Section 11.5 Alternating Series and Absolute Convergence

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Section 11.5 Alternating Series and Absolute Convergence

14. Test the series for convergence or divergence.

∞

P

n=1

(−1)ⁿ⁻¹arctann Solution:

SECTION 11.5 ALTERNATING SERIES AND ABSOLUTE CONVERGENCE ¤ 1091 12. = 

2^  0for  ≥ 1. {^} is decreasing for  ≥ 2 since  2^

0

= 2^−  2^ln 2

(2^)² = 1 −  ln 2

2^  0for   1

ln 2 ≈ 14.

Also, lim

→∞= lim

→∞

 2^

= limH

→∞

1

2^ln 2 = 0. Thus, the series ^∞

=1(−1)^ 

2^ converges by the Alternating Series Test.

13. lim

→∞= lim

→∞^2= ⁰= 1, so lim

→∞(−1)^⁻¹^2does not exist. Thus, the series ^∞

=1(−1)^⁻¹^2diverges by the Test for Divergence.

14. lim

→∞= lim

→∞arctan  = ^₂, so lim

→∞(−1)^⁻¹arctan does not exist. Thus, the series ^∞

=1(−1)^⁻¹arctan diverges by the Test for Divergence.

15. = sin

 +¹₂



1 +√ = (−1)^

1 +√. Now = 1

1 +√  0 for  ≥ 0, {^} is decreasing, and lim__→∞= 0, so the series

∞

=0

sin

 +¹₂



1 +√ converges by the Alternating Series Test.

16.  =  cos 

2^ = (−1)^ 

2^ = (−1)^. {^} is decreasing for  ≥ 2 since (2^−)⁰= (−2^−ln 2) + 2^− = 2^−(1 −  ln 2)  0 for   1

ln 2[≈ 14]. Also, lim__→∞= 0since

lim→∞

 2^

= limH

→∞

1

2^ln 2 = 0. Thus, the series ^∞

=1

 cos 

17. ^∞

=1(−1)^sin

. = sin

  0for  ≥ 2 and sin

 ≥ sin 

 + 1, and lim

→∞sin

 = sin 0 = 0, so the series converges by the Alternating Series Test.

18. ^∞

=1(−1)^cos

. lim

→∞cos

 = cos(0) = 1, so lim

→∞(−1)^cos

 does not exist and the series diverges by the Test for Divergence.

19.  = ²

5^  0for  ≥ 1. {^} is decreasing for  ≥ 2 since

² 5^

0

= 5^· 2 − ²5^ln 5

(5^)² =  5^(2 −  ln 5)

(5^)² = (2 −  ln 5)

5^  0for   2

ln 5 ≈ 12. Also,

lim→∞= lim

→∞

² 5^

= limH

→∞

2

5^ln 5

= limH

→∞

2

5^(ln 5)² = 0. Thus, the series ^∞

=1(−1)^²

20. =

√ + 1 −√



1 ·

√ + 1 +√

√ 

 + 1 +√

 = √( + 1) − 

 + 1 +√

 = 1

√ + 1 +√

  0for  ≥ 1. {^} is decreasing and

lim→∞= 0, so the series ^∞

=1(−1)^√

 + 1 −√

converges by the Alternating Series Test.

18. Test the series for convergence or divergence.

∞

P

n=1

(−1)ⁿcos^π_n Solution:

SECTION 11.5 ALTERNATING SERIES ¤ 145 17. ^∞

=1(−1)^sin 



. = sin 



 0for  ≥ 2 and sin 



≥ sin

 

 + 1



, and lim

→∞sin 



= sin 0 = 0, so the

series converges by the Alternating Series Test.

18. ^∞

=1(−1)^cos 



. lim

→∞cos 



= cos(0) = 1, so lim

→∞(−1)^cos 



does not exist and the series diverges by the Test

for Divergence.

19. ^

! =  ·  · · · 

1 · 2 · · ·  ≥  ⇒ lim

→∞

^

! = ∞ ⇒ lim

→∞

(−1)^^

! does not exist. So the series ^∞

=1(−1)^^

! diverges by the Test for Divergence.

20. =

√ + 1 −√



1 ·

√ + 1 +√

√ 

 + 1 +√

 = √( + 1) − 

 + 1 +√

 = 1

√ + 1 +√

  0for  ≥ 1. {^} is decreasing and

lim→∞= 0, so the series ^∞

=1(−1)^√

 + 1 −√

converges by the Alternating Series Test.

21. The graph gives us an estimate for the sum of the series

∞

=1

(−08)^

! of −055.

