k෇1



2–30 Determine whether the series is absolutely convergent,

conditionally convergent, or divergent.

ϱ



͚



n



͑1.1͒ n

n4



͑Ϫ3͒ n

͑2n ϩ 1͒!

n!

100 n



ϱ



10.



͚ ͑Ϫ1͒



n



n෇1

ϱ



͑Ϫ1͒n e 1͞n

n3



12.



10 n

͑n ϩ 1͒4 2nϩ1



14.



͚



n෇1

ϱ



͚



n෇1



n

sn 3 ϩ 2



sin 4n

4n

n 10

͑Ϫ10͒ nϩ1



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_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 762



762



CHAPTER 11

ϱ



15.



͚



n෇1

ϱ



17.



͚



n෇2

ϱ



19.



͚



n෇1



16.



͑Ϫ1͒n

ln n



18.



n ϩ1

2n 2 ϩ 1







n෇1

ϱ



25.



͚



n෇1



20.



ϱ



22.



͚



n෇2



n2



1

n



͚



n෇1



n



ϱ



24.



͚



n෇1

ϱ



n 100 100 n

n!



27. 1 Ϫ



͚

ϱ



ͩ ͪ

͚ͩ

ͪ

ϱ



n෇1



n෇1



2



͚



͚

ϱ



cos͑n␲͞3͒

n!



n෇1



23.



ϱ



͑Ϫ1͒ n arctan n

n2



ϱ



21.



INFINITE SEQUENCES AND SERIES



26.



͚



n෇1



3 Ϫ cos n

n 2͞3 Ϫ 2

n!

nn

͑Ϫ2͒

nn



29.



͚



n෇1



(b) Deduce that lim n l ϱ x n͞n! ෇ 0 for all x.



ͩ ͪ

Ϫ2n

nϩ1



5n



͑2n͒!

͑n!͒ 2



n



n෇1



38. Let ͸ a n be a series with positive terms and let rn ෇ a nϩ1 ͞a n.



Suppose that lim n l ϱ rn ෇ L Ͻ 1, so ͸ a n converges by the

Ratio Test. As usual, we let Rn be the remainder after n terms,

that is,

Rn ෇ a nϩ1 ϩ a nϩ2 ϩ a nϩ3 ϩ и и и

(a) If ͕rn ͖ is a decreasing sequence and rnϩ1 Ͻ 1, show, by

summing a geometric series, that



2



2n

n!



2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒

n!



͚ ͑Ϫ1͒



͑n!͒2

͑kn͒!



37. (a) Show that ͸ϱn෇0 x n͞n! converges for all x.

n



1ؒ3ؒ5

1ؒ3ؒ5ؒ7

1ؒ3

ϩ

Ϫ

ϩ иии

3!

5!

7!

1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒

ϩ ͑Ϫ1͒ nϪ1

ϩ иии

͑2n Ϫ 1͒!



ϱ



30.



ϱ



͚



n෇1



2ؒ6

2 ؒ 6 ؒ 10

2 ؒ 6 ؒ 10 ؒ 14

2

28.

ϩ

ϩ

ϩ

ϩ иии

5

5ؒ8

5 ؒ 8 ؒ 11

5 ؒ 8 ؒ 11 ؒ 14

ϱ



36. For which positive integers k is the following series convergent?



Rn ഛ



a nϩ1

1 Ϫ rnϩ1



(b) If ͕rn ͖ is an increasing sequence, show that

Rn ഛ



a nϩ1

1ϪL



39. (a) Find the partial sum s5 of the series ͸ϱn෇1 1͑͞n2 n͒. Use Exer-



cise 38 to estimate the error in using s5 as an approximation

to the sum of the series.

(b) Find a value of n so that sn is within 0.00005 of the sum.

Use this value of n to approximate the sum of the series.



2 n n!

5 ؒ 8 ؒ 11 ؒ и и и ؒ ͑3n ϩ 2͒



40. Use the sum of the first 10 terms to approximate the sum of



the series

ϱ



31. The terms of a series are defined recursively by the equations



a1 ෇ 2



a nϩ1 ෇



5n ϩ 1

an

4n ϩ 3



Determine whether ͸ a n converges or diverges.



a1 ෇ 1



a nϩ1 ෇



sn



an



33–34 Let ͕bn͖ be a sequence of positive numbers that converges



to 12. Determine whether the given series is absolutely convergent.

