Calculus early transcendental functions 6th edition larson solutions manual

Calculus Early Transcendental Functions 6th Edition Larson Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/calculus-early-transcendental-functions-6th-edition-larson-solutions-manual/

C H A P T E R 9 Infinite Series Section 9.1

Sequences............................................................................................859

Section 9.2

Series and Convergence .....................................................................869

Section 9.3

The Integral Test and p-Series ...........................................................880

Section 9.4

Comparisons of Series........................................................................890

Section 9.5

Alternating Series ...............................................................................897

Section 9.6

The Ratio and Root Tests...................................................................905

Section 9.7

Taylor Polynomials and Approximations .........................................918

Section 9.8

Power Series .......................................................................................930

Section 9.9

Representation of Functions by Power Series...................................943

Section 9.10

Taylor and Maclaurin Series ..............................................................952

Review Exercises .......................................................................................................970 Problem Solving .........................................................................................................983

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

C H A P T E R Infinite Series

9

Section 9.1 Sequences 1. an = 3n 1

a1 = 3 = 3 a2 = 32 = 9 a3 = 33 = 27 a4 = 34 = 81 a5 = 35 = 243

( ) a = ( − 52 ) = − 52 4 a = ( − 52 ) = 25 8 a = ( − 52 ) = − 125 16 a = ( − 52 ) = 625 32 a = ( − 52 ) = − 3125

2. an = − 52

n

1

1

2

5. an = ( −1)

n +1 ⎛ 2 ⎞

⎜ ⎟ ⎝n⎠

2 = 2 1 2 a2 = − = −1 2 2 a3 = 3 2 1 a4 = − = − 4 2 2 a5 = 5 a1 =

2

4

2 1 − 2 n n a1 = 2 + 2 − 1 = 3

5

a2 = 2 + 1 −

3

3

4

5

3. an = sin a1 = sin

nπ 2

π 2

=1

a2 = sin π = 0 3π = −1 2 a4 = sin 2π = 0 a3 = sin

a5 = sin 4. an =

5π =1 2

3n n + 4 3(1)

3 = a1 = 1+ 4 5 3( 2) 6 = =1 a2 = 2 + 4 6 3(3) 9 = a3 = 3+ 4 7 3( 4) 12 3 = = a4 = 4 + 4 8 2 3(5) 15 5 = = a5 = 5+ 4 9 3

6. an = 2 +

1 11 = 4 4 2 1 23 a3 = 2 + − = 3 9 9 2 1 39 a4 = 2 + − = 4 16 16 2 1 59 a5 = 2 + − = 5 25 25

7. a1 = 3, ak + 1 = 2( ak − 1) a2 = 2( a1 − 1) = 2(3 − 1) = 4 a3 = 2( a2 − 1) = 2( 4 − 1) = 6 a4 = 2( a3 − 1) = 2(6 − 1) = 10 a5 = 2( a4 − 1) = 2(10 − 1) = 18 8. a1 = 6, ak + 1 = 13 ak 2

a2 = 13 a12 = 13 (62 ) = 12

a3 = 13 a2 2 = 13 (122 ) = 48

a4 = 13 a32 = 13 ( 482 ) = 768 a5 = 13 a4 2 = 13 (768) = 196,608 2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

859

860

Chapter 9

9. an =

Infinite Series

10 10 10 , a1 = = 5, a2 = 1+1 3 n +1

Matches (c).

10. an =

10n 10 20 , a1 = = 5, a2 = n +1 2 3

Matches (a).

5n 2 = 5 + 2

21. lim

n → ∞ n2

2⎞ ⎛ 22. lim ⎜ 6 + 2 ⎟ = 6 + 0 = 6 n → ∞⎝ n ⎠ 2n

23. lim

n→∞

11. an = ( −1) , a1 = −1, a2 = 1, a3 = −1, …

n +1 2

2

= lim

1 + (1 n )

n→∞

2

=

2 = 2 1

n

Matches (d).

12. an =

(−1)n , a = −1 = −1, a = 1 . n

1

1

2

⎛ 2⎞ 24. lim cos⎜ ⎟ = 1 n→∞ ⎝n⎠ 25.

7

2

Matches (b).

13. an = 3n − 1

0

a5 = 3(5) − 1 = 14

11 0

The graph seems to indicate that the sequence converges to 4. Analytically,

a6 = 3(6) − 1 = 17 Add 3 to preceding term.

lim an = lim

n→∞

14. an = 3 + 5n

a6 = 3 + 5(6) = 33

26.

n→∞

4n + 1 4x + 1 = lim = 4. x→∞ n x

2

a7 = 3 + 5(7) = 38 Add 5 to preceding term.

−1

15. an + 1 = 2an , a1 = 5

−1

a5 = 2( 40) = 80

The graph seems to indicate that the sequence converges to 0. Analytically,

a6 = 2(80) = 160 Multiply the preceding term by 2.

16. an = − 13 an −1 , a1 = 6

( ) 2 a = − 13 ( 27 ) = − 812

12

lim an = lim

n→∞

27.

1

n → ∞ n3 2

= lim

1

x → ∞ x3 2

= 0.

2

2 a5 = − 13 − 92 = 27

0

11

6

Multiply the preceding term by − 13 .

17.

(n + 1)! = n!(n + 1) = n + 1 n!

n!

18.

n! n!

1 = = (n + 2)! (n + 2)(n + 1)n! (n + 2)(n + 1)

19.

(2n − 1)! = (2n − 1)!

1 = (2n + 1)! ( 2n − 1)!( 2n)( 2n + 1) 2n(2n + 1)

20.

−2

The graph seems to indicate that the sequence diverges.

Analytically, the sequence is

{an} = {1, 0, −1, 0, 1, … }.

So, lim an does not exist. n→∞

(2n + 2)! = (2n)!(2n + 1)(2n + 2) (2n)!

(2n)! = ( 2n + 1)( 2n + 2)

© 2014 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 9.1 28.

3

40. an = n sin

−1 −1

The graph seems to indicate that the sequence converges to 2. Analytically,

1 n

lim x sin

x →∞

sin (1 x) 1 = lim x →∞ x 1x = lim

1⎞ ⎛ lim an = lim ⎜ 2 − n ⎟ = 2 − 0 = 2. n→∞ ⎝ 4 ⎠ 5

(−1 x 2 ) cos(1 x) −1 x 2

x →∞

n→∞

n→∞ n + 2

861

1 Let f ( x) = x sin . x

12

29. lim

Sequences

1 = cos 0 x = 1 (L'Hôpital's Rule) = lim cos x →∞

= 0, converges

or,

5⎞ ⎛ 30. lim ⎜ 8 + ⎟ = 8 + 0 = 8, converges n→∞ ⎝ n⎠ n ⎞ n⎛ 31. lim ( −1) ⎜ ⎟ n→∞ n ⎝ + 1⎠

lim

sin (1 x)

lim n sin

n→∞

does not exist (oscillates between −1 and 1), diverges.

= lim

1x

x→∞

y → 0+

sin ( y ) y

= 1. Therefore,

1 = 1, converges. n

41. lim 21 n = 20 = 1, converges n→∞

32. lim

1 + ( −1) n2

n→∞

n

= 0, converges

10n 2 + 3n + 7 10 + 3 n + 7 n 2 = lim 2 n→∞ n→∞ 2n − 6 2 − 6 n2 =

10 = 5, converges 2

3

n→∞

35. lim

n→∞

n = 1, converges n +1

ln ( n3 ) 2n

3 ln ( n) = lim n→∞ 2 n 3⎛1⎞ = lim ⎜ ⎟ = 0, converges n→∞ 2 ⎝ n ⎠

(L’Hôpital’s Rule) n

5n ⎛5⎞ 36. lim n = lim ⎜ ⎟ = ∞, diverges n→∞ 3 n→∞ ⎝ 3 ⎠

(n + 1)! = lim n + 1 = ∞, diverges ( ) n →∞ n →∞

37. lim

38. lim

n→∞

43. lim

n→∞

n!

1 = 0, converges n → ∞ n( n − 1)

np = 0, converges n → ∞ en

converges (because (sin n) is bounded) 44. lim

n→∞

cos π n = 0, converges n2

45. an = − 4 + 6n 46. an =

1 n!

47. an = n 2 − 3 48. an = 49. an =

(−1)n −1 n2

n +1 n + 2

50. an = ( 2n)!, n = 1, 2, 3, …

51. an = 1 +

39. lim

( p > 0, n ≥ 2)

= 0, converges

sin n 1 = lim (sin n) = 0, n→∞ n n

n!

(n − 2)! = lim

−1

n → ∞ 3n

n→∞

33. lim

34. lim 3

42. lim − 3− n = lim

52. an =

1 n +1 = n n

n

(n + 1)(n + 2)

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

862

Chapter 9

53. an = 4 −

Infinite Series

1 1 < 4− = an + 1 , n n +1

57. an =

6 3x . . Then f ′( x) = x + 2 ( x + 2) 2

n

58. an =

= an + 1

( 32 ) < ( 32 ) n

n +1

= an + 1

Monotonic; lim an = ∞, not bounded

So, f is increasing which implies {an} is increasing.

n→∞

55. an = ne−n 2

⎛ nπ ⎞ 59. an = sin ⎜ ⎟ ⎝ 6 ⎠ a1 = 0.500

a1 = 0.6065

a2 = 0.8660

an < 3, bounded

n +1

Monotonic; an ≤ 23 , bounded

Monotonic; an < 4, bounded 54. Let f ( x) =

( 23 ) > ( 23 )

a2 = 0.7358

a3 = 1.000

a3 = 0.6694

a4 = 0.8660

Not monotonic; an ≤ 0.7358, bounded

Not monotonic; an ≤ 1, bounded

( )

56. an = − 23

cos n n a1 = 0.5403

n

60. an =

a1 = − 23

a2 = −0.2081

a2 = 94

a3 = −0.3230

8 a3 = − 27

a4 = −0.1634

Not monotonic; an 61. (a) an = 7 +

7 +

2 ≤ , bounded 3

1 n

Not monotonic; an ≤ 1, bounded

(b)

10

1 ≤ 8 ⇒ {an}, bounded n

an = 7 +

1 1 > 7 + = an + 1 ⇒ {an}, monotonic n n +1

0

1⎞ ⎛ lim ⎜ 7 + ⎟ = 7 n⎠

Therefore, {an} converges.

62. (a) an = 5 −

5−

2 n

11 0

n → ∞⎝

(b)

5

2 ≤ 5 ⇒ {an}, bounded n

an = 5 −

2 2 < 5− = an + 1 ⇒ {an}, monotonic n n +1

Therefore, {an} converges.

0

11 0

2⎞ ⎛ lim ⎜ 5 − ⎟ = 5 − 0 = 5 n⎠

n → ∞⎝

© 2014 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 9.1

63. (a) an =

1⎛ 1⎞ ⎜1 − n ⎟ 3⎝ 3 ⎠

(b)

1⎛ 1⎞ 1 ⇒ {an}, bounded ⎜1 − n ⎟ < 3⎝ 3 ⎠ 3

1⎛ 1⎞ 1⎛ 1 ⎞ an = ⎜1 − n ⎟ < ⎜1 − n + 1 ⎟ 3⎝ 3 ⎠ 3⎝ 3 ⎠

−1

12 −0.1

Therefore, {an} converges.

2+

863

0.4

= an + 1 ⇒ {an}, monotonic

64. (a) an = 2 +

Sequences

⎡1 ⎛ 1 ⎞⎤ 1 lim ⎢ ⎜1 − n ⎟⎥ = 3 ⎠⎦ 3 ⎣3⎝

n→∞

1 5n

1 < 3 ⇒ {an}, bounded 5n

an = 2 +

1 1 > 2 + n + 1 = an + 1 ⇒ {an}, monotonic 5n 5

Therefore, {an} converges. (b)

5

0

11 0

1⎞ ⎛ lim ⎜ 2 + n ⎟ = 2 + 0 = 2 5 ⎠

n → ∞⎝

65. {an} has a limit because it is a bounded, monotonic

sequence. The limit is less than or equal to 4, and greater than or equal to 2.

2 ≤ lim an ≤ 4 n→∞

66. The sequence {an} could converge or diverge. If {an} is

increasing, then it converges to a limit less than or equal to 1. If {an} is decreasing, then it could converge

(example: an = 1 n) or diverge (example: an = −n).

r ⎞ ⎛ 67. An = P⎜1 + ⎟ 12 ⎠ ⎝

n

r ⎞ ⎛ (a) Because P > 0 and ⎜1 + ⎟ > 1, the sequence 12 ⎝ ⎠ diverges. lim An = ∞ n→∞

0.055 ⎞ ⎛ (b) P = 10,000, r = 0.055, An = 10,000⎜1 + ⎟ 12 ⎠ ⎝ A0 = 10,000

A1 = 10,045.83 A2 = 10,091.88 A3 = 10,138.13 A4 = 10,184.60 A5 = 10,231.28 A6 = 10,278.17 A7 = 10,325.28 A8 = 10,372.60 A9 = 10,420.14 A10 = 10,467.90

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

n

864

Chapter 9

Infinite Series

68. (a) An = 100( 401)(1.0025n − 1)

74. Pn = 25,000(1.045)

A0 = 0

P1 = $26,125.00

A1 = 100.25

P2 = $27,300.63

A2 = 200.75

P3 = $28,529.15

A3 = 301.50

P4 = $29,812.97

A4 = 402.51

P5 = $31,154.55

A5 = 503.76

n

75. an = n n = n1 n

A6 = 605.27 (b) A60 = 6480.83

a1 = 11 1 = 1

(c) A240 = 32,912.28

a2 =

69. No, it is not possible. See the “Definition of the Limit of a sequence”. The number L is unique.

70. (a) A sequence is a function whose domain is the set of positive integers.

2 ≈ 1.4142

a3 =

3

3 ≈ 1.4422

a4 =

4

4 ≈ 1.4142

a5 =

5

5 ≈ 1.3797

a6 =

6

6 ≈ 1.3480

(b) A sequence converges if it has a limit. See the definition.

Let y = lim n1 n .

(c) A sequence is monotonic if its terms are nondecreasing, or nonincreasing.

ln n 1n ⎛1 ⎞ = lim = 0 ln y = lim ⎜ ln n ⎟ = lim n→∞ ⎝ n n → ∞ n → ∞ n n ⎠

(d) A sequence is bounded if it is bounded below (an ≥ N for some N ) and bounded above

(an ≤ M for some M ).

71. (a) an = 10 −

1 n

(b) Impossible. The sequence converges by Theorem 9.5.

3n (c) an = 4n + 1

n→∞

Because ln y = 0, you have y = e0 = 1. Therefore,

lim n n = 1.

n→∞

1⎞ ⎛ 76. an = ⎜1 + ⎟ n⎠ ⎝ a1 = 2.0000

n

a2 = 2.2500 a3 ≈ 2.3704

(d) Impossible. An unbounded sequence diverges.

72. The graph on the left represents a sequence with alternating signs because the terms alternate from being above the x-axis to being below the x-axis.