8= (08)^

8! ≈ 0000 004, so

∞

=1

(−08)^

! ≈ ⁷=

7

=1

(−08)^

!

≈ −08 + 032 − 00853 + 001706 − 0002 731 + 0000 364 − 0000 042 ≈ −05507 Adding 8to 7does not change the fourth decimal place of 7, so the sum of the series, correct to four decimal places, is −05507.

22. The graph gives us an estimate for the sum of the series

∞

=1(−1)^⁻¹  8^ of 01.

6= 6

8⁶ ≈ 0000 023, so

∞

=1(−1)^⁻¹

8^≈ ⁵=

5

=1(−1)^⁻¹  8^

≈ 0125 − 003125 + 0005 859 − 0000 977 + 0000 153 ≈ 00988

Adding 6to 5does not change the fourth decimal place of 5, so the sum of the series, correct to four decimal places, is 00988.

49. Show that the series P(−1)ⁿ⁻¹b_n, where b_n = 1/n if n is odd and b_n = 1/n² if n is even, is divergent. Why does the Alternating Series Test not apply?

Solution:

SECTION 11.5 ALTERNATING SERIES ¤ 147 30.

∞

=1

(−1)^⁻¹

4^ ≈ ⁶= 1 4− 1

2 · 4²+ 1

3 · 4³ − 1

4 · 4⁴ + 1

5 · 4⁵ − 1

6 · 4⁶ ≈ 0223136. Adding ⁷= 1

7 · 4⁷ ≈ 0000 0087 to ⁶ does not change the fourth decimal place of 6, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is 02231.

31. ^∞

=1

(−1)^⁻¹

 = 1 −1

2+ 1 3−1

4+ · · · + 1 49 − 1

50+ 1 51− 1

52+ · · · . The 50th partial sum of this series is an underestimate, since ^∞

=1

(−1)^⁻¹

 = 50+

 1 51 − 1

52

 +

 1 53− 1

54



+ · · · , and the terms in parentheses are all positive.

The result can be seen geometrically in Figure 1.

32. If   0, 1

( + 1)^ ≤ 1

^ ({1^} is decreasing) and lim_

→∞

1

^ = 0, so the series converges by the Alternating Series Test.

If  ≤ 0, lim_

→∞

(−1)^⁻¹

^ does not exist, so the series diverges by the Test for Divergence. Thus, ^∞

=1

(−1)^⁻¹

^ converges ⇔   0.

33. Clearly = 1

 +  is decreasing and eventually positive and lim

→∞= 0for any . So the series ^∞

=1

(−1)^

 +  converges (by the Alternating Series Test) for any  for which every is defined, that is,  +  6= 0 for  ≥ 1, or  is not a negative integer.

34. Let () = (ln )^

 . Then ⁰() = (ln )^⁻¹( − ln )

²  0if   ^so  is eventually decreasing for every . Clearly

lim→∞

(ln )^

 = 0if  ≤ 0, and if   0 we can apply l’Hospital’s Rule [[ + 1]] times to get a limit of 0 as well. So the series

∞

=2(−1)^⁻¹(ln )^

 converges for all  (by the Alternating Series Test).

35. 

2=

1(2)²clearly converges (by comparison with the -series for  = 2). So suppose that

(−1)^⁻¹

converges. Then by Theorem 11.2.8(ii), so does 

(−1)^⁻¹+ 

= 2

1 +¹₃ +¹₅+ · · ·

= 2 1

2 − 1. But this diverges by comparison with the harmonic series, a contradiction. Therefore,

(−1)^⁻¹must diverge. The Alternating Series Test does not apply since {^} is not decreasing.

36. (a) We will prove this by induction. Let  () be the proposition that 2= 2− ^.  (1) is the statement 2= 2− ¹, which is true since 1 − ¹2 =

1 +¹₂

− 1. So suppose that  () is true. We will show that  ( + 1) must be true as a consequence.

2+2− ^+1=



2+ 1

2 + 1 + 1 2 + 2



−



+ 1

 + 1



= (2− ^) + 1

2 + 1 − 1 2 + 2

= 2+ 1

2 + 1 − 1

2 + 2 = 2+2

which is  ( + 1), and proves that 2= 2− ^for all .

50. Use the following steps to show that

∞

X

n=1

(−1)ⁿ⁻¹ n = ln 2

Let hn and sn be the partial sums of the harmonic and alternating harmonic series.