ϱ



͚



n෇1



bnn cos n␲

n



ϱ



34.



͚



n෇1



͑Ϫ1͒ n n!

n b1 b 2 b 3 и и и bn

n



35. For which of the following series is the Ratio Test inconclusive



(that is, it fails to give a definite answer)?

ϱ



(a)



͚



n෇1

ϱ



(c)



͚



n෇1



ϱ



1

n3



(b)



͚



n෇1



͑Ϫ3͒



ϱ



nϪ1



sn



(d)



͚



n෇1



n

2n

sn

1 ϩ n2



n

2n



Use Exercise 38 to estimate the error.

41. Prove the Root Test. [Hint for part (i): Take any number r such



Խ Խ



Determine whether ͸ a n converges or diverges.



33.



n෇1



that L Ͻ r Ͻ 1 and use the fact that there is an integer N such

n

that s

a n Ͻ r whenever n ജ N .]



32. A series ͸ a n is defined by the equations



2 ϩ cos n



͚



42. Around 1910, the Indian mathematician Srinivasa Ramanujan



discovered the formula

1

2 s2





9801



ϱ



͚



n෇0



͑4n͒!͑1103 ϩ 26390n͒

͑n!͒ 4 396 4n



William Gosper used this series in 1985 to compute the first

17 million digits of ␲.

(a) Verify that the series is convergent.

(b) How many correct decimal places of ␲ do you get if you

use just the first term of the series? What if you use two

terms?

43. Given any series ͸ a n , we define a series ͸ aϩn whose terms are



all the positive terms of ͸ a n and a series ͸ aϪn whose terms

are all the negative terms of ͸ a n. To be specific, we let

aϩn ෇



Խ Խ



an ϩ an

2



aϪn ෇



Խ Խ



a n Ϫ an

2



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_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 763



SECTION 11.7



Notice that if a n Ͼ 0, then aϩn ෇ a n and aϪn ෇ 0, whereas if

a n Ͻ 0, then aϪn ෇ a n and aϩn ෇ 0.

(a) If ͸ a n is absolutely convergent, show that both of the

series ͸ aϩn and ͸ aϪn are convergent.

(b) If ͸ a n is conditionally convergent, show that both of the

series ͸ aϩn and ͸ aϪn are divergent.



763



Take just enough positive terms aϩn so that their sum is greater

than r. Then add just enough negative terms aϪn so that the

cumulative sum is less than r. Continue in this manner and use

Theorem 11.2.6.]

45. Suppose the series ͸ a n is conditionally convergent.



(a) Prove that the series ͸ n 2 a n is divergent.

(b) Conditional convergence of ͸ a n is not enough to determine whether ͸ na n is convergent. Show this by giving an

example of a conditionally convergent series such that

͸ na n converges and an example where ͸ na n diverges.



44. Prove that if ͸ a n is a conditionally convergent series and



r is any real number, then there is a rearrangement of ͸ a n

whose sum is r. [Hints: Use the notation of Exercise 43.



11.7



STRATEGY FOR TESTING SERIES



Strategy for Testing Series

We now have several ways of testing a series for convergence or divergence; the problem

is to decide which test to use on which series. In this respect, testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given

series, but you may find the following advice of some use.

It is not wise to apply a list of the tests in a specific order until one finally works. That

would be a waste of time and effort. Instead, as with integration, the main strategy is to

classify the series according to its form.

1. If the series is of the form



͸ 1͞n p, it is a p-series, which we know to be convergent



if p Ͼ 1 and divergent if p ഛ 1.

2. If the series has the form



Խ Խ



͸ ar nϪ1 or ͸ ar n, it is a geometric series, which converges



Խ Խ



if r Ͻ 1 and diverges if r ജ 1. Some preliminary algebraic manipulation may

be required to bring the series into this form.

