section 11.5 alternating series - math.ntu.edu.tmathcal/download/108/hw/11.5.pdf · section...

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Section 11.5 Alternating Series 10. = 3 +2 0 for 1. {} is decreasing for 2 since 3 +2 0 = ( 3 + 2) 12 (1) · 1 2 ( 3 + 2) 12 (3 2 ) 3 +2 2 = 1 2 ( 3 + 2) 12 [2( 3 + 2) 3 3 ] ( 3 + 2) 1 = 4 3 2( 3 + 2) 32 0 for 3 4 16. Also, lim →∞ = lim →∞ 3 +2 2 = lim →∞ 1 +2 2 =0. Thus, the series =1 (1) 3 +2 converges by the Alternating Series Test. 17. =1 (1) sin . = sin 0 for 2 and sin sin +1 , and lim →∞ sin = sin 0 = 0, so the series converges by the Alternating Series Test. 20. = +1 1 · +1+ +1+ = ( + 1) +1+ = 1 +1+ 0 for 1. {} is decreasing and lim →∞ =0, so the series =1 (1) +1 converges by the Alternating Series Test. 23. The series =1 (1) +1 6 satisfies (i) of the Alternating Series Test because 1 ( + 1) 6 1 6 and (ii) lim →∞ 1 6 =0, so the series is convergent. Now 5 = 1 5 6 =0000064 000005 and 6 = 1 6 6 000002 000005, so by the Alternating Series Estimation Theorem, =5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to get the sum to the desired accuracy.) 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 1 2 = 1+ 1 2 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 . (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. 1

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Page 1: Section 11.5 Alternating Series - math.ntu.edu.tmathcal/download/108/HW/11.5.pdf · SECTION 11.5ALTERNATING SERIES¤ 1007 17. S" q=1 (3 1)q sin e q=sin A 0IRUq D2DQG +1 DQGlim q

Section 11.5 Alternating Series

1006 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

10. =√

3 + 2 0 for ≥ 1. {} is decreasing for ≥ 2 since

3 + 2

0=

(3 + 2)12(1)− · 12(3 + 2)−12(32)√

3 + 22 =

12(3 + 2)−12[2(3 + 2)− 33]

(3 + 2)1=

4− 3

2(3 + 2)32 0 for

3√

4 ≈ 16. Also, lim→∞

= lim→∞

√3 + 2

√2

= lim→∞

1+ 22

= 0. Thus, the series∞=1

(−1)√

3 + 2

converges by the Alternating Series Test.

11. =2

3 + 4 0 for ≥ 1. {} is decreasing for ≥ 2 since

2

3 + 4

0=

(3 + 4)(2)− 2(32)

(3 + 4)2=

(23 + 8− 33)

(3 + 4)2=

(8− 3)

(3 + 4)2 0 for 2. Also,

lim→∞

= lim→∞

1

1 + 43= 0. Thus, the series

∞=1

(−1)+1 2

3 + 4converges by the Alternating Series Test.

12. = − =

0 for ≥ 1. {} is decreasing for ≥ 1 since (−)0 = (−−) + − = −(1− ) 0 for

1. Also, lim→∞

= 0 since lim→∞

H= lim

→∞1

= 0. Thus, the series

∞=1

(−1)+1− converges by the Alternating

Series Test.

13. =

10 0 for ≥ 1. {} is decreasing for ≥ 1 since

10

0=

10(1)− · 10 ln 10

(10)2=

10(1− ln 10)

(10)2=

1− ln 10

10 0 for 1− ln 10 0 ⇒ ln 10 1 ⇒

1

ln 10≈ 04. Also, lim

→∞ = lim

→∞

10= lim

→∞

10H= lim

→∞

10 ln 10= 0. Thus, the series

∞=1

(−1)

10

converges by the Alternating Series Test.

14. =1

0 for ≥ 1. {} is decreasing since

1

0=

· 1(−12)− 1 · 12

=−1(1 + )

3 0 for

0. Also, lim→∞

= 0 since lim→∞

1 = 1. Thus, the series∞=1

(−1)−1 1

converges by the Alternating Series Test.

15. =sin+ 1

2

1 +√

=(−1)

1 +√. Now =

1

1 +√ 0 for ≥ 0, {} is decreasing, and lim

→∞ = 0, so the series

∞=0

sin+ 1

2

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[≈14]. Also, lim

→∞ = 0 since

lim→∞

2H= lim

→∞1

2 ln 2= 0. Thus, the series

∞=1

cos

2converges by the Alternating Series Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.5 ALTERNATING SERIES ¤ 1007

17.∞=1

(−1) sin

. = sin

0 for ≥ 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 +√ 0 for ≥ 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

(−08)

!of −055.

8 =(08)

8!≈ 0000 004, so

∞=1

(−08)

!≈ 7 =

7=1

(−08)

!

≈ −08 + 032− 00853 + 001706− 0002 731 + 0000 364− 0000 042 ≈ −05507

Adding 8 to 7 does not change the fourth decimal place of 7, so the sum of the series, correct to four decimal places,

is −05507.

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

∞=1

(−1)−1

8of 01.