(a) Show that s2n= h2n− hn. (b) From Exercise 11.3.46 we have

hn− ln n → γ as n → ∞ and therefore

h_2n− ln(2n) → γ as n → ∞ Use these facts together with part (a) to show that s2n → ln 2 as n → ∞.

Solution:

SECTION 11.5 ALTERNATING SERIES ¤ 147 30.

∞

=1

(−1)^⁻¹

4^ ≈ ⁶= 1 4− 1

2 · 4² + 1

3 · 4³ − 1

4 · 4⁴ + 1

5 · 4⁵ − 1

6 · 4⁶ ≈ 0223136. Adding ⁷= 1

7 · 4⁷ ≈ 0000 0087 to ⁶ does not change the fourth decimal place of 6, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is 02231.

31. ^∞

=1

(−1)^⁻¹

 = 1 − 1

2+1 3−1

4 + · · · + 1 49 − 1

50 + 1 51 − 1

52 + · · · . The 50th partial sum of this series is an underestimate, since ^∞

=1

(−1)^⁻¹

 = 50+

 1 51− 1

52

 +

 1 53 − 1

54



+ · · · , and the terms in parentheses are all positive.

The result can be seen geometrically in Figure 1.

32. If   0, 1

( + 1)^ ≤ 1

^ ({1^} is decreasing) and lim_

→∞

1

^ = 0, so the series converges by the Alternating Series Test.

If  ≤ 0, lim_

→∞

(−1)^⁻¹

^ does not exist, so the series diverges by the Test for Divergence. Thus, ^∞

=1

(−1)^⁻¹

^ converges ⇔   0.

33. Clearly = 1

 +  is decreasing and eventually positive and lim

→∞= 0for any . So the series ^∞

=1

(−1)^

 +  converges (by the Alternating Series Test) for any  for which every is defined, that is,  +  6= 0 for  ≥ 1, or  is not a negative integer.

34. Let () = (ln )^

 . Then ⁰() = (ln )^⁻¹( − ln )

²  0if   ^so  is eventually decreasing for every . Clearly

lim→∞

(ln )^

 = 0if  ≤ 0, and if   0 we can apply l’Hospital’s Rule [[ + 1]] times to get a limit of 0 as well. So the series

∞

=2(−1)^⁻¹(ln )^

 converges for all  (by the Alternating Series Test).

35. 

2=

1(2)²clearly converges (by comparison with the -series for  = 2). So suppose that

(−1)^⁻¹

converges. Then by Theorem 11.2.8(ii), so does 

(−1)^⁻¹+ 

= 2

1 +¹₃+¹₅ + · · ·

= 2 1

2 − 1. But this diverges by comparison with the harmonic series, a contradiction. Therefore,

(−1)^⁻¹must diverge. The Alternating Series Test does not apply since {^} is not decreasing.

36. (a) We will prove this by induction. Let  () be the proposition that 2= 2− ^.  (1) is the statement 2= 2− ¹, which is true since 1 − ¹2 =

1 +¹₂

− 1. So suppose that  () is true. We will show that  ( + 1) must be true as a consequence.

2+2− ^+1=



2+ 1

2 + 1+ 1 2 + 2



−



+ 1

 + 1



= (2− ^) + 1

2 + 1− 1 2 + 2

= 2+ 1

2 + 1− 1

2 + 2 = 2+2

which is  ( + 1), and proves that 2= 2− ^for all .

(2)

148 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

(b) We know that 2− ln(2) →  and ^− ln  →  as  → ∞. So

2= 2− ^= [2− ln(2)] − (^− ln ) + [ln(2) − ln ], and

lim→∞2=  −  + lim_

→∞[ln(2) − ln ] = lim_

→∞(ln 2 + ln  − ln ) = ln 2.

11.6 Absolute Convergence and the Ratio and Root Tests

1. (a) Since lim

→∞



+1





 = 8  1, part (b) of the Ratio Test tells us that the series

is divergent.

(b) Since lim

→∞



+1





 = 08  1, part (a) of the Ratio Test tells us that the series

is absolutely convergent (and therefore convergent).

(c) Since lim

→∞



+1





 = 1, the Ratio Test fails and the series

might converge or it might diverge.

2. = 1

√  0for  ≥ 1, {^} is decreasing for  ≥ 1, and lim__→∞= 0, so ^∞

=1

(−1)√^⁻¹

 converges by the Alternating Series Test. To determine absolute convergence, note that ^∞

=1

√1

 diverges because it is a -series with  = ¹₂ ≤ 1. Thus, the series ^∞

=1

(−1)√^⁻¹

 is conditionally convergent.