73. (a) An = (0.8) 4,500,000,000

a4 ≈ 2.4414 a5 ≈ 2.4883 a6 ≈ 2.5216 n

1⎞ ⎛ lim ⎜1 + ⎟ = e n→∞ ⎝ n⎠

n

(b) A1 = $3,600,000,000 A2 = $2,880,000,000 A3 = $2,304,000,000 A4 = $1,843,200,000

(c) lim An = lim (0.8) ( 4.5) = 0, converges n

n→∞

n→∞

77. Because lim sn = L > 0,

n→∞

there exists for each ε > 0, an integer N such that sn − L < ε for every n > N .

Let ε = L > 0 and you have, sn − L < L, − L < sn − L < L, or 0 < sn < 2 L for each n > N .

© 2014 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 9.1 78. (a) an = 0.072n 2 + 0.02n + 5.8

Sequences

865

(b) For 2020, n = 20: a20 = $35 trillion

20

0

12 0

79. True

81. True

80. True

82. False. Let an = ( −1) and bn = ( −1) n

n +1

then {an} and

{bn} diverge. But {an + bn} = {(−1) + (−1) n

}

n +1

converges to 0.

83. an + 2 = an + an + 1

(a) a1 = 1

a7 = 8 + 5 = 13

a2 = 1

a8 = 13 + 8 = 21

a3 = 1 + 1 = 2

a9 = 21 + 13 = 34

a4 = 2 + 1 = 3

a10 = 34 + 21 = 55

a5 = 3 + 2 = 5

a11 = 55 + 34 = 89

a6 = 5 + 3 = 8

a12 = 89 + 55 = 144

(b) bn =

an + 1 ,n ≥ 1 an

(c) 1 +

1 1 =1+ bn −1 an an − 1 =1+

an − 1 a + an −1 a = n = n +1 = bn an an an

⎛ 1 ⎞ (d) If lim bn = ρ , then lim ⎜1 + ⎟ = ρ. n→∞ n→∞ bn −1 ⎠ ⎝

Because lim bn = lim bn −1 , you have n→∞

13 = 1.625 8 21 b7 = ≈ 1.6154 13 34 b8 = ≈ 1.6190 21 55 b9 = ≈ 1.6176 34 89 b10 = ≈ 1.6182 55

1 =1 1 2 b2 = = 2 1 3 b3 = = 1.5 2 5 b4 = ≈ 1.6667 3 8 b5 = = 1.6 5

b6 =

b1 =

n→∞

1 + (1 ρ ) = ρ .

ρ + 1 = ρ2 0 = ρ2 − ρ − 1

ρ =

1± 1+ 4 1± 5 = 2 2

Because an , and therefore bn , is positive,

ρ =

1+

5 2

≈ 1.6180.

84. Let f ( x) = sin (π x)

lim sin (π x) does not exist.

x→∞

an = f ( n) = sin (π n) = 0 for all n lim an = 0, coverges

n→∞

85. (a) a1 =

2 ≈ 1.4142

a2 =

2+

2 ≈ 1.8478

a3 =

2+

2+

2 ≈ 1.9616

a4 =

2+

2+

2+

2 ≈ 1.9904

a5 =

2+

2+

2+

2+

(b) an =

2 + an −1 ,

n ≥ 2, a1 =

2 ≈ 1.9976 2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

866

Chapter 9

Infinite Series

(c) First use mathematical induction to show that an ≤ 2; clearly a1 ≤ 2. So assume ak ≤ 2. Then ak + 2 ≤ 4 ak + 2 ≤ 2 ak + 1 ≤ 2.

Now show that {an} is an increasing sequence. Because an ≥ 0 and an ≤ 2,

(an − 2)(an + 1) ≤ 0 an2 − an − 2 ≤ 0 an2 ≤ an + 2 an ≤

an + 2

an ≤ an + 1 .

Because {an} is a bounding increasing sequence, it converges to some number L, by Theorem 9.5. lim an = L ⇒

n→∞

2 + L = L ⇒ 2 + L = L2 ⇒ L2 − L − 2 = 0

⇒ ( L − 2)( L + 1) = 0 ⇒ L = 2

( L ≠ −1)

86. (a) Use mathematical induction to show that

1+

an ≤

1 + 4k .

2

[Note that if k = 2, and an ≤ 3, and if k = 6, then an ≤ 3. ] Clearly, a1 =

k ≤

1 + 4k 1+ ≤ 2

1 + 4k .

2

Before proceeding to the induction step, note that 2 + 2 1 + 4k + 4k = 2 + 2 1 + 4k + 4 k 1+

1 + 4k 1 + 2 1 + 4 k + 1 + 4k + k = 2 4

1+

⎡1 + 1 + 4k + k = ⎢ 2 ⎣⎢

1 + 4k ⎤ ⎥ 2 ⎦⎥

1+

1 + 4k 1+ + k = 2

1 + 4k .

2

So assume an ≤ an + k ≤ an + k ≤ an + 1 ≤

1+

1+ 1+ 1+

2

1 + 4k . Then 2

1 + 4k + k 2 1 + 4k + k 2 1 + 4k .

2

{an} is increasing because ⎛ 1+ ⎜⎜ an − ⎝

1 + 4k ⎞⎛ 1− ⎟⎜ ⎟⎜ an − 2 ⎠⎝

1 + 4k ⎞ ⎟⎟ ≤ 0 2 ⎠

an2 − an − k ≤ 0 an2 ≤ an + k an ≤

an + k

an ≤ an + 1. © 2014 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 9.1

Sequences

867

(b) Because {an} is bounded and increasing, it has a limit L. (c) lim an = L implies that n→∞

L =

k + L ⇒ L2 = k + L ⇒ L2 − L − k = 0 1 ± 1 + 4k .

2

⇒ L =

Because L > 0, L = 87. (a)

1+

1 + 4k .

2

y

(d)

y = lnx

2.0 1.5

n

< n n! < e1 − (1 n)

1.0 0.5

2

n

∫1

(n + 1) nn < n! < n −1 e en

3

...

4

1

x

< e1 − (1 n)

n

ln x dx < ln 2 + ln 3 +

n) = ln ( n!)

n→∞

e

1

=

1 e

(n + 1)1+ (1 n) = lim (n + 1) (n + 1)1 n

y 2.0

n + 1) n

1 + (1 n)

lim

lim

(n + 1)(

(n + 1) n! < n ne

n → ∞ e1 − (1 n)

+ ln n

= ln (1 ⋅ 2 ⋅ 3

(b)

n

n +1

ne

n

n→∞

e

1 = (1) e 1 = e

y = lnx

1.5 1.0 0.5

n 2

n +1

∫1 (c)

3

4

...

By the Squeeze Theorem, lim

x n+1

n→∞

ln x dx > ln 2 + ln 3 +

+ ln n = ln ( n!)

(e) n = 20:

n = 50:

∫ ln x dx = x ln x − x + C n

20

20! ≈ 0.4152 20

50

50! ≈ 0.3897 50

100

∫ 1 ln x dx = n ln n − n + 1 = ln n − n + 1

n = 100:

From part (a): ln n n − n + 1 < ln ( n!)

1 ≈ 0.3679 e

n

n

eln n − n + 1 < n!

n!

1 = . n e

100! ≈ 0.3799 100

nn < n! e n −1 n +1

∫1

ln x dx = ( n + 1) ln ( n + 1) − ( n + 1) + 1 = ln ( n + 1)

n +1

−n

From part (b): ln(n + 1) n + 1 − n > ln ( n!) eln(n + 1)

n +1 − n

(n + 1) en

> n!

n +1

> n!

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

868

Chapter 9

Infinite Series

88. For a given ε > 0, you must find M > 0 such that

an − L =

1 < ε n3

whenever n > M . That is, 13

⎛1⎞ or n > ⎜ ⎟ . ε ⎝ε ⎠ 1

n3 >

So, let ε > 0 be given. Let M be an integer satisfying M > (1 ε ) . For n > M , you have 13

13

⎛1⎞ n > ⎜ ⎟ ⎝ε ⎠ 1 n3 >

n

{a2 n} = {( −1)2 n} = {1, 1, 1, 1, …} converges

91. If {an} is bounded, monotonic and nonincreasing, then

a1 ≥ a2 ≥ a3 ≥ ≥ an ≥ . Then − a1 ≤ − a2 ≤ − a3 ≤ ≤ − an ≤ is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Because {− an} converges,

92. Define an =

1 1 ⇒ 3 − 0 < ε.

3 n n

So, lim

{an} = {( −1) } = {−1, 1, −1, 1, …} diverges

then so does {an}.

ε

ε >

90. Answers will vary. Sample answer:

1

= 0.

n → ∞ n3

xn + 1 + xn −1 , xn

n ≥ 1.

xn + 12 − xn xn + 2 = 1 = xn2 − xn −1xn + 1 ⇒ xn + 1 ( xn + 1 + xn −1 ) = xn ( xn + xn + 2 )

89. For a given ε > 0, you must find M > 0 such that

xn + 1 + xn −1 x + xn = n+2 xn xn + 1

an − L = r n ε whenever n > M . That is,

an = an + 1

n ln r < ln (ε ) or ln (ε )

n >

ln r

Therefore, a1 = a2 = … = a. So, xn + 1 = an xn − xn −1 = axn − xn −1.

(because ln r < 0 for r < 1 ).

So, let ε > 0 be given. Let M be an integer satisfying M >

ln (ε ) ln r

.

For n > M , you have n >

ln (ε ) ln r

n ln r < ln (ε ) ln r

n

< ln (ε )

r

n

< ε

r − 0 < ε. n

So, lim r n = 0 for −1 < r < 1. n→∞

© 2014 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 9.2

Series and Convergence

869

93. Tn = n! + 2n

Use mathematical induction to verify the formula.

T0 = 1 + 1 = 2 T1 = 1 + 2 = 3 T2 = 2 + 4 = 6

Assume Tk = k! + 2k . Then

Tk + 1 = ( k + 1 + 4)Tk − 4( k + 1)Tk −1 + ( 4( k + 1) − 8)Tk − 2 = ( k + 5) ⎡⎣k! + 2k ⎤⎦ − 4( k + 1)(( k − 1)! + 2k −1 ) + ( 4k − 4)(( k − 2)! + 2k − 2 ) = ⎡⎣( k + 5)( k )( k − 1) − 4( k + 1)( k − 1) + 4( k − 1)⎤⎦ ( k − 2)! + ⎡⎣( k + 5)4 − 8( k + 1) + 4( k − 1)⎤⎦ 2k − 2 = ⎡⎣k 2 + 5k − 4k − 4 + 4⎤⎦ ( k − 1)! + 8 ⋅ 2k − 2 = ( k + 1)! + 2k + 1.

By mathematical induction, the formula is valid for all n.

Section 9.2 Series and Convergence 1. S1 = 1 S 2 = 1 + 14

5. S1 = 3 = 1.2500

S 2 = 3 + 32 = 4.5

S3 = 1 + 14 + 91 ≈ 1.3611

S3 = 3 + 32 + 34 = 5.250

1 S 4 = 1 + 14 + 91 + 16 ≈ 1.4236

S 4 = 3 + 32 + 34 + 83 = 5.625

1 + 1 ≈ 1.4636 S5 = 1 + 14 + 91 + 16 25

3 = 5.8125 S5 = 3 + 32 + 34 + 83 + 16

6. S1 = 1

2. S1 = 16 ≈ 0.1667

S 2 = 16 + 16 ≈ 0.3333 3 ≈ 0.4833 S3 = 16 + 16 + 20 3 2 + 15 ≈ 0.6167 S 4 = 16 + 16 + 20 3 2 + 5 ≈ 0.7357 + 15 S5 = 16 + 16 + 20 42

S3 = 1 − 12 + 16 ≈ 0.6667 1 ≈ 0.6250 S 4 = 1 − 12 + 16 − 24 1 1 + 120 ≈ 0.6333 S5 = 1 − 12 + 16 − 24 ∞

3. S1 = 3

7. ∑

n=0

S 2 = 3 − 92 = −1.5 S3 = 3 − 92 + 27 = 5.25 4 S 4 = 3 − 92 + 27 − 81 = −4.875 4 8 S5 = 3 − 92 + 27 − 81 + 243 = 10.3125 4 8 16

( 76 )

n

Geometric series r = 76 > 1 Diverges by Theorem 9.6 ∞

8. ∑ 4( −1.05)

4. S1 = 1

n

n=0

S 2 = 1 + 12 = 23 S3 = 1 + 12 + 14

S 2 = 1 − 12 = 0.5

Geometric series =

7 4

23 S 4 = 1 + 12 + 14 + 16 = 12

r = −1.05 = 1.05 > 1 Diverges by Theorem 9.6

49 S5 = 1 + 12 + 14 + 16 + 18 = 24

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

870

Chapter 9

Infinite Series

∞

∞

n n =1 n + 1

9. ∑

n

lim

n→∞ n + 1

14. ∑

n!

n =1 2

=1≠ 0

lim

n

n!

n → ∞ 2n

Diverges by Theorem 9.9

= ∞

Diverges by Theorem 9.9 ∞

∞

n n = 1 2n + 3

15. ∑

10. ∑

n=0

( 56 )

n

n 1 lim = ≠ 0 n → ∞ 2n + 3 2

Geometric series with r = 56 < 1

Diverges by Theorem 9.9

Converges by Theorem 9.6 ∞

n2 2 n =1 n + 1

n=0

2

Converges by Theorem 9.6

Diverges by Theorem 9.9

∞

∞

17. ∑ (0.9)

n

12. ∑

lim

n2 + 1

n→∞

n

n=0

n2 + 1 n

n =1

n

Geometric series with r = − 12 < 1

n2 =1 ≠ 0 n→∞ n + 1 lim

( )

16. ∑ 2 − 12

∞

11. ∑

Geometric series with r = 0.9 < 1

1

= lim

1 + (1 n 2 )

n→∞

=1≠ 0

Converges by Theorem 9.6 ∞

18. ∑ ( −0.6)

Diverges by Theorem 9.9

n

n=0

∞

2n + 1 13. ∑ n + 1 n =1 2

Geometric series with r = −0.6 < 1 Converges by Theorem 9.6

2n + 1 1 + 2− n 1 = lim = ≠ 0 + 1 n n→∞ 2 n→∞ 2 2 lim

Diverges by Theorem 9.9 ∞

∞ 1 1 ⎞ ⎛ 1 ⎞ ⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞ ⎛1 = ∑⎜ − ⎟ = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + 1 1 2 ⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠ ⎝ 4 5⎠ + + n n n n ( ) ⎠ ⎝ n =1 n =1 ⎝

19. ∑ ∞

⎛

1

, Sn = 1 −

1 n +1

1 ⎞

S n = lim ⎜1 − ∑ n(n + 1) = nlim ⎟ =1 →∞ n→∞ ⎝ n + 1⎠

n =1

∞ ⎛ ⎞ ⎛1 1 1 1 1⎞ ⎛ 1 1⎞ ⎛ 1 1 ⎞ ⎛1 1⎞ ⎛1 1⎞ = ∑⎜ − − ⎟⎟ = ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +⎜ ⎟ + ⎜ 2( n + 2) ⎠ ⎝ 2 6 ⎠ ⎝ 4 8 ⎠ ⎝ 6 10 ⎠ ⎝ 8 12 ⎠ ⎝ 10 14 ⎠ n = 1 n( n + 2) n = 1 ⎝ 2n ∞

20. ∑ ∞

⎡1

1

1

1

⎤

1

1

1

3

⎛1

1⎞

⎛1

S n = lim ⎢ + − − + = ⎥ = ∑ n(n + 2) = nlim →∞ n→∞ ⎢2 4 2( n + 1) 2( n + 2) ⎥ 2 4 4

⎣

n =1

21. (a)

∞

∞

6

⎛1

⎦

1

⎞

⎡⎛

1⎞

⎛1

1⎞

1⎞

∑ n(n + 3) = 2∑ ⎜⎝ n − n + 3 ⎟⎠ = 2⎢⎣⎜⎝1 − 4 ⎟⎠ + ⎜⎝ 2 − 5 ⎟⎠ + ⎜⎝ 3 − 6 ⎟⎠ + ⎜⎝ 4 − 7 ⎟⎠ + n =1

n =1

⎤ ⎥ ⎦

⎛ ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎞ 1 1⎞ 11 ⎛ + + ≈ 3.667 ⎜ Sn = 2 ⎢1 + + − ⎜ ⎟⎥ ⎟ = 2⎜1 + + ⎟ = + + + n n n 2 3 1 2 3 2 3 3 ⎝ ⎠⎦ ⎠ ⎝ ⎠ ⎣ ⎝ (b)

n

5

10

20

50

100

Sn

2.7976

3.1643

3.3936

3.5513

3.6078

© 2014 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 9.2

Series and Convergence

871

5

(c)

0

11 0

(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.