3. If the series has a form that is similar to a p-series or a geometric series, then



one of the comparison tests should be considered. In particular, if a n is a rational

function or an algebraic function of n (involving roots of polynomials), then the

series should be compared with a p-series. Notice that most of the series in Exercises 11.4 have this form. (The value of p should be chosen as in Section 11.4 by

keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if ͸ a n has some negative

terms, then we can apply the Comparison Test to ͸ a n and test for absolute

convergence.



Խ Խ



4. If you can see at a glance that lim n l ϱ a n



0, then the Test for Divergence should



be used.

5. If the series is of the form



͸ ͑Ϫ1͒nϪ1bn or ͸ ͑Ϫ1͒nbn , then the Alternating Series



Test is an obvious possibility.

6. Series that involve factorials or other products (including a constant raised to the



nth power) are often conveniently tested using the Ratio Test. Bear in mind that

nϩ1͞a n l 1 as n l ϱ for all p-series and therefore all rational or algebraic

functions of n. Thus the Ratio Test should not be used for such series.



Խa



Խ



7. If a n is of the form ͑bn ͒n, then the Root Test may be useful.

8. If a n ෇ f ͑n͒, where



x1ϱ f ͑x͒ dx is easily evaluated, then the Integral Test is effective



(assuming the hypotheses of this test are satisfied).



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_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 764



764



CHAPTER 11



INFINITE SEQUENCES AND SERIES



In the following examples we don’t work out all the details but simply indicate which

tests should be used.

ϱ



v



EXAMPLE 1



nϪ1

2n ϩ 1



͚



n෇1



Since a n l



0 as n l ϱ, we should use the Test for Divergence.



1

2



ϱ



͚



EXAMPLE 2



n෇1



sn 3 ϩ 1

3n ϩ 4n 2 ϩ 2

3



Since a n is an algebraic function of n, we compare the given series with a p-series. The

comparison series for the Limit Comparison Test is ͸ bn , where

bn ෇

ϱ



v



EXAMPLE 3



n 3͞2

1

sn 3



෇ 3͞2

3n 3

3n 3

3n



͚ ne



Ϫn 2



n෇1



Since the integral x1ϱ xeϪx dx is easily evaluated, we use the Integral Test. The Ratio Test

also works.

2



ϱ



͚ ͑Ϫ1͒



n



EXAMPLE 4



n෇1



n3

n ϩ1

4



Since the series is alternating, we use the Alternating Series Test.

ϱ



v



EXAMPLE 5



͚



k෇1



2k

k!



Since the series involves k!, we use the Ratio Test.

ϱ



EXAMPLE 6



͚



n෇1



1

2 ϩ 3n



Since the series is closely related to the geometric series ͸ 1͞3 n, we use the Comparison

Test.



Exercises



11.7



1–38 Test the series for convergence or divergence.

ϱ



1.



͚



n෇1



1

n ϩ 3n



ϱ



n

3. ͚ ͑Ϫ1͒

nϩ2

n෇1

n



ϱ



5.



ϱ



8.



k෇1



2 Ϫk



e



n



n

4. ͚ ͑Ϫ1͒ 2

n ϩ2

n෇1



1

nsln n



͚k



͑2n ϩ 1͒

n 2n

n



ϱ



ϱ



ϱ



ϱ



6.



͚



͚



n෇1



͚



n෇2



9.



2.



n 2 2 nϪ1

͑Ϫ5͒ n



n෇1



7.



ϱ



͚



n෇1

ϱ



͚



k෇1

ϱ



10.



1

2n ϩ 1

2 k k!

͑k ϩ 2͒!



͚ne



n෇1



2 Ϫn 3



11.



͚



n෇1

ϱ



13.



͚



n෇1

ϱ



15.



͚



k෇1

ϱ



17.



͚



n෇1

ϱ



18.



͚



n෇2



ͩ



1

1

ϩ n

n3

3



ͪ



ϱ



12.



͚



k෇1

ϱ



3n n2

n!



14.



2 kϪ1 3 kϩ1

kk



16.



1

ksk 2 ϩ 1



͚



sin 2n

1 ϩ 2n



ϱ



n2 ϩ 1

n3 ϩ 1



n෇1



͚



n෇1



1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒

2 ؒ 5 ؒ 8 ؒ и и и ؒ ͑3n Ϫ 1͒

͑Ϫ1͒ nϪ1

sn Ϫ 1



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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