6 =6

86≈ 0000 023, so

∞=1

(−1)−1

8≈ 5 =

5=1

(−1)−1

8

≈ 0125− 003125 + 0005 859− 0000 977 + 0000 153 ≈ 00988

Adding 6 to 5 does not change the fourth decimal place of 5, so the sum of the series, correct to four decimal places,

is 00988.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.5 ALTERNATING SERIES ¤ 1007

17.∞=1

(−1) sin

. = sin

0 for ≥ 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 +√ 0 for ≥ 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

(−08)

!of −055.

8 =(08)

8!≈ 0000 004, so

∞=1

(−08)

!≈ 7 =

7=1

(−08)

!

≈ −08 + 032− 00853 + 001706− 0002 731 + 0000 364− 0000 042 ≈ −05507

Adding 8 to 7 does not change the fourth decimal place of 7, so the sum of the series, correct to four decimal places,

is −05507.

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

∞=1

(−1)−1

8of 01.

6 =6

86≈ 0000 023, so

∞=1

(−1)−1

8≈ 5 =

5=1

(−1)−1

8

≈ 0125− 003125 + 0005 859− 0000 977 + 0000 153 ≈ 00988

Adding 6 to 5 does not change the fourth decimal place of 5, so the sum of the series, correct to four decimal places,

is 00988.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1008 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

23. The series∞=1

(−1)+1

6satisfies (i) of the Alternating Series Test because

1

(+ 1)6

1

6and (ii) lim

→∞1

6= 0, so the

series is convergent. Now 5 =1

56= 0000064 000005 and 6 =

1

66≈ 000002 000005, so by the Alternating Series

Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to get the

sum to the desired accuracy.)

24. The series∞=1

(− 13)

=

∞=1

(−1)1

3satisfies (i) of the Alternating Series Test because

1

(+ 1)3+1

1

3and

(ii) lim→∞

1

3= 0, so the series is convergent. Now 5 =

1

5 · 35≈ 00008 00005 and 6 =

1

6 · 36≈ 00002 00005,

so by the Alternating Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to

add the first 5 terms to get the sum to the desired accuracy.)

25. The series∞=1

(−1)−1

22satisfies (i) of the Alternating Series Test because

1

(+ 1)22+1

1

22and (ii) lim

→∞1

22= 0,

so the series is convergent. Now 5 =1

5225= 000125 00005 and 6 =

1

6226≈ 00004 00005, so by the Alternating

Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to

get the sum to the desired accuracy.)

26. The series∞=1

− 1

=

∞=1

(−1)1

satisfies (i) of the Alternating Series Test because

1

(+ 1)+1

1

and

(ii) lim→∞

1

= 0, so the series is convergent. Now 5 =

1

55= 000032 000005 and 6 =

1

66≈ 000002 000005, so

by the Alternating Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add

the first 5 terms to get the sum to the desired accuracy.)

27. 4 =1

8!=

1

40,320≈ 0000 025, so

∞=1

(−1)

(2)!≈ 3 =

3=1

(−1)

(2)!= −1

2+

1

24− 1

720≈ −0459 722

Adding 4 to 3 does not change the fourth decimal place of 3, so by the Alternating Series Estimation Theorem, the sum of

the series, correct to four decimal places, is −04597.

28. 6 =1

36 · 6! =1

524,880≈ 0000 001 9, so

∞=1

(−1)

3!≈ 5 =

5=1

(−1)

3!= − 1

3+ 1

18− 1

162+ 1

1944− 1

29,160 ≈ −0283 471

Adding 6 to 5 does not change the fourth decimal place of 5, so by the Alternating Series Estimation Theorem, the sum of

the series, correct to four decimal places, is −02835.

29.∞=1

(−1)−2 ≈ 5 = − 1

2+

2

4− 3

6+

4

8− 5

10≈ −0105 025. Adding 6 = 612 ≈ 0000 037 to 5 does not

change the fourth decimal place of 5, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four

decimal places, is −01050.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.5 ALTERNATING SERIES ¤ 1009

30.∞=1

(−1)−1

4≈ 6 =

1

4− 1

2 · 42+

1

3 · 43− 1

4 · 44+

1

5 · 45− 1

6 · 46≈ 0223136. Adding 7 =

1

7 · 47≈ 0000 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 02231.

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

→∞ = 0 for 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 0 if so is eventually decreasing for every . Clearly

lim→∞

(ln)

= 0 if ≤ 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)2 clearly 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 +

= 21 + 1

3+ 1

5+ · · · = 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 + 1

2

− 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 .

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1010 ¤ 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

= 08 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√ 0 for ≥ 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 = 1

2≤ 1. Thus, the

series∞=1

(−1)−1

is conditionally convergent.

3. =1

5+ 1 0 for ≥ 0, {} is decreasing for ≥ 0, and lim

→∞ = 0, so

∞=0

(−1)

5+ 1converges 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+ 1diverges by the Limit Comparison Test with the

harmonic series. Thus, the series∞=0

(−1)

5+ 1is conditionally convergent.

4. 0 1

3 + 1

1

3for ≥ 1 and

∞=1

1

3is a convergent -series ( = 3 1), so

∞=1

1

3 + 1converges by comparison and

the series∞=1

(−1)

3 + 1is absolutely convergent.

5. 0

sin2

1

2for ≥ 1 and

∞=1

1

2is a convergent geometric series ( = 1

2 1), so

∞=1

sin2

converges bycomparison and the series

∞=1

sin

2is 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.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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