22. (a)

∞

∞

4

⎛1

⎞

1

∑ n(n + 4) = ∑ ⎜⎝ n − n + 4 ⎟⎠ n =1

n =1

⎛ 1 ⎞ ⎛1 1 ⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛1 1 ⎞ 1 ⎞ ⎛1 = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + 5 2 6 3 7 4 8 5 9 6 10 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 1 1 1 25 =1+ + + = ≈ 2.0833 2 3 4 12 (b)

(c)

n

5

10

20

50

100

Sn

1.5377

1.7607

1.9051

2.0071

2.0443

3

0

11 0

(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.

23. (a)

∞

∑ 2(0.9)

n −1

n =1

(b)

(c)

∞

= ∑ 2(0.9) = n

n=0

2 = 20 1 − 0.9

n

5

10

20

50

100

Sn

8.1902

13.0264

17.5685

19.8969

19.9995

22

0

11

0

(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.

24. (a)

∞

⎛ 1⎞

∑ 10⎜⎝ − 4 ⎟⎠

n −1

n =1

(b)

(c)

n

∞ 10 ⎛ 1⎞ = ∑ 10⎜ − ⎟ = = 8 1 − ( −1 4) ⎝ 4⎠ n=0

n

5

10

20

50

100

Sn

8.0078

7.99999

8.0000

8.0000

8.0000

11

0

11 7

(d) The terms of the series decrease in magnitude rapidly. So, the sequence of partial sums approaches the sum rapidly. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

872

Chapter 9

Infinite Series

n

∞ 5 ⎛ 2⎞ = 15 25. ∑ 5⎜ ⎟ = − 3 1 (2 3) n=0 ⎝ ⎠ n

∞ 1 5 ⎛ 1⎞ 26. ∑ ⎜ − ⎟ = = − − 5 1 1 5 6 ( ) ⎠ n=0 ⎝ ∞

∞ 4 1 ⎞ ⎛1 = 2∑ ⎜ − ⎟ n + 2⎠ n = 1 n ( n + 2) n =1 ⎝ n

27. ∑

⎡⎛ 1⎞ ⎛ 1 1⎞ ⎛1 1⎞ S n = 2 ⎢⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + 3⎠ ⎝ 2 4⎠ ⎝ 3 5⎠ ⎣⎝ ∞

⎛

4

1

1 ⎞ ⎛1 1 ⎞⎤ 1 1 1 ⎞ ⎛ 1 ⎛ +⎜ − − ⎟+⎜ − ⎟⎥ = 2⎜1 + − ⎟ 2 n + 1 n + 2⎠ ⎝ n − 1 n + 1 ⎠ ⎝ n n + 2 ⎠⎦ ⎝ 1

1

⎞

Sn = lim 2⎜1 + − − ∑ n(n + 2) = nlim ⎟ = 3 n →∞ ⎝ →∞ 2 n + 1 n + 2⎠ n =1 ∞

28. ∑

n =1

1

(2n + 1)(2n + 3)

Sn =

=

1 ∞ ⎛ 1 1 ⎞ − ∑⎜ ⎟ 2 n = 1 ⎝ 2 n + 1 2n + 3 ⎠

1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎜ − ⎟+⎜ − ⎟+⎜ − ⎟+ 2 ⎢⎣⎝ 3 5 ⎠ ⎝ 5 7 ⎠ ⎝ 7 9 ⎠

∞

1⎛ 1

1

1 ⎞⎤ 1⎛ 1 1 ⎞ ⎛ 1 +⎜ − ⎟⎥ = ⎜ − ⎟ 2 ⎝ 3 2n + 3 ⎠ ⎝ 2 n + 1 2 n + 3 ⎠⎦ 1

⎞

1

S n = lim ⎜ − ∑ (2n + 1)(2n + 3) = nlim ⎟ = n →∞ 2 ⎝ 3 →∞ 2n + 3 ⎠ 6 n =1

n

n

∞ 8 ⎛ 3⎞ 29. ∑ 8⎜ ⎟ = = 32 1 − (3 4) n=0 ⎝ 4 ⎠

∞ ∞ ∞ n n ⎛3⎞ ⎛8⎞ 32. ∑ ⎡(0.3) + (0.8) ⎤ = ∑ ⎜ ⎟ + ∑ ⎜ ⎟ ⎣ ⎦ n=0 n = 0 ⎝ 10 ⎠ n = 0 ⎝ 10 ⎠

n

∞

9 27 ⎛ 1⎞ 30. ∑ 9⎜ − ⎟ = = 1 − ( −1 3) 4 n=0 ⎝ 3⎠ n

∞ ∞ ∞ 1⎞ ⎛1 ⎛1⎞ ⎛1⎞ 31. ∑ ⎜ n − n ⎟ = ∑ ⎜ ⎟ − ∑ ⎜ ⎟ 3 ⎠ n=0 ⎝ 2 ⎠ n=0 ⎝ 2 n=0 ⎝ 3 ⎠

=

n

=

1 1 + 1 − (3 10) 1 − (8 10)

=

10 45 +5 = 7 7

n

∞

1 1 3 1 − = 2− = 1 − (1 2) 1 − (1 3) 2 2

33. Note that sin (1) ≈ 0.8415 < 1. The series ∑ ⎣⎡sin (1)⎦⎤

n

n =1

is geometric with r = sin (1) < 1. So, ∞

∞

sin (1)

∑ ⎡⎣sin(1)⎤⎦ = sin(1)∑ ⎡⎣sin (1)⎤⎦ = 1 − sin(1) ≈ 5.3080. n =1

n

n

n

n=0

n

1 1 = ∑ 2 9 3 2 3 1 3k + 2) k + k − k − ( )( k =1 k =1

34. S n = ∑

n 1 ⎤ 1 n ⎡ 1 1 ⎤ ⎡ 1 = ∑⎢ − = ∑⎢ − ⎥ 9 3 9 6 3 3 1 3 k k k k − + − + 2 ⎥⎦ ⎦ k =1 ⎣ k =1 ⎣

=

1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1⎞ ⎜ − ⎟+⎜ − ⎟+⎜ − ⎟+ 3 ⎢⎣⎝ 2 5 ⎠ ⎝ 5 8 ⎠ ⎝ 8 11 ⎠

lim Sn = lim

n→∞

1⎛ 1 ⎜ −

n → ∞ 3⎝ 2

1 ⎞⎤ 1⎛ 1 1 ⎞ ⎛ 1 +⎜ − ⎟⎥ = ⎜ − ⎟ 3 1 3 2 3 2 3 n n n − + + 2⎠ ⎝ ⎠⎦ ⎝

1 ⎞ 1 ⎟ = 3n + 2 ⎠ 6

© 2014 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 9.2 ∞

4⎛1⎞ ⎜ ⎟ 10 ⎝ 10 ⎠ n=0

35. (a) 0.4 = ∑

n

40. (a) 0.215 =

(b) Geometric series with a = S =

∞

36 ⎛ 1 ⎞ ⎜ ⎟ n = 0 100 ⎝ 100 ⎠

81 ⎛ 1 ⎞ 37. (a) 0.81 = ∑ ⎜ ⎟ n = 0 100 ⎝ 100 ⎠

∞

n

3 1 and r = 200 100

n

n=0

Geometric series with r = 1.075

36 1 and r = 100 100

Diverges by Theorem 9.6 ∞

3n n = 1 1000

42. ∑

Geometric series with r = 3 > 1.

n

(b) Geometric series with a =

873

1 a 1 3 200 71 + = + = 5 1− r 5 99 100 330

41. ∑ (1.075)

a 36 100 36 4 = = = 1− r 1 − (1 100) 99 11 ∞

S =

S =

n

(b) Geometric series with a =

∞ 1 3 ⎛ 1 ⎞ + ∑ ⎜ ⎟ 5 n = 0 200 ⎝ 100 ⎠

(b) Geometric series with a =

a 4 10 4 = = 1− r 1 − (1 10) 9

36. (a) 0.36 = ∑

S =

4 1 and r = 10 10

Series and Convergence

Diverges by Theorem 9.6 ∞

n + 10 n +1 10 n =1

43. ∑

81 1 and r = 100 100

a 81 100 81 9 = = = 1− r 1 − (1 100) 99 11

lim

n + 10

n → ∞ 10n + 1

=

1 ≠ 0 10

Diverges by Theorem 9.9 n

∞

∞

1 ⎛ 1 ⎞ ⎛ 1 ⎞ 38. (a) 0.01 = ∑ ⎜ ∑ ⎜⎝ 100 ⎟⎠ ⎟ = 100 100 ⎝ ⎠ n =1 n=0 (b) 0.01 =

∞

44. ∑

∞

3⎛ 1 ⎞ ⎜ ⎟ n = 0 40 ⎝ 100 ⎠

lim

4n + 1

n → ∞ 3n − 1

=

4 ≠ 0 3

Diverges by Theorem 9.9

n

(b) Geometric series with a =

4n + 1

n = 1 3n − 1

1 1 1 100 1 ⋅ = ⋅ = 100 1 − (1 100) 100 99 99

39. (a) 0.075 = ∑

S =

n

3 1 and r = 40 100

a 3 40 5 = = 1− r 99 100 66

∞ 1 ⎞ ⎛1 45. ∑ ⎜ − ⎟ + n n 2⎠ n =1 ⎝

1⎞ ⎛ 1 1⎞ ⎛1 1⎞ ⎛ S n = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + 3⎠ ⎝ 2 4⎠ ⎝ 3 5⎠ ⎝ ∞

⎛1

1

⎞

⎛

1 ⎞ ⎛1 1 ⎞ 1 1 1 ⎛ 1 +⎜ − − ⎟+⎜ − ⎟ = 1+ − 2 n +1 n + 2 ⎝ n − 1 n + 1⎠ ⎝ n n + 2 ⎠ 1

1

1

⎞

3

− S n = lim ⎜1 + − ∑ ⎜⎝ n − n + 2 ⎟⎠ = nlim ⎟ = , converges →∞ n→∞ ⎝ 2 2 n + 1 n + 2⎠ n =1

∞ 1 ⎞ ⎛ 1 − 46. ∑ ⎜ ⎟ n + 2⎠ n =1 ⎝ n + 1

⎛ 1 1⎞ ⎛1 1⎞ Sn = ⎜ − ⎟ + ⎜ − ⎟ + ⎝ 2 3⎠ ⎝ 3 4⎠ ∞

⎛

1

1

⎞

1 ⎞ 1 1 ⎛ 1 +⎜ − − ⎟ = 2 n+ 2 ⎝ n + 1 n + 2⎠ ⎛1

1

⎞

1

Sn = lim ⎜ − ∑ ⎜⎝ n + 1 − n + 2 ⎟⎠ = nlim ⎟ = , converges n→∞ ⎝ 2 →∞ 2 n + 2⎠ n =1

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

874

Chapter 9

Infinite Series

∞

n ⎛ k + 1⎞ 54. S n = ∑ ln⎜ ⎟ k =1 ⎝ k ⎠

3n 3 n =1 n

47. ∑

= (ln 2 − ln1) + (ln 3 − ln 2) +

(ln 2)2 3n = lim (ln n)3 3n = ∞ = lim 6n

n→∞

∞ 3 ⎛1⎞ = 3 ∑ ⎜ ⎟ , convergent n S n=0 n=0 ⎝ 5⎠

56. lim an = 5 means that the limit of the sequence {an} is n→∞

1 Geometric series with r = 5

5. ∞

n 49. Because n > ln ( n), the terms an = do not ln ( n) ∞

n diverges. ln ( n) n=2

approach 0 as n → ∞. So, the series ∑

∑ an = a1 + a2 +

partial sums is 5.

57. The series given by ∞

∑ ar n = a + ar + ar 2 +

r ≥ 1. ∞

58. If lim an ≠ 0, then ∑ an diverges. n→∞

51. For k ≠ 0, k

= e k ≠ 0.

59. (a)

(b)

52. ∑ e n =1

∞

(c)

∑ ak = ak + ak +

n =1

n

⎛1⎞ = ∑ ⎜ ⎟ converges because it is geometric n =1 ⎝ e ⎠

∞

∑ ak = a1 + a2 + a3 +

These are the same. The third series is different, unless a1 = a2 = = a is constant.

n

∞

∞

∑ an = a1 + a2 + a3 + k =1

n

n→∞

∞ k⎤ ⎡ So, ∑ ⎢1 + ⎥ diverges. n⎦ ⎣ n =1

n =1

n =1

For k = 0, lim (1 + 0) = 1 ≠ 0.

−n

,a ≠ 0

series converges to a (1 − r ). The series diverges if

∞ ⎛1⎞ Because lim Sn diverges, ∑ ln ⎜ ⎟ diverges. n→∞ n =1 ⎝ n ⎠

∞

+ ar n +

is a geometric series with ratio r. When 0 < r < 1, the

− ln ( n)

n nk ⎡⎛ k⎞ k⎞ ⎤ ⎛ lim ⎜1 + ⎟ = lim ⎢⎜1 + ⎟ ⎥ n→∞ ⎝ n→∞ ⎝ n⎠ n ⎠ ⎥⎦ ⎢⎣

= 5 means that the limit of the

n =1

n=0

n n ⎛1⎞ 50. S n = ∑ ln ⎜ ⎟ = ∑ − ln ( k ) k =1 ⎝ k ⎠ k =1

= 0 − ln 2 − ln 3 −

Diverges 55. See definitions on page 595.

n

∞

+ (ln(n + 1) − ln n)

= ln(n + 1) − ln(1) = ln(n + 1)

6

n→∞

(by L’Hôpital’s Rule); diverges by Theorem 9.9 48. ∑

⎛ n + 1⎞ + ln⎜ ⎟ ⎝ n ⎠

⎛ 2⎞ ⎛ 3⎞ = ln⎜ ⎟ + ln⎜ ⎟ + ⎝1⎠ ⎝ 2⎠

3n (ln 2)3n = lim 3 n→∞ n n→∞ 3n 2 lim

60. (a) Yes, the new series will still diverge.