3. = 1

5 + 1  0for  ≥ 0, {^} is decreasing for  ≥ 0, and lim__→∞= 0, so ^∞

=0

(−1)^

5 + 1 converges by the Alternating Series Test. To determine absolute convergence, choose = 1

to get

lim→∞





= lim

→∞

1

1(5 + 1) = lim

→∞

5 + 1

 = 5  0, so ^∞

=1

1

5 + 1 diverges by the Limit Comparison Test with the harmonic series. Thus, the series ^∞

=0

(−1)^

5 + 1is conditionally convergent.

4. 0  1

³+ 1  1

³ for  ≥ 1 and ^∞

=1

1

³ is a convergent -series ( = 3  1), so ^∞

=1

1

³+ 1converges by comparison and the series

∞

=1

(−1)^

³+ 1is absolutely convergent.

5. 0 



sin  2^



  1

2^ for  ≥ 1 and ^∞

=1

1

2^ is a convergent geometric series ( = ¹₂  1), so

∞

=1



sin  2^



 converges by

comparison and the series

∞

=1

sin 

2^ is absolutely convergent.

6. lim

→∞



+1





 = lim→∞



 (−3)^+1

[2( + 1) + 1]!· (2 + 1)!

(−3)^



 = lim→∞



(−3) 1

(2 + 3)(2 + 2)



 = 3 lim→∞

1

(2 + 3)(2 + 2)

= 3(0) = 0  1 so the series ^∞

=0

(−3)^

(2 + 1)! is absolutely convergent by the Ratio Test.

53. Suppose the seriesP an is conditionally convergent.

(a) Prove that the seriesP n²an is divergent.

(b) Conditional convergence ofP an is not enough to determine whetherP nan is convergent. Show this by giving an example of a conditionally convergent series such thatP nan converges and an example whereP nan diverges.

Solution:

SECTION 11.5 ALTERNATING SERIES AND ABSOLUTE CONVERGENCE ¤ 1097

(b) We know that 2− ln(2) →  and ^− ln  →  as  → ∞. So

2= 2− ^ = [2− ln(2)] − (^− ln ) + [ln(2) − ln ], and

lim→∞2=  −  + lim__→∞[ln(2) − ln ] = lim__→∞(ln 2 + ln  − ln ) = ln 2.

51. (a) Sinceis absolutely convergent, and since⁺ ≤ |^| and⁻ ≤ |^| (because ⁺^ and ⁻ each equal either or 0), we conclude by the Direct Comparison Test that both

⁺and

⁻ must be absolutely convergent.

Or: Use Theorem 11.2.8.

(b) We will show by contradiction that both

⁺ and

⁻ must diverge. For suppose that

⁺ converged. Then so would

⁺ −¹2by Theorem 11.2.8. But

⁺ −¹2

= 1

2(+ |^|) −¹2

= ¹₂

|^|, which diverges because

is only conditionally convergent. Hence,

⁺can’t converge. Similarly, neither can

⁻. 52. Let

be the rearranged series constructed in the hint. [This series can be constructed by virtue of the result of Exercise 51(b).] This series will have partial sums that oscillate in value back and forth across . Since lim

→∞= 0 (by Theorem 11.2.6), and since the size of the oscillations |^− | is always less than |^| because of the way

was constructed, we have that

= lim

→∞= .

53. Suppose that

is conditionally convergent.

(a)

²is divergent: Suppose

²converges. Then lim

→∞²= 0by Theorem 6 in Section 11.2, so there is an integer   0 such that    ⇒ ²|^|  1. For   , we have |^|  1

², so 

|^| converges by comparison with the convergent series 



1

². In other words,

converges absolutely, contradicting the assumption that

is conditionally convergent. This contradiction shows that

²diverges.

Remark: The same argument shows that

^diverges for any   1.

(b) ^∞

=2

(−1)^

 ln  is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely



by the Integral Test, since the function () = 1

 ln is continuous, positive, and decreasing on [2 ∞) and

 ∞ 2



 ln  = lim

→∞

  2



 ln  = lim

→∞



ln(ln ) 2= ∞



. Setting = (−1)^

 ln  for  ≥ 2, we find that

∞

=2

= ^∞

=2

(−1)^

ln  converges by the Alternating Series Test.

It is easy to find conditionally convergent series

such that

diverges. Two examples are ^∞

=1

(−1)^⁻¹

 and

∞

=1

(−1)^⁻¹

√ , both of which converge by the Alternating Series Test and fail to converge absolutely because

|^| is a

series with  ≤ 1. In both cases,

diverges by the Test for Divergence.

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