(b) Yes, the new series will converge.

with r =

1 < 1. e

∞

n =1

53. lim arctan n = n→∞

∞

π 2

≠ 0

So, ∑ arctan n diverges. n =1

∞

61. ∑ (3 x) = (3 x) ∑ (3x) n

n

n=0

Geometric series: converges for 3 x < 1 ⇒ x <

1 3

∞ 1 3x 1 n f ( x ) = (3 x ) ∑ (3 x ) = (3 x ) , x < = 1 3 x 1 3 x 3 − − n=0

© 2014 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 9.2 ∞ ⎛ 2⎞ 62. ∑ ⎜ ⎟ n=0 ⎝ x ⎠

Series and Convergence

875

n

Geometric series: converges for 2 < 1 ⇒ x > 2 ⇒ x < − 2 or x > 2 x n

∞ 1 x ⎛ 2⎞ , x > 2 or x < − 2 f ( x) = ∑ ⎜ ⎟ = = 1 2 2 − − x x x ( ) n=0 ⎝ ⎠ ∞

∞

63. ∑ ( x − 1) = ( x − 1) ∑ ( x − 1) n

n =1

n

n=0

Geometric series: converges for x − 1 < 1 ⇒ 0 < x < 2 ∞

f ( x) = ( x − 1) ∑ ( x − 1)

n

n=0

1 x −1 = ( x − 1) = , 1 − ( x − 1) 2− x ∞ ⎛ x − 2⎞ 64. ∑ 5⎜ ⎟ 3 ⎠ n=0 ⎝

0 < x < 2

67. (a) x is the common ratio.

n

∞

= ∑ xn =

(b) 1 + x + x 2 +

n=0

Geometric series: converges for

x − 2 < 1 ⇒ x − 2 < 3 ⇒ −1 < x < 5 3 n

∞ ⎛ x − 2⎞ f ( x) = ∑ 5⎜ ⎟ 3 ⎠ n=0 ⎝ 5 5 = = x − 2 ⎛ ⎞ (3 − x + 2) 3 1−⎜ ⎟ ⎝ 3 ⎠ 15 = , −1 < x < 5 5− x

(c) y1 =

1 , 1− x

x <1

1 1− x

y 2 = S3 = 1 + x + x 2 y3 = S5 = 1 + x + x 2 + x3 + x 4

Answers will vary.

3

f

S5 S3

− 1.5

1.5 0

∞

∞

65. ∑ ( −1) x n = ∑ ( − x) n

n=0

n

n=0

Geometric series: converges for − x < 1 ⇒ x < 1 ⇒ −1 < x < 1 ∞

f ( x) = ∑ ( − x) = n

n=0

∞

∞

1 , 1+ x

66. ∑ ( −1) x 2 n = ∑ ( − x 2 ) n

n=0

−1 < x < 1

n

n=0

Geometric series: converges for − x 2 < 1 ⇒ −1 < x < 1 ∞

f ( x) = ∑ (− x 2 ) = n=0

n

1 1 = , −1 < x < 1 2 1 + x2 1 − (− x )

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

876

Chapter 9

Infinite Series

⎛ x⎞ 68. (a) ⎜ − ⎟ is the common ratio. ⎝ 2⎠ (b) 1 −

2

n −1

n

2 , = 2+ x

72. V (t ) = 475,000(1 − 0.3) = 475,000(0.7) n

n

V (5) = 475,000(0.7) = $79,833.25 5

x < 2

2 (c) y1 = 2+ x

∞

73. ∑ 200(0.75) = 800 million dollars i

i=0

x x2 + 2 4 x x2 x3 x4 y3 = S5 = 1 − + − + 2 4 8 16

y 2 = S3 = 1 −

Answers will vary.

∞

74. ∑ 200(0.60) = 500 million dollars i

i=0

75. D1 = 16 D2 = 0.81(16) + 0.81(16) = 32(0.81) up

5

f

1 − 0.95

= 160,000 ⎣⎡1 − 0.95n ⎦⎤ , n > 0

⎛ x⎞ = ∑ ⎜− ⎟ n=0 ⎝ 2 ⎠ 1 = 1 − ( − x 2)

x x x + − + 2 4 8

i

i =0

∞

3

8000 ⎡⎣1 − 0.95n ⎤⎦

71. ∑ 8000(0.95) =

2

S3 −5

down

D3 = 16(0.81) + 16(0.81) = 32(0.81)

S5

2

2

5

D = 16 + 32(0.81) + 32(0.81) + 2

−5

∞

= −16 + ∑ 32(0.81) = −16 +

1 < 0.0001 69. n( n + 1)

n=0

0 < n 2 + n − 10,000 −1 ± 12 − 4(1)( −10,000) 2

Choosing the positive value for n you have n ≈ 99.5012. The first term that is less than 0.0001 is n = 100. n

⎛1⎞ ⎜ ⎟ < 0.0001 ⎝8⎠ 10,000 < 8n

This inequality is true when n = 5. This series converges at a faster rate.

70.

32 1 − 0.81

≈ 152.42 feet

10,000 < n 2 + n

n =

n

1 < 0.0001 2n 10,000 < 2n

76. The ball in Exercise 75 takes the following times for each fall.

s1 = −16t 2 + 16

s1 = 0 if t = 1

s2 = −16t + 16(0.81) 2

s2 = 0 if t = 0.9

2

s3 = −16t + 16(0.81)

2

s3 = 0 if t = (0.9)

2

sn = −16t 2 + 16(0.81)

n −1

sn = 0 if t = (0.9)

n −1

Beginning with s2 , the ball takes the same amount of time to bounce up as it takes to fall. The total elapsed time before the ball comes to rest is ∞

∞

t = 1 + 2∑ (0.9) = −1 + 2 ∑ (0.9) n =1

= −1 +

n

n

n=0

2 = 19 seconds.

1 − 0.9

This inequality is true when n = 14.

(0.01)

n

< 0.0001

10,000 < 10n This inequality is true when n = 5. This series converges at a faster rate.

© 2014 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 9.2

77. P( n) =

1⎛ 1 ⎞ ⎜ ⎟ 2⎝ 2 ⎠

n

78. P( n) =

2

P ( 2) =

1⎛ 1 ⎞ 1 ⎜ ⎟ = 2⎝ 2 ⎠ 8

∞

n

1⎛ 1 ⎞

12

∑ 2 ⎜⎝ 2 ⎟⎠ = 1 − (1 2) = 1

1⎛ 2 ⎞ ⎜ ⎟ 3⎝ 3 ⎠

Series and Convergence

n

2

P ( 2) =

1⎛ 2 ⎞ 4 ⎜ ⎟ = 3⎝ 3 ⎠ 27

∞

n

1⎛ 2 ⎞

877

13

∑ 3⎜⎝ 3 ⎟⎠ = 1 − (2 3) = 1

n=0

n=0

79. (a)

∞

n

n

∞ 1⎛ 1 ⎞ 1 1 ⎛1⎞ ∑ ⎜⎝ 2 ⎟⎠ = ∑ 2 ⎜⎝ 2 ⎟⎠ = 2 (1 − (1 2)) = 1 n =1 n=0

(b) No, the series is not geometric. (c)

n

⎛1⎞ ∑ n⎜⎝ 2 ⎟⎠ = 2 n =1

n

1 ∞ ⎛1⎞ 1 1 4 = ∑ ⎜ ⎟ = 2 n=0 ⎝ 8 ⎠ 2 1 − (1 8) 7

80. Person 1:

1 1 1 + 4 + 7 + 2 2 2

Person 2:

1 1 1 + 5 + 8 + 22 2 2

=

1 ∞ ⎛1⎞ 1 1 2 ∑ ⎜ ⎟ = 4 1 − (1 8) = 7 4 n=0 ⎝ 8 ⎠

Person 3:

1 1 1 + 6 + 9 + 23 2 2

=

1 ∞ ⎛1⎞ 1 1 1 ∑ ⎜ ⎟ = 8 1 − (1 8) = 7 8 n=0 ⎝ 8 ⎠

Sum:

∞

=

n

n

4 2 1 + + =1 7 7 7

81. (a) 64 + 32 + 16 + 8 + 4 + 2 = 126 in.2

(b)

n

∞

64 ⎛1⎞ ∑ 64⎜⎝ 2 ⎟⎠ = 1 − (1 2) = 128 in.2 n=0 Note: This is one-half of the area of the original square

16 in.

16 in.

82. (a) sin θ = sin θ = sin θ =

Yy1 z x1 y1 Yy1 x1 y2 x1 y1

⇒ Yy1 = z sin θ ⇒ x1 y1 = Yy1 sin θ = z sin 2 θ ⇒ x1 y2 = x1 y1 sin θ = z sin 3 θ

Total: z sin θ + z sin 2 θ + z sin 3 θ +

(b) If z = 1 and θ =

π 6

, then total =

= z

sin θ 1 − sin θ

12 = 1.

1 − (1 2)

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

878

Chapter 9

Infinite Series i

100,000 ⎡1 − 1.06 ⎤ ⎛ 1 ⎞ ( ) = 100,000 ∑ ⎜⎝ 1.06 ⎟⎠ = 1.06 ⎢ 1 − 1.06 ⎥ (n = 20, r = 1.06 ) ≈ $1,146,992.12 1.06

20

1 83. ∑ 100,000 1.06 n =1

19

n

−20

⎣

i =0

−1

−1

⎦

The $2,000,000 sweepstakes has a present value of $1,146,992.12. After accruing interest over the 20-year period, it attains its full value.

12 t ⎡ r ⎞ ⎤ ⎛ 1 1 − + P ⎢ ⎥ ⎜ ⎟ n 12t −1 12 ⎠ ⎦⎥ ⎝ r ⎞ ⎢ ⎛ 84. ∑ P⎜1 + ⎟ = ⎣ r ⎞ 12 ⎠ ⎛ ⎝ n=0 1 − ⎜1 + ⎟ 12 ⎠ ⎝

n −1

85. w = ∑ 0.01( 2) = i

0.01(1 − 2n )

= 0.01( 2n − 1)

1− 2

i=0

(a) When n = 29: w = $5,368,709.11 (b) When n = 30: w = $10,737,418.23

12 t r ⎞ ⎤ ⎛ 12 ⎞ ⎡⎛ ⎛ = P⎜ − ⎟ ⎢⎜1 − ⎜1 + ⎟ ⎥ 12 ⎠ ⎥ ⎝ r ⎠ ⎢⎣⎜⎝ ⎝ ⎦

(c) When n = 31: w = $21,474,836.47

12 t ⎤ r ⎞ ⎛ 12 ⎞ ⎡⎛ = P⎜ ⎟ ⎢⎜1 + ⎟ − 1⎥ 12 ⎠ ⎝ r ⎠ ⎣⎢⎝ ⎦⎥

( ∑ P(e ) =

12t −1

n

r 12

P 1 − (e r 12 )

n=0

12t

1 − er 12

) = P(e − 1) rt

e r 12 − 1

()

()

2 2 2 86. Surface area = 4π (1) + 9⎛⎜ 4π 13 ⎞⎟ + 92 ⋅ 4π 19 + ⎝ ⎠

87. P = 45,

r = 0.03,

t = 20

12( 20) ⎤ 0.03 ⎞ ⎛ 12 ⎞ ⎡⎛ − 1⎥ ≈ $14,773.59 (a) A = 45⎜ ⎟ ⎢⎜1 + ⎟ 12 ⎠ ⎝ 0.03 ⎠ ⎣⎢⎝ ⎦⎥

(b) A = 88. P = 75,

(

) ≈ $14,779.65

45 e0.03(20) − 1 e

0.03 12

−1

r = 0.055,

t = 25

12( 25) ⎤ 0.055 ⎞ ⎛ 12 ⎞ ⎡⎛ − 1⎥ ≈ $48,152.81 (a) A = 75⎜ ⎟ ⎢⎜1 + ⎟ 0.055 12 ⎝ ⎠ ⎢⎣⎝ ⎠ ⎥⎦

(b) A =

(

) ≈ $48,245.07

75 e0.055(25) − 1 e0.055 12 − 1

89. P = 100,

= 4(π + π +

∞ 1 1 = 0, but ∑ diverges. n→∞ n n =1 n

91. False. lim 92. True

∞ ⎛ a ⎞ 93. False; ∑ ar n = ⎜ ⎟ − a ⎝1 − r ⎠ n =1

The formula requires that the geometric series begins with n = 0.

94. True

lim

90. P = 30,

t = 35

1 ≠ 0 1000

9 9 + + 103 104

= 0.74 +

9 ∞ ⎛1⎞ ∑⎜ ⎟ 103 n = 0 ⎝ 10 ⎠

(

= 0.74 +

9 1 ⋅ 103 1 − (1 10)

) ≈ $91,503.32

100 e0.04(35) − 1 e0.04 12 − 1 r = 0.06,

) ≈ $114,227.18

n

9 10 ⋅ 103 9 1 = 0.74 + = 0.75 100 = 0.74 +

t = 50

30 e0.06(50) − 1 e0.06 12 − 1

=

0.74999 … = 0.74 +

12(50) ⎤ 0.06 ⎞ ⎛ 12 ⎞ ⎡⎛ 1 + − 1⎥ ≈ 113,615.73 (a) A = 30⎜ ⎢ ⎟⎜ ⎟ 12 ⎠ ⎝ 0.06 ⎠ ⎢⎣⎝ ⎥⎦

(b) A =

)

95. True

r = 0.04,

(

n

(

n → ∞ 1000 n + 1

12(35) ⎤ 0.04 ⎞ ⎛ 12 ⎞ ⎡⎛ 1 + − 1⎥ ≈ $91,373.09 (a) A = 100⎜ ⎢ ⎟⎜ ⎟ 12 ⎠ ⎝ 0.04 ⎠ ⎢⎣⎝ ⎥⎦

(b) A =

)=∞

96. True

© 2014 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 9.2 ∞

∞

n=0

n=0

Series and Convergence

879

98. If ∑ ( an + bn ) converged, then

97. Let ∑ an = ∑ 1 and ∑bn = ∑ ( −1).

∑ ( an + bn ) − ∑ an = ∑ bn would converge, which is a contradiction. So, ∑ ( an + bn ) diverges.

Both are divergent series. ∞

∞

n=0

n=0

∑(an + bn ) = ∑ ⎡⎣1 + (−1)⎤⎦ = ∑ [1 − 1] = 0 99. (a)

1 1 a − an + 1 an + 2 1 − = n+3 = = an +1an + 2 an + 2 an + 3 an +1an + 2 an + 3 an +1an + 2 an + 3 an +1an + 3 n

1 a k = 0 k +1ak + 3

(b) S n = ∑

n ⎡ 1 ⎤ 1 = ∑⎢ − ak + 2 ak + 3 ⎥⎦ k = 0 ⎣ ak +1ak + 2

⎡ 1 1 ⎤ ⎡ 1 1 ⎤ = ⎢ − + − + a2 a3 ⎥⎦ ⎢⎣ a2 a3 a3a4 ⎥⎦ ⎣ a1a2 ∞

1

n=0

n + 1an + 3

∑a

⎡ 1 ⎤ 1 1 1 1 + ⎢ − = − = 1− an + 2 an + 3 ⎥⎦ a1a2 an + 2 an + 3 an + 2 an + 3 ⎣ an +1an + 2

⎡ ⎤ 1 = lim S n = lim ⎢1 − ⎥ =1 n→∞ n→∞ a a + + 2 3 n n ⎣ ⎦

100. Let {S n} be the sequence of partial sums for the convergent series ∞

S n = L and because ∑ an = L. Then nlim →∞ n =1

∞

Rn =

∑ ak = L − Sn ,

k = n +1

you have lim Rn = lim ( L − S n ) = lim L − lim S n = L − L = 0.

n →∞

101.

n →∞

1 1 1 + 2 + 3 + r r r

n →∞

n →∞

n

∞ 1⎛ 1 ⎞ 1r 1 = ∑ ⎜ ⎟ = = − −1 r r r r 1 1 ( ) ⎝ ⎠ n=0

⎛ ⎞ 1 < 1⎟ ⎜ since r ⎝ ⎠

This is a geometric series which converges if 1 < 1 ⇔ r > 1. r 102. The entire rectangle has area 2 because the height is 1 1 1 and the base is 1 + + + = 2. The squares all lie 2 4 inside the rectangle, and the sum of their areas is 1+

1 1 1 + 2 + 2 + 22 3 4 ∞

1 < 2.

2 n =1 n

So, ∑

.

103. The series is telescoping:

6k

n

Sn = ∑

k =1

(3

k +1

− 2

k +1

)(3k − 2k )

n ⎡ 3k ⎤ 3k + 1 = ∑⎢ k − k +1 k k +1 ⎥ − 2 ⎦ 3 k =1 ⎣ 3 − 2

= 3−

3n + 1 − 2n +1

n +1

3

lim Sn = 3 − 1 = 2

n→∞

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

880

Chapter 9

Infinite Series

104. f (1) = 0, f ( 2) = 1, f (3) = 2, f ( 4) = 4, … ⎧⎪n 2 4, n even In general: f ( n) = ⎨ 2 ⎪⎩( n − 1) 4, n odd.

(See below for a proof of this.) x + y and x − y are either both odd or both even. If both even, then f ( x + y) − f ( x − y) =

( x + y )2 − ( x − y )2 = xy.

4

4

If both odd, f ( x + y) − f ( x − y) =

( x + y )2 − 1 − ( x − y )2 − 1 = xy.

4

4

Proof by induction that the formula for f ( n) is correct. It is true for n = 1. Assume that the formula is valid for k. If k is even, then f ( k ) = k 2 4 and f ( k + 1) = f ( k ) +

(k + 1) − 1. k k2 k k 2 + 2k = + = = 2 4 2 4 4 2

The argument is similar if k is odd.

Section 9.3 The Integral Test and p-Series ∞

∞

1 n =1 n + 3

4. ∑ 3− n

1. ∑

n =1

Let f ( x) =

1 1 , f ′( x) = − < 0 for x ≥ 1. x +3 ( x + 3)2

f is positive, continuous, and decreasing for x ≥ 1. ∞

1

∞

∫ 1 x + 3 dx = ⎡⎣ln( x + 3)⎤⎦1 = ∞ So, the series diverges by Theorem 9.10.

Let f ( x) =

1 , f ′( x) = − (ln 3)3− x < 0 for x ≥ 1.

3x

f is positive, continuous, and decreasing for x ≥ 1. ∞

⎡ −1 ⎤ 1 1 ∫ 1 3x dx = ⎢⎢(ln 3) 3x ⎥⎥ = 3 ln 3 ⎣ ⎦1 ∞

So, the series converges by Theorem 9.10. ∞

5. ∑ e − n

∞

2 2. ∑ n = 1 3n + 5

n =1

Let f ( x) = e− x , f ′( x) = − e − x < 0 for x ≥ 1.

2 Let f ( x) = .

3x + 5

f is positive, continuous, and decreasing for x ≥ 1.

f is positive, continuous, and decreasing for x ≥ 1. ∞

2

⎡2

⎤

∞

∫ 1 3x + 5 dx = ⎢⎣ 3 ln(3x + 5)⎥⎦1 = ∞ So, the series diverges by Theorem 9.10.

−x

dx = ⎡⎣−e − x ⎤⎦

∞

1

=

1 e

So, the series converges by Theorem 9.10. ∞

6. ∑ ne −n 2 n =1

∞

1 n 2 n =1

3. ∑

Let f ( x) = xe− x 2 , f ′( x) =

1 , f ′( x) = − (ln 2)2− x < 0 for x ≥ 1.

2x f is positive, continuous, and decreasing for x ≥ 1.

Let f ( x) =

∞

⎡ −1 ⎤ 1 1 ∫ 1 2 x dx = ⎢⎢(ln 2) 2 x ⎥⎥ = 2 ln 2 ⎣ ⎦1 ∞

∞

∫1 e

2 − x < 0 for x ≥ 3.

2e x 2

f is positive, continuous, and decreasing for x ≥ 3 ∞

∫ 3 xe

−x 2

dx = ⎡⎣−2( x + 2)e− x 2 ⎤⎦

∞ 3

= 10e−3 2

So, the series converges by Theorem 9.10.

So, the series converges by Theorem 9.10. © 2014 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 9.3 ∞

∞

1 n =1 n + 1

7. ∑

n = 1 2n + 1

Let 1 2x < 0 for x ≥ 1. , f ′( x) = − 2 2 2 x +1 ( x + 1)

f ( x) =

f is positive, continuous, and decreasing for x ≥ 1.

1 π ∞ ∫ 1 x 2 + 1 dx = [arctan x]1 = 4 ∞

n =1

f ( x) =

1 2 , f ′( x) = − < 0 for x ≥ 1.

2x + 1 2 x ( + 1)2

f is positive, continuous, and decreasing for x ≥ 1.

1

∞

∫ 1 2 x + 1 dx = ⎡⎣ln

2 x + 1⎤⎦

∞

1

= ∞

So, the series diverges by Theorem 9.10.

So, the series converges by Theorem 9.10. ∞

881

1

8. ∑

2

Let

9. ∑

The Integral Test and p-Series

ln ( n + 1) n +1 ln ( x + 1)

Let f ( x) =

x +1

1 − ln ( x + 1)

, f ′( x) =

( x + 1)2

< 0 for x ≥ 2.

f is positive, continuous, and decreasing for x ≥ 2.

(

2 ∞

) dx = ⎡⎢ ⎡⎣ln( x + 1)⎤⎦ ⎤⎥

∞ ln x + 1

∫1

x +1

⎢ ⎣

⎥ ⎦1

2

= ∞

So, the series diverges by Theorem 9.10. ∞

∞

ln n n n=2

10. ∑

12. ∑

n =1 n

Let f ( x ) =

ln x 2 − ln x , f ′( x) = .

2 x3 2 x

∫2

dx = ⎡⎣2 x

∞

x (ln x − 2)⎤⎦ = ∞ 2

So, the series diverges by Theorem 9.10.

∞

11. ∑

n =1

n

( n + 1)

Let f ( x) = f ′( x) = −

1+ 2 2x

32

,

x

( x + 1)

2

f ′( x) = < 0.

∫1

x

(

1

dx = ⎡2 ln ⎣ x +1

)

(

arctan x , x2 + 1

1 − 2 x arctan x

( x 2 + 1)

2

< 0 for x ≥ 1.

f is positive, continuous, and decreasing for x ≥ 1.

f is positive, continuous, and decreasing for x ≥ 1. ∞

arctan n n2 + 1

Let f ( x) =

( x + 1)

∞

x 2 + 3⎤ = ∞ ⎦1

So, the series diverges by Theorem 9.10.

n =1

1

x

x

∫ 1 x 2 + 3 dx = ⎣⎡ln 13. ∑

1

x 3 − x2 ′ , f x = < 0 for x ≥ 2. ( ) x2 + 3 ( x 2 + 3)

f is positive, continuous, and decreasing for x ≥ 2

∞

∞

n +3

Let f ( x) =

f is positive, continuous, and decreasing for x > e 2 ≈ 7.4. ∞ ln x

2

)

∞

x +1⎤ = ∞ ⎦1

∫1

∞

⎡ (arctan x)2 ⎤ 3π 2 ⎥ = dx = ⎢ 2 2 32 x +1 ⎣⎢ ⎦⎥1

∞ arctan x

So, the series converges by Theorem 9.10.

So, the series diverges by Theorem 9.10.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

882

Chapter 9

Infinite Series

∞

∞

ln n 3 n=2 n

14. ∑

18. ∑ ln x 1 − 3 ln x , f ′( x) = . x3 x4

Let f ( x ) =

⎡

∞ ln x

∫ 2 x3 dx = ⎢⎣− =

(2 ln x + 1) ⎤

∞

∞ x + 2

[Note: lim

= 1 ≠ 0, so the series diverges.]

∞

∞

4n 2 +1 2 n n =1

ln n 2 n =1 n

19. ∑

Let f ( x ) =

ln x 1 − 2 ln x , f ′( x) = . x2 x3

Let f ( x) =

f is positive, continuous, and decreasing for x > e1 2 ≈ 1.6.

f is positive, continuous, and decreasing for x ≥ 1.

∞

⎡−(ln x + 1) ⎤ dx = ⎢ ⎥ =1 x ⎣ ⎦1

∞

1 ln n n=2 n

∞

20. ∑

n =1

1 2 ln x + 1 , f ′( x) = − .

32 x ln x 2 x 2 (ln x)

f is positive, continuous, and decreasing for x ≥ 2. ∞ 1 ∫ 2 x ln x dx = ⎡⎣2 ln x ⎤⎦ 2 = ∞ ∞

n =1

(2n + 3)

∞

1 n + 2 1 −1 , f ′( x ) = < 0.

32 x + 2 2( x + 2)

Let f ( x ) =

f is positive, continuous, and decreasing for x ≥ 1. ∞

1

∞

∫ 1 ( x + 2)1 2 dx = ⎡⎣2 x + 2 ⎤⎦1 = ∞

So, the series diverges by Theorem 9.10.

1

2

So, the series diverges by Theorem 9.10.

∞

∞

4x

∫ 1 2 x 2 + 1 dx = ⎡⎣ln(2 x + 1)⎤⎦1 = ∞

16. ∑

Let f ( x ) =

−4( 2 x 2 − 1) 4x ′ = < 0 , f x ( ) 2 2 x2 + 1 (2 x 2 + 1)

for x ≥ 1.

So, the series converges by Theorem 9.10.

So, the series diverges by Theorem 9.10. ∞

n 4 n +1 n =1

21. ∑

3

Let f ( x ) = ( 2 x + 3) , f ′( x) =

−6

−3

(2 x + 3)

4

< 0

f is positive, continuous, and decreasing for x ≥ 1. ∞

n + 2

n→∞ n + 1

15. ∑

17. ∑

∞

So, the series diverges by Theorem 9.10.

2 ln 2 + 1 16

So, the series converges by Theorem 9.10.

∫1 x2

f is positive, continuous, and decreasing for x ≥ 1.

∫ 1 x + 1 dx = ⎡⎣x + ln( x + 1)⎤⎦1 = ∞

⎥ ⎦2

4 x4

−1 x+2 1 < 0 = 1+ , f ′( x) = 2 x +1 x +1 ( x + 1)

Let f ( x ) =

f is positive, continuous, and decreasing for x > 2.

∞ ln x

n + 2

n =1 n + 1

⎡

∞

⎤ 1 ⎥ = 2 100 ⎢⎣ 4( 2 x + 3) ⎥⎦1

−3 ∫ 1 (2 x + 3) dx = ⎢

−1

So, the series converges by Theorem 9.10.

Let f ( x) =

x 1 − 3x 4 , f ′( x) = < 0 for x > 1.

2 x +1 ( x 4 + 1) 4

f is positive, continuous, and decreasing for x > 1. ∞

x π ⎡1 2 ⎤ ∫ 1 x 4 + 1 dx = ⎢⎣ 2 arctan( x )⎥⎦1 = 8 ∞

So, the series converges by Theorem 9.10.

© 2014 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 9.3 ∞

∞ n n = ∑ 2 2 n = 1 n + 2n + 1 n = 1 ( n 2 + 1)

22. ∑

Let f ( x) =

( x2 + 1)

2

2

The function f is not decreasing for x ≥ 1.

−(3 x 2 − 1)

, f ′( x) =

( x 2 + 1)

< 0 for

3

f is positive, continuous, and decreasing for x ≥ 1.

Let f ( x) =

1 . x3

f is positive, continuous, and decreasing for x ≥ 1. ∞

1 1 ⎡ 1 ⎤ ∫ 1 x3 dx = ⎢⎣− 2 x 2 ⎥⎦1 = 2 ∞

So, the series converges by Theorem 9.10.

Converges by Theorem 9.10

n k −1 k n =1 n + c ∞

23. ∑

∞

30. ∑

Let x x k −1 , f ′( x) = k x + c

k −2

⎡⎣c( k − 1) − x k ⎤⎦ < 0

( x + c) k

2

1

12 n =1 n

Let f ( x ) =

1 = x1 2

1 . x

f is positive, continuous, and decreasing for x ≥ 1.

for x > k c( k − 1) .

1

∞

12 ∞

∫ 1 x1 2 dx = ⎣⎡2 x ⎦⎤1 = ∞

f is positive, continuous, and decreasing for x > k c( k − 1) .

Diverges by Theorem 9.10

x k −1

⎡1

⎤

∞

∫ 1 x k + c dx = ⎢⎣ k ln( x + c)⎥⎦1 = ∞ ∞

1 3 n n =1

∞

⎡ −1 ⎤ 1 ∫ 1 x 2 + 1 2 dx = ⎢⎢ 2( x 2 + 1) ⎥⎥ = 4 ( ) ⎣ ⎦1 x

f ( x) =

∞

29. ∑

x ≥ 1.

∞

k

∞

1 14 n n =1

31. ∑

So, the series diverges by Theorem 9.10.

Let f ( x ) =

∞

24. ∑ n k e − n

1 −1 , f ′( x) = < 0 for x ≥ 1 x1 4 4 x5 4

f is positive, continuous, and decreasing for x ≥ 1.

n =1

∞

x k −1 ( k − x ) xk Let f ( x) = x , f ′( x) = < 0 for x > k . e ex

⎡ 4 x3 4 ⎤ 1 ∫ 1 x1 4 dx = ⎢⎣ 3 ⎥⎦ = ∞ 1

f is positive, continuous, and decreasing for x > k .

Diverges by Theorem 9.10

Use integration by parts. ∞

∫1

k −x

x e

883

⎛ sin x ⎞ 28. Let f ( x ) = ⎜ ⎟ , f ( n ) = an . ⎝ x ⎠

4

x

The Integral Test and p-Series

∞ ⎡⎣− x k e − x ⎤⎦ 1

dx =

∞

∞

+ k∫ x

k −1 − x

1

e

k ( k − 1) k 1 + + + e e e

=

dx

+

So, the series converges by Theorem 9.10.

25. Let f ( x) =

∞

(−1) , f n = a . ( ) n x

x

The function f is not positive for x ≥ 1.

k! e

32. ∑

The function f is not positive for x ≥ 1.

27. Let f ( x ) =

2 + sin x , f ( n ) = an . x

5

Let f ( x) =

1 . x5

f is positive, continuous, and decreasing for x ≥ 1. ∞

1 1 ⎡ 1 ⎤ ∫ 1 x5 dx = ⎢⎣− 4 x4 ⎥⎦1 = 4 ∞

Converges by Theorem 9.10 ∞

26. Let f ( x ) = e− x cos x, f ( n) = an .

1

n =1 n

33. ∑ 5 n =1

∞ 1 1 = ∑ 15 n n n =1

Divergent p-series with p =

1 <1 5

The function f is not decreasing for x ≥ 1.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

884

Chapter 9

Infinite Series

∞

∞

3 53 n =1 n

34. ∑

36. ∑

1

n =1 n

Convergent p-series with p =

5 >1 3

23

Divergent p-series with p =

∞

2 <1 3

∞

1 32 n n =1

1 1.04 n n =1

35. ∑

37. ∑

Convergent p-series with p =

Convergent p-series with p = 1.04 > 1

3 >1 2

∞

1 π n n =1

38. ∑

Convergent p-series with p = π > 1 39. (a)

n

5

10

20

50

100

Sn

3.7488

3.75

3.75

3.75

3.75

The partial sums approach the sum 3.75 very rapidly.

11

0

11 0

(b)

n

5

10

20

50

100

Sn

1.4636

1.5498

1.5962

1.6251

1.635

The partial sums approach the sum π 2 6 ≈ 1.6449 slower than the series in part (a).

8

0

12 0

N

1 1 1 1 =1+ + + + 2 3 4 n =1 n

40. ∑

(a)

+

1 > M N

41. Let f be positive, continuous, and decreasing for x ≥ 1 and an = f ( n). Then, ∞

M

2

4

6

8

N

4

31

227

1674

∞

∑ an and ∫ 1 f ( x) dx n =1

(b) No. Because the terms are decreasing (approaching zero), more and more terms are required to increase the partial sum by 2.

either both converge or both diverge (Theorem 9.10).

See Example 1, page 620. ∞

42. A series of the form ∑

1

n =1 n

p

is a p-series, p > 0.

The p-series converges if p > 1 and diverges if 0 < p ≤ 1.

© 2014 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 9.3 43. Your friend is not correct. The series ∞

1

1

1

∑ n = 10,000 + 10,001 +

n = 10,000

is the harmonic series, starting with the 10,000th term, and therefore diverges.

6

7

1

n =1

∞

1

(

n = 2 n ln n

44. ∑ an ≥ ∫ f ( x) dx ≥ ∑ an n=2

)p

If p = 1, then the series diverges by the Integral Test. If p ≠ 1,

y

∞

1

∞

∫ 2 x(ln x) p dx = ∫ 2 (ln x)

1

2

45. (a)

3

4

5

6

∞

⎡ (ln x)− p + 1 ⎤ ⎥ . dx = ⎢ x ⎢⎣ − p + 1 ⎥⎦ 2

−p 1

Converges for − p + 1 < 0 or p > 1

x 1

885

46. Answers will vary. Sample answer: The graph of the partial sums of the first series seems to be increasing without bound; therefore, the series diverges. The graph of the partial sums of the second series seems to be approaching a limit; therefore the series converges.

47. ∑

7

The Integral Test and p-Series

7

y

1

x 1

∞

∑

n =1

2

3

4

∞ 1 1 > ∫ dx 1 n x

The area under the rectangle is greater than the area under the curve.

Because ∫ ∞

∑

n =1

(b)

∞

1

1 dx = ⎡⎣2 x

x ⎤⎦

∞

= ∞, diverges,

1

1 diverges. n

y

1

x 1

∞

1

2

3

∞

4

1

∑ n2 < ∫ 1 x 2 dx

n=2

The area under the rectangles is less than the area under the curve.

Because ∫ ∞

1

∞

1 ⎡ 1⎤ dx = ⎢− ⎥ = 1, converges, 1 x2 ⎣ x ⎦1 ∞

⎛

∞

1⎞

⎝

n =1

⎠

∑ n2 converges ⎜ and so does ∑ n2 ⎟.

n=2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

886

Chapter 9 ∞

48. ∑

n=2

Infinite Series

ln n np

If p = 1, then the series diverges by the Integral Test. If p ≠ 1, ∞

∞ ln x

∞

∫ 2 x p dx = ∫ 2 x

−p

⎡ x − p +1 ⎤ ln x dx = ⎢ ⎡−1 + (− p + 1) ln x⎤⎦⎥ . (Use integration by parts.)

2⎣ ⎢⎣ ( − p + 1) ⎥⎦ 2

Converges for − p + 1 < 0 or p > 1 ∞

49. ∑

53.

n

(

n =1 1 + n 2

)

y

f(1) = a1

p

f(2) = a2

1

∞

n diverges (see Example 1). Let If p = 1, ∑ n2 1 + n =1 x

f ( x) =

(1 + x 2 )

p

,p ≠ 1

1 − ( 2 p − 1) x 2

f ′( x) =

(1 + x ) 2

p +1

1

.

N N+1

⎡ ⎤ 1 ⎢ ⎥ dx = ∫1 1 + x2 p ⎢ x 2 + 1 p −1 2 − 2 p ⎥ ( ) ( ) ) ⎣( ⎦1

For p > 1, this integral converges. For 0 < p < 1, it diverges. ∞

n = N +1

n = N +1

Because p > 0, the series diverges for all values of p. n

⎛3⎞ 51. ∑ ⎜ ⎟ , Geometric series. n =1 ⎝ p ⎠

∞

(

So, 0 ≤ Rn ≤ ∫

∞

∞ N

f ( x) dx

f ( x) dx

N

3 <1⇒ p > 3 ⇒ p > 3 p

)⎦

n = 3 n ln n ⎡ln ln n ⎤

∞ N

SN ≤ S ≤ SN + ∫ N

N

n =1

n =1

f ( x) dx ∞ N

f ( x) dx ∞

∑ an ≤ S ≤ ∑ an + ∫ N f ( x) dx

1 ⎣

≤ ∫

0 ≤ S − SN ≤ ∫

n =1

52. ∑

∞

∑ an = aN +1 + aN + 2 +

RN = S − S N =

54. From Exercise 53, you have:

p

Converges for

∞

∑ an > 0

RN = S − S N =

x

50. ∑ n(1 + n 2 )

+ aN

n =1

∞

55. S5 = 1 +

1 1 1 1 + 2 + 2 + 2 ≈ 1.4636 22 3 4 5

0 ≤ R5 ≤ ∫

p

series diverges by the Integral Test.

If p ≠ 1, ∞

⎡ ⎡ln (ln x)⎤ − p + 1 ⎤ ⎦ ⎢⎣ ⎥ . dx = ∫ 3 x ln x ⎡ln(ln x)⎤ p ⎢ −p + 1 ⎥ ⎣ ⎦ ⎣ ⎦3 1

∞

1 1 ⎡ 1⎤ = 0.2 dx = ⎢− ⎥ = 5 x2 5 ⎣ x ⎦5 ∞

∞

If p = 1, then ∞ ∞ 1 ∫ 3 x ln x ⎡ln(ln x)⎤ dx = ⎡⎣ln(ln(ln x))⎤⎦3 = ∞, so the ⎣ ⎦

∞

x

...

N

For a fixed p > 0, p ≠ 1, f ′( x) is eventually negative. f

∞

2

S N = ∑ an = a1 + a2 +

is positive, continuous, and eventually decreasing. ∞

f(N + 1) = aN + 1 f(N) = aN

1.4636 ≤ ∑

1

n =1 n

56. S6 = 1 +

2

1 + 25

0 ≤ R6 ≤ ∫ ∞

≤ 1.4636 + 0.2 = 1.6636

+

1 ≈ 1.0368 65 ∞

1 ⎡ 1 ⎤ dx = ⎢− 4 ⎥ ≈ 0.0002 6 x5 ⎣ 4x ⎦6 ∞

1.0368 ≤ ∑

1

n =1 n

5

≤ 1.0368 + 0.0002 = 1.0370

This converges for − p + 1 < 0 ⇒ p > 1.

So, the series converges for p > 1, and diverges for 0 < p ≤ 1. © 2014 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 9.3

The Integral Test and p-Series

887

1 1 1 1 1 1 1 1 1 1 + + + + + + + + + ≈ 0.9818 2 5 10 17 26 37 50 65 82 101 ∞ π 1 ∞ dx = [arctan x]10 = − arctan 10 ≈ 0.0997 0 ≤ R10 ≤ ∫ 10 x 2 + 1 2 ∞ 1 ≤ 0.9818 + 0.0997 = 1.0815 0.9818 ≤ ∑ 2 n +1 n =1

57. S10 =

1

58. S10 =

2(ln 2)

+

3

1 3(ln 3)

3

+

1 4(ln 4)

3

+

+

1 11(ln 11)

3

≈ 1.9821 ∞

⎡ ⎤ 1 1 ⎢− ⎥ = = ≈ 0.0870 0 ≤ R10 ≤ ∫ dx 3 2 3 10 ⎢ 2⎡ln ( x + 1)⎤ ⎥ 2(ln 11) ( x + 1)⎡⎣ln( x + 1)⎤⎦ ⎦ ⎦10 ⎣ ⎣ 1

∞

∞

1

n =1

(n + 1)⎡⎣ln(n + 1)⎤⎦

1.9821 ≤ ∑

3

≤ 1.9821 + 0.0870 = 2.0691

1 2 3 4 + 4 + 9 + 16 ≈ 0.4049 e e e e

59. S 4 =

∞

∞ 2 e −16 ⎡ 1 2⎤ 0 ≤ R4 ≤ ∫ xe − x dx = ⎢− e − x ⎥ = ≈ 5.6 × 10−8 4 2 ⎣ 2 ⎦4 ∞

0.4049 ≤ ∑ ne − n ≤ 0.4049 + 5.6 × 10−8 2

n =1

1 1 1 1 + 2 + 3 + 4 ≈ 0.5713 e e e e

60. S 4 =

∞

∞

4

4

0 ≤ R4 ≤ ∫ e − x dx = ⎡⎣−e − x ⎤⎦ ∞

∞

N

1 1 ⎡ 1 ⎤ dx = ⎢− 3 ⎥ = < 0.001 3N 3 x4 ⎣ 3x ⎦ N

N ≥ 16

1 < 0.003 N3

64. RN ≤ ∫

N > 333.33 3

∞ N

1 ∞ dx = [arctan x]N x2 + 1

N > 6.93

=

N ≥ 7 62. 0 ≤ RN ≤ ∫

N

−1 2

∞ N

1 ⎡ 2 ⎤ dx = ⎢− 1 2 ⎥ = x3 2 ⎣ x ⎦N

−arctan N < 0.001 −

2 < 0.001 N

arctan N >

< 0.0005

N ≥ 4,000,000 ∞

π 2

π 2

− arctan N < 0.001

π 2

− 0.001

⎛π ⎞ N > tan ⎜ − 0.001⎟ 2 ⎝ ⎠ N ≥ 1000

N > 2000

65. (a)

2 < 0.001 eN 2

N > ln 2000 2 N > 2 ln 2000 ≈ 15.2

∞

∞

=

e N 2 > 2000

n=0

N

∞

2 < 0.001 eN 2

≈ 0.0183

0.5713 ≤ ∑ e − n ≤ 0.5713 + 0.0183 = 0.5896

61. 0 ≤ RN ≤ ∫

∞

N

63. RN ≤ ∫ e − x 2 dx = ⎡⎣−2e − x 2 ⎤⎦

∞

1

1

∑ n1.1 . This is a convergent p-series with p = 1.1 > 1. ∑ n ln n is a divergent series. Use the Integral Test.

n=2

n=2

f ( x) = ∞

1

1 is positive, continuous, and decreasing for x ≥ 2. x ln x ∞

∫ 2 x ln x dx = ⎡⎣ln ln x ⎤⎦ 2 = ∞ © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

888

Chapter 9

(b)

6

Infinite Series

1

1

1

1

1

1

∑ n1.1 = 21.1 + 31.1 + 41.1 + 51.1 + 61.1 ≈ 0.4665 + 0.2987 + 0.2176 + 0.1703 + 0.1393

n=2

6

1

1

1

1

1

1

∑ n ln n = 2 ln 2 + 3 ln 3 + 4 ln 4 + 5 ln 5 + 6 ln 6 ≈ 0.7213 + 0.3034 + 0.1803 + 0.1243 + 0.0930

n=2

For n ≥ 4, the terms of the convergent series seem to be larger than those of the divergent series.

1 1 < n1.1 n ln n

(c)

n ln n < n1.1 ln n < n 0.1 This inequality holds when n ≥ 3.5 × 1015. Or, n > e 40 . Then ln e40 = 40 < (e 40 ) 66. (a)

0.1

= e 4 ≈ 55.

∞

⎡ x − p +1 ⎤ 1 1 ∫ 10 x p dx = ⎢⎣ − p + 1⎥⎦ = ( p − 1)10 p −1 , p > 1 10 ∞

1 xp

(b) f ( x) =

∞

1

R10 ( p ) = ∑

n = 11 n

p

≤ Area under the graph of f over the interval [10, ∞)

(c) The horizontal asymptote is y = 0. As n increases, the error decreases.

67. (a) Let f ( x) = 1 x. f is positive, continuous, and decreasing on [1, ∞).

Sn − 1 ≤ ∫

n 1

x S n − 1 ≤ ln n 1

dx

So, S n ≤ 1 + ln n. Similarly, Sn ≥ ∫

n +1 1

1

x

dx = ln ( n + 1).

So, ln ( n + 1) ≤ S n ≤ 1 + ln n. y

1

1 2

1

2

3 ... n n−1

x n+1

(b) Because ln ( n + 1) ≤ Sn ≤ 1 + ln n, you have ln ( n + 1) − ln n ≤ Sn − ln n ≤ 1. Also, because ln x is an increasing function, ln ( n + 1) − ln n > 0 for n ≥ 1. So, 0 ≤ S n − ln n ≤ 1 and the sequence {an} is bounded. (c) an − an + 1 = [S n − ln n] − ⎡⎣S n + 1 − ln ( n + 1)⎤⎦ = ∫

n +1 1 n

x

dx −

1 ≥ 0 n +1

So, an ≥ an + 1 and the sequence is decreasing. (d) Because the sequence is bounded and monotonic, it converges to a limit, γ . (e) a100 = S100 − ln 100 ≈ 0.5822 ( Actually γ ≈ 0.577216.)

© 2014 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 9.3

The Integral Test and p-Series

889

∞ ∞ ∞ ⎛ n2 − 1⎞ (n + 1)(n − 1) = ∞ ⎡ln n + 1 + ln n − 1 − 2 ln n⎤ 1⎞ ⎛ 68. ∑ ln ⎜1 − 2 ⎟ = ∑ ln ⎜ ⎟ = ∑ ln ∑⎣ ( ) ( ) ⎦ 2 n ⎠ n2 ⎝ n=2 n=2 n=2 n=2 ⎝ n ⎠

(

) (

) (

) (

= ln 3 + ln 1 − 2 ln 2 + ln 4 + ln 2 − 2 ln 3 + ln 5 + ln 3 − 2 ln 4 + ln 6 + ln 4 − 2 ln 5

(

) (

) (

)

+ ln 7 + ln 5 − 2 ln 6 + ln 8 + ln 6 − 2 ln 7 + ln 9 + ln 7 − 2 ln 8 + ∞

∞

69. ∑ x ln n ∞

∞

(a) x = 1: ∑ 1ln n = ∑ 1, diverges n=2

(b) x =

1 ∞ ⎛1⎞ : ∑⎜ ⎟ e n=2 ⎝ e ⎠

p-series with p =

∞ ∞ 1 = ∑ e − ln n = ∑ , diverges n=2 n=2 n −p

⇔ ln x = − p.

∞

∞

∞

∞

n=2

n=2

n=2

n=2

1

∑ xln n = ∑ e− p ln n = ∑ n− p = ∑ n p

This series converges for p > 1 ⇒ x < ∞

1 . e

1 x n =1 n

n =1

f is positive, continuous, and decreasing for x ≥ 1. ⎤

∞

Let f ( x) =

1 x

x2 − 1

π

π

∫ 2 x x 2 − 1 dx = [arcsec x]2 = 2 − 3 Converges by Theorem 9.10

()

n=0

n

76. ∑ (1.042) is geometric with r = 1.042 > 1. Diverges n

n=0

by Theorem 9.6. ∞

77. ∑

n n +1 2

n n2 + 1

= lim

n→∞

1 1 + (1 n 2 )

=1≠ 0

Because these are both convergent p-series, the difference is convergent.

.

1

∞

75. ∑ 23

∞ ∞ ∞ 1⎞ 1 1 ⎛1 78. ∑ ⎜ 2 − 3 ⎟ = ∑ 2 − ∑ 3 n n n n ⎝ ⎠ n =1 n =1 n =1

f is positive, continuous, and decreasing for x ≥ 2. ∞

Diverges by Theorem 9.11

Diverges by Theorem 9.9

diverges by Theorem 9.10.

n2 − 1

p-series with p = 0.95

n→∞

1 So, the series ∑ n = 1 3n − 2

n=2 n

0.95

lim

∞

72. ∑

1

n =1 n

n =1

∫ 1 3x − 2 dx = ⎢⎣3 ln 3x − 2 ⎥⎦1 = ∞

1

∞

74. 3∑

∞

1 −3 71. Let f ( x) = , f ′( x) = < 0 for x ≥ 1 3x − 2 ( 3 x − 2) 2

⎡1

Converges by Theorem 9.11

Converges by Theorem 9.6

Converges for x > 1 by Theorem 9.11

1

5 4

Geometric series with r = 23

∞

70. ξ ( x) = ∑ n − x = ∑

∞

1 54 n n =1

= ∑

n=2

ln n

(c) Let x be given, x > 0. Put x = e

∞

4

n n =1 n

n=2

= −ln 2

∞

1

73. ∑

)

∞

∞ 1⎞ ⎛ 79. ∑ ⎜1 + ⎟ n⎠ ⎝ n =1

n

n

1⎞ ⎛ lim ⎜1 + ⎟ = e ≠ 0 n→∞ ⎝ n⎠

Fails nth-Term Test Diverges by Theorem 9.9

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

890

Chapter 9

Infinite Series ∞

∞

80. ∑ ln ( n)

82. ∑

n=2

n=2

lim ln ( n) = ∞

Let f ( x) =

n→∞

Diverges by Theorem 9.9 ∞

81. ∑

)3

Let f ( x) =

1 x(ln x)

3

∞

1 ∞ 1 ⎡ ln x ⎤ dx = ⎢− 2 ⎥ + ∫ dx 2 2 x3 x3 ⎣ 2x ⎦2

∞ ln x

∫2

.

∞

f is positive, continuous, and decreasing for x ≥ 2.

1

∞

ln x . x3

f is positive, continuous, and decreasing for x ≥ 2 since 1 − 3 ln x < 0 for x ≥ 2. f ′( x) = x4

1

(

n = 2 n ln n

ln n n3

=

ln 2 ⎡ 1 ⎤ + ⎢− 2 ⎥ 8 ⎣ 4x ⎦2

=

ln 2 1 + ( Use integration by parts.)

8 16

−3 1

∞

∫ 2 x(ln x)3 dx = ∫ 2 (ln x) x dx ∞

⎡ (ln x)−2 ⎤ = ⎢ ⎥ ⎢⎣ −2 ⎥⎦ 2

Converges by Theorem 9.10. See Exercise 34. ∞

⎡ 1 ⎤ 1 = ⎢− ⎥ = 2 2 2(ln 2) ⎢⎣ 2(ln x) ⎥⎦ 2 Converges by Theorem 9.10. See Exercise 47.

Section 9.4 Comparisons of Series an

∞

1. (a)

6 6 6 ∑ n 3 2 = 1 + 23 2 + n =1

; S1 = 6

5

∞

6 6 6 ∑ n 3 2 + 3 = 4 + 23 2 + 3 + n =1 ∞

an = 6 n 3/2

6

6

3 ; S1 = 2

6 6 = + + ∑ 2 1 1.5 2 4.5 n + 0.5 n =1 n

; S1 =

an =

4

n

6 n 2 + 0.5

3

6 ≈ 4.9 1.5

2

an =

6 n 3/2 + 3

6

8

1 n 2

4

10

3 ⎛ ⎞ > 1⎟. (b) The first series is a p-series. It converges ⎜ p = 2 ⎝ ⎠ (c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. So, the other two series converge. (d) The smaller the magnitude of the terms, the smaller the magnitude of the terms of the sequence of partial sums.

Sn

n

Σ

6

k = 1 k 3/2

12

n

Σ

10

k=1 k

6 k 2 + 0.5

8 6 4

n

Σ

2

6

k = 1 k 3/2 + 3

n 2

4

6

8

10

© 2014 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 9.4 ∞

2 = 2+ n

∑

2. (a)

n =1

2 + 2

S1 = 2

an

an =

Comparisons of Series

891

2 n − 0.5

4

∞

∑

2 2 = + 0.5 n − 0.5

∞

4 = n + 0.5

n =1

∑

n =1

4 + 1.5

2 + 2 − 0.5 4 + 2.5

S1 = 4

an =

3

4 n + 0.5 an = 2 n

2

S1 ≈ 3.3

1

1 ⎛ ⎞ < 1⎟. (b) The first series is a p-series. It diverges ⎜ p = 2 ⎝ ⎠

n 2

4

6

8

10

(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. So, the other two series diverge. (d) The larger the magnitude of the terms, the larger the magnitude of the terms of the sequence of partial sums.

Sn

Σ

20 16

Σ

2 n − 0.5

4 n + 0.5

12 8

Σ 2n

4

n 2

3.

4

6

8

10

1 1 > > 0 for n ≥ 1 2n − 1 2n Therefore, 1 ∑ 2n − 1 n =1

n

∞

4n

∑ 5n + 3

n=0

diverges by comparison with the divergent p-series 1 ∞ 1 ∑ .

2 n =1 n 1 1 < 3n 2 + 2 3n 2

converges by comparison with the convergent geometric series n

∞

⎛ 4⎞ ∑ ⎜⎝ 5 ⎟⎠ . n=0 7. For n ≥ 3,

Therefore, ∞

4n ⎛ 4⎞ < ⎜ ⎟ n 5 +3 ⎝5⎠

Therefore,

∞

4.

6.

1

∑ 3n2 + 2

ln n 1 > > 0. n +1 n +1

Therefore,

n =1

∞

ln n

converges by comparison with the convergent p-series

∑n +1

1∞ 1 ∑ .

3 n =1 n 2

diverges by comparison with the divergent series

1 > n −1

5.

∞

1 for n ≥ 2 n

Therefore, ∞

∑

n=2

n =1

1

∑ n + 1. n =1

∞

Note: ∑

1

n =1 n + 1

diverges by the Integral Test.

1 n −1

diverges by comparison with the divergent p-series ∞

∑

n=2

1 . n

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

892

Chapter 9 1

8.

<

n +1 3

Infinite Series 1 n3 2

Therefore, ∞

∞

∞

1 ∑ n3 2 . n =1

diverges by a limit comparison with the divergent p-series ∞

9. For n > 3,

n

n =1

converges by comparison with the convergent p-series

1 1 > > 0. n2 n!

1

∑ n. n =1

14. lim

Therefore, 1 ∑ n = 0 n!

14

n

= lim

5 ⋅ 4n = 5 +1

n → ∞ 4n

Therefore, ∞

converges by comparison with the convergent p-series ∞

5

∑ 4n + 1

n =1

1 ∑ n2 . n =1

converges by a limit comparison with the convergent geometric series

1 1 > 4 43 n − 1 4 n

∑ ⎜⎝ 4 ⎟⎠ .

∞

⎛1⎞

∞

15. lim

1 ∑ 43 n − 1 n =1

en

2

=1

1 n +1 2

∞

1

∑ n. n =1

1

16. lim

n=0 e

n→∞

converges by comparison with the convergent geometric series

(2n + 1) (5n + 1) = lim 2n + 1 ⋅ 5n = 1 (2 5)n

n → ∞ 5n

+1

2n

Therefore, ∞

n

∞

n +1

diverges by a limit comparison with the divergent p-series

1 ≤ n e

∑ n2

2n + 1

∑ 5n + 1

⎛1⎞ ∑ ⎜⎝ e ⎟⎠ . n=0

n =1

n

3n ⎛ 3⎞ > ⎜ ⎟ for n ≥ 1 n 2 − 1 ⎝ 2⎠

Therefore, ∞

∞

n=0

Therefore, ∞

n 2

Therefore,

∑

1 ∞ 1 ∑ .

4 n =1 4 n 11. 0 <

n2 + 1 = lim n→∞ 1n

1

n→∞

diverges by comparison with the divergent p-series

1

n

n =1

Therefore,

12.

5 ( 4n + 1)

n→∞

∞

10.

1n

n→∞

n2 =1 n → ∞ n2 + 1

= lim

∑ n2 + 1

n3 + 1

n =1

n ( n 2 + 1)

Therefore,

1

∑

13. lim

converges by a limit comparison with the convergent geometric series ∞

n

⎛ 2⎞ ∑⎜ ⎟ . n =1 ⎝ 5 ⎠

3n

∑ 2n − 1

n =1

diverges by comparison with the divergent geometric series ∞

⎛ 3⎞

n

∑ ⎜⎝ 2 ⎟⎠ .

n =1

© 2014 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 9.4 2n 2 − 1 2n5 − n3 2 = 17. lim 3n + 23n + 1 = lim 5 n→∞ n → ∞ 1n 3n + 2n + 1 3 5

Therefore,

(nk −1 ) (nk + 1) = lim

21. lim

nk =1 n → ∞ nk + 1

1n

n→∞

Therefore, n k −1

∞

2n 2 − 1 ∑ 3n5 + 2n + 1 n =1

n =1

converges by a limit comparison with the convergent p-series

diverges by a limit comparison with the divergent p-series ∞

1

∑ n. n =1

1

∑ n3 . n =1

1 n 2 ( n + 3)

18. lim

893

∑ nk + 1

∞

∞

Comparisons of Series

1n

n→∞

3

n3 =1 n → ∞ n ( n + 3)

= lim

2

Therefore, ∞

⎛1⎞

∑ sin⎜⎝ n ⎟⎠

Therefore, ∞

(−1 n2 ) cos(1 n) = lim cos⎛ 1 ⎞ = 1 sin(1 n) = lim ⎜ ⎟ n →∞ n →∞ n →∞ −1 n 2 1n ⎝n⎠

22. lim

n =1

1

∑ n2 (n + 3) n =1

diverges by a limit comparison with the divergent p-series

converges by a limit comparison with the convergent p-series

∑ n.

∞

1

∑ n3 .

1

n =1

∞ 13

∞ n 1 = ∑ 23 n n n =1

23. ∑

n =1

19. lim

∞

(

1 n n +1 2

1n

n→∞

2

) = lim

n→∞

n =1

n2 n n2 + 1

Diverges; =1

p-series with p =

Therefore, ∞

1

n =1 n

n2 + 1

∑

converges by a limit comparison with the convergent p-series

∞ ⎛ 4⎞ 24. ∑ 5⎜ − ⎟ n=0 ⎝ 3 ⎠

1 ∑ n2 . n =1

Diverges;

∞

n ⎡( n + 1)2n −1 ⎤⎦ n = lim =1 20. lim ⎣ n→∞ n→∞ n + 1 1 ( 2n −1 ) Therefore, ∞

n ∑ (n + 1)2n −1 n =1

converges by a limit comparison with the convergent geometric series ⎛1⎞ ∑ ⎜⎝ 2 ⎟⎠ n =1

n

Geometric series with r = −

∞

∞

2 3

n −1

.

25. ∑

n =1 5

n

4 4 = >1 3 3

1 +1

Converges; Direct comparison with convergent geometric series ∞

⎛1⎞

∑ ⎜⎝ 5 ⎟⎠

n

n =1

∞

1 n=3 n − 8

26. ∑

3

∞

1 3 n n=3

Converges; limit comparison with ∑ ∞

2n n = 1 3n − 2

27. ∑

Diverges; nth-Term Test lim

2n

n → ∞ 3n − 2

=

2 ≠ 0 3

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

894

Chapter 9

Infinite Series

∞ 1 ⎞ ⎛ 1 ⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞ − 28. ∑ ⎜ ⎟ = ⎜ − ⎟ +⎜ − ⎟ +⎜ − ⎟ + n + 2⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠ ⎝ 4 5⎠ n =1 ⎝ n + 1

=

1 2

Converges; telescoping series ∞

n

n =1

(n2 + 1)

29. ∑

∞

n2 3 n =1 n + 1

36. ∑

2

diverges because the degree of the numerator is only one less than the degree of the denominator.

Converges; Integral Test ∞

3 n = 1 n( n + 3)

30. ∑

⎛ n3 ⎞ n4 1 37. lim n⎜ 4 = lim = ≠ 0 ⎟ 4 n→∞ n → ∞ 5n + 3 + 5 n 3 5 ⎝ ⎠

Converges; telescoping series

∞

n3 diverges. n =1 5n + 3

Therefore, ∑

∞

1 ⎞ ⎛1 ∑ ⎜⎝ n − n + 3 ⎟⎠ n =1 an = lim nan . By given conditions lim nan is n→∞ 1 n n→∞ n→∞

31. lim

n → ∞ ln n

∑ an

diverges by a limit comparison with the p-series

39.

1

∑ n.

32. If j < k − 1, then k − j > 1. The p-series with p = k − j converges and because

P ( n) Q( n) 1 nk − j

∞

= L > 0, the series ∑

40.

n→∞

diverges.

1 1 1 + + + 200 400 600

∞

1 n 200 n =1

= ∑

( )

n =1 Q n

P( n)

1 1 1 + + + 200 210 220

∞

n = 0 200 + 10n

41.

∞ 1 1 1 1 1 + + + = ∑ 2 201 204 209 216 n = 1 200 + n

converges

n =1

1 1 1 1 + + + + 201 208 227 264

diverges by the Limit Comparison Test.

converges

∑ Q( n) 1 2 3 4 5 + + + + + 2 5 10 17 26

42.

∞

n , +1 n n =1

= ∑

2

which diverges because the degree of the numerator is only one less than the degree of the denominator.

1 1 1 1 1 + + + + + 34.

3 8 15 24 35

∞

1 , = ∑ 2 n=2 n − 1

which converges because the degree of the numerator is two less than the degree of the denominator. ∞

1 n +1 n =1

35. ∑

3

converges because the degree of the numerator is three less than the degree of the denominator.

1

= ∑

diverges

P( n)

converges by the Limit Comparison Test. Similarly, if j ≥ k − 1, then k − j ≤ 1 which implies that

33.

= lim n = ∞ ≠ 0

diverges, (harmonic)

n =1

∞

1

n = 2 ln n

n =1

lim

1

n→∞ 1 n

∞

∞

n→∞

= lim

Therefore, ∑

finite and nonzero. Therefore,

∞

n

38. lim

4

∞

1 + n3 200 n =1

= ∑

43. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms.

44. See Theorem 9.12, page 612. One example is ∞

1

1

1

∑ n2 + 1 converges because n2 + 1 < n2 and n =1

∞

1

∑ n2 converges (p-series). n =1

45. See Theorem 9.13, page 614. One example is ∞

∑

n=2

∞

∑

n=2

1 1 n −1 diverges because lim = 1 and n→∞ 1 n n −1 1 diverges (p-series). n

© 2014 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 9.4

Comparisons of Series

895

46. This is not correct. The beginning terms do not affect the convergence or divergence of a series. In fact,

1 1 + + 1000 1001

47. (a)

∞

1

n = 1000

∞

1 1 + + 4 9

and 1 +

∞

∑ n diverges (harmonic)

=

1 converges ( p -series).

2 n =1 n

= ∑ ∞

1

1

∑ 2 n − 1 2 = ∑ 4 n 2 − 4n + 1 n =1

(

)

n =1

converges because the degree of the numerator is two less than the degree of the denominator. (See Exercise 32.) (b)

n

5

10

20

50

100

Sn

1.1839

1.2087

1.2212

1.2287

1.2312

1

π2

∞

∑ 2n − 1 2 = 8 − S2 ≈ 0.1226

(c)

n=3

(

∞

(d)

π2

1

∑ 2n − 1 2 = 8 − S9 ≈ 0.0277

n = 10

48.

)

(

)

∞

1.0

54. False. ∑ an could converge or diverge.

Terms of Σ an

0.8

n =1

n=1

0.6

∞

∞

For example, let ∑ bn = ∑

Terms of 2 Σ an

0.4

n =1

n=1

n =1

1 , which diverges. n

0.2 n 4

8

12

16

20

For 0 < an < 1, 0 < an 2 < an < 1.

So, the lower terms are those of Σ an 2 .

49. False. Let an =

1 1 and bn = 2 . 0 < an ≤ bn and both 3 n n

∞

∞ 1 1 ∑ n3 and ∑ n2 converge. n =1 n =1

∞

0 <

1 < n

1 1 and ∑ diverges, but n n =1 n

0 <

1 < n2

∞ 1 1 and ∑ 2 converges. n n n =1

∞

55. Because ∑ bn converges, lim bn = 0. There exists N n→∞

n =1

such that bn < 1 for n > N . So, anbn < an for ∞

n > N and ∑ anbn converges by comparison to the

50. True

n =1

∞

convergent series ∑ an .

51. True

i =1

52. False. Let an = 1 n, bn = 1 n, cn = 1 n . Then, 2

∞

an ≤ bn + cn , but ∑ cn converges. n =1

53. True

∞

56. Because ∑ an converges, then n =1

∞

∞

n =1

n =1

∑ an an = ∑ an 2 converges by Exercise 55.

1 1 and ∑ 3 both converge, and therefore, so does n2 n 1 ⎛ 1 ⎞⎛ 1 ⎞ = ∑ 5. ∑ ⎜⎝ n 2 ⎟⎜ 3⎟ n ⎠⎝ n ⎠

57. ∑

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

896

Chapter 9

58. ∑

Infinite Series

1 converge, and therefore, so does n2 ⎛1⎞

2

62. (a) Let ∑ an = ∑

1

∑ ⎜⎝ n 2 ⎟⎠ = ∑ n4 .

lim

(b) Let ∑ an = ∑ lim

an

n→∞ b n

whenever n > M . So, an < bn for n > M . From the Comparison Test, Σ an converges. = ∞ and Σ bn diverges. From the

definition of limit of a sequence, there exists M > 0 such that

1 1 , and ∑ bn = ∑ , diverges. n ln n n

= lim

n → ∞ ln n

By Exercise 60, ∑

= ∞

1 diverges. ln n

63. Because lim an = 0, the terms of Σ sin ( an ) are positive n→∞

for sufficiently large n. Because lim

sin ( an )

= 1 and ∑ an

an

n→∞

an >1 bn

converges, so does Σ sin ( an ).

for n > M . So, an > bn for n > M . By the Comparison Test, Σ an diverges.

61. (a) Let ∑ an = ∑

1

(n + 1)

3

, and ∑ bn = ∑

∞

1 n n ⎡ + 1)⎤⎦ 2 ( n =1 ⎣

= ∑ ∞

= ∑

(

2

)

n =1 n n + 1

Because Σ1 n 2 converges, and

1 ⎡( n + 1) ⎤ an n2 ⎦ = lim = lim ⎣ = 0 2 n→∞ b n→∞ n→∞ n + 1 3 1 (n ) n ( ) 3

∞

1

n =1

(n + 1)

By Exercise 59, ∑

3

converges.

1 1 , and ∑ bn = ∑ n , π nπ n

(b) Let ∑ an = ∑ converges. an

+ n

1 , n2

lim

n→∞ b n

∞

1 + + 1 2 n =1

64. ∑

converges.

lim

n→∞

ln n diverges. n =1 n

an −0 <1 bn

an

1n

n→∞

∞

From the definition of limit of a sequence, there exists M > 0 such that

n→∞ b n

(ln n) n = lim ln n = ∞

= lim

By Exercise 60, ∑

an = 0 and Σ bn converges. n→∞ b n

59. Suppose lim

60. Suppose lim

an

n→∞ b n

ln n 1 , and ∑ bn = ∑ , diverges. n n

= lim

n→∞

1

(

nπ n

1 (π n ) ∞

By Exercise 59, ∑

n =1

) = lim 1 = 0 n→∞

n

1 converges. nπ n

lim

2 ⎡⎣n( n + 1)⎤⎦

n→∞

1 (n ) 2

1

∑1 + 2 +

+ n

2n 2 = 2, n → ∞ n( n + 1)

= lim

converges.

65. First note that f ( x) = ln x − x1 4 = 0 when

x ≈ 5503.66. That is, ln n < n1 4 for n > 5504 which implies that ln n 1 < 5 4 for n > 5504. n3 2 n ∞

1 is a convergent p-series, 54 n n =1

Because ∑ ∞

ln n

∑ n3 2 n =1

converges by direct comparison.

© 2014 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 9.5 66. The series diverges. For n > 1,

n

1 ( n + 1)

⎛ 1 ⎞ , then an1 (n + 1) ≥ ⎜ n + 1 ⎟ 2n +1 ⎝2 ⎠ an n ( n + 1) = 1 (n + 1) ≤ 2an . an an 1

If an ≥

< 2

1 1 > n1 n 2 1 1 > 2n n(n + 1) n Because ∑

897

67. Consider two cases:

n < 2n 1n

Alternating Series

⎛ 1 ⎞ , then an n (n + 1) ≤ ⎜ n + 1 ⎟ n +1 2 ⎝2 ⎠ 1 n ( n + 1) combining, an ≤ 2 an + n .

2 1

If an ≤

1 1 diverges, so does ∑ (n + 1) n .

2n n

n ( n + 1)

=

1 , and 2

=

1 , and 2n

∞ ∞ 1⎞ ⎛ Because ∑ ⎜ 2an + n ⎟ converges, so does ∑ an n (n + 1) 2 ⎠ n =1 ⎝ n =1 by the Comparison Test.

Section 9.5 Alternating Series ∞

1. ∑

(−1)n −1 = π ≈ 0.7854

n =1 2n − 1

(a)

(b)

4

n

1

2

3

4

5

6

7

8

9

10

Sn

1

0.6667

0.8667

0.7238

0.8349

0.7440

0.8209

0.7543

0.8131

0.7605

1.1

0

11 0.6

(c) The points alternate sides of the horizontal line y =

π

that represents the sum of the series.

4 The distance between successive points and the line decreases.

(d) The distance in part (c) is always less than the magnitude of the next term of the series.

(−1)n −1 = 1 ≈ 0.3679 e n = 1 ( n − 1)! ∞

2. ∑

(a)

(b)

n

1

2

3

4

5

6

7

8

9

10

Sn

1

0

0.5

0.3333

0.375

0.3667

0.3681

0.3679

0.3679

0.3679

2

0

11 0

(c) The points alternate sides of the horizontal line y =

1 that represents the sum of the series. e

The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next series.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

898

Chapter 9 ∞

3. ∑

Infinite Series

(−1)n −1 = π 2 ≈ 0.8225 n2

n =1

(a)

(b)

12

n

1

2

3

4

5

6

7

8

9

10

Sn

1

0.75

0.8611

0.7986

0.8386

0.8108

0.8312

0.8156

0.8280

0.8180

1.1

0

11 0.6

(c) The points alternate sides of the horizontal line y =

π2 12

that represents the sum of the series.

The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next term in the series.

(−1)n −1 = sin 1 ≈ 0.8415 () n = 1 ( 2 n − 1)! ∞

4. ∑

(a)

(b)

n

1

2

3

4

5

6

7

8

9

10

Sn

1

0.8333

0.8417

0.8415

0.8415

0.8415

0.8415

0.8415

0.8415

0.8415

2

0

11 0

(c) The points alternate sides of the horizontal line y = sin (1) that represents the sum of the series.

The distance between successive points and the line decreases. (d) The distance in part (c) is always less than the magnitude of the next series. ∞

5. ∑

(−1)n +1

∞

7. ∑

n +1

n =1

n =1

1 1 < = an n + 2 n +1 1 lim an = lim = 0 n→∞ n→∞ n + 1

(−1)n 3n 1

1 = an 3n

an + 1 =

an + 1 =

Converges by Theorem 9.14

Converges by Theorem 9.14

∞

6. ∑

(−1)

n =1

lim

n +1

n +1

3 1 = 0 lim n → 0 3n

<

n

∞ ⎛ −1 ⎞ (Note: ∑ ⎜ ⎟ is a convergent geometric series) n =1 ⎝ 3 ⎠

n

3n + 2 n

n → ∞ 3n + 2

=

1 3

Diverges by nth-Term test

© 2014 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 9.5 ∞

8. ∑

(−1)n

∞

9. ∑

en

n =1

1 1 < n = an en +1 e

lim

1

4n + 1

lim

5n − 1

n → ∞ 4n + 1

=

5 4

Diverges by nth-Term test

= 0

n → ∞ en

899

(−1)n (5n − 1)

n =1

an + 1 =

Alternating Series

Converges by Theorem 9.14 n

∞ ⎛ −1 ⎞ (Note: ∑ ⎜ ⎟ is a convergent geometric series) n =1 ⎝ e ⎠

∞

10. ∑

(−1)

n +1

n

n =1 n + 5 2

Let f ( x) =

−( x 2 − 5) x ′ , f x = < 0 for x ≥ 3 ( ) 2 x2 + 5 ( x 2 + 5)

So, an + 1 < an for n ≥ 3 lim

n → ∞ n2

n = 0 +5

Converges by Theorem 9.14

( −1) n ln (n + 1) n =1 n

∞

11. ∑

n = ∞ n → ∞ ln ( n + 1)

n

n→∞

( −1)n n = 1 ln ( n + 1)

1 < n +1

1 = an n

1 = 0 n

Converges by Theorem 9.14

∞

12. ∑

∞

1 1 an + 1 = < = an ln ( n + 2) ln ( n + 1) 1

n =1

lim

Diverges by nth-Term test

(

( −1)n

an + 1 =

lim

lim

∞

13. ∑

)

n → ∞ ln n + 1

14. ∑

(−1)n +1 n 2

n =1

n2 =1 n→∞ n + 4 lim

= 0

n2 + 4 2

Diverges by nth-Term test

Converges by Theorem 9.14

(−1)n +1(n + 1) ln ( n + 1) n =1 ∞

15. ∑

lim

n +1

(

)

n → ∞ ln n + 1

= lim

n→∞ 1

1

(n + 1)

= lim ( n + 1) = ∞ n→∞

Diverges by the nth-Term Test

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

900

Chapter 9 ∞

16. ∑

Infinite Series

(−1)n +1 ln (n + 1) n +1

n =1

ln ( n + 1)

<

n +1 (n + 1) + 1 ln ( n + 1) 1 ( n + 1) lim = lim = 0 n +1

n→∞

lim

∞

21. ∑

n=0

n!

an + 1 = lim

1

n → ∞ n!

n

lim

<

n for n ≥ 2 n + 2

= 0

Converges by Theorem 9.14 ∞

22. ∑

(−1)n +1 3

n =1

Converges by Theorem 9.14

(−1)n

n +1

n→∞ n + 2

1 = 0 n→∞ n

∞

n

(n + 1) + 2

n

lim

19. ∑

= 0

n + 2

an + 1 =

1 1 < = an n +1 n

an + 1 =

(−1)n +1

n =1

Diverges by the nth-Term Test ∞ −1 ( ) 1 cos nπ = ∑ n n =1 n n =1

(2n + 1)!

Converges by Theorem 9.14

∞ ∞ ⎡ ( 2n − 1)π ⎤ n +1 17. ∑ sin ⎢ ⎥ = ∑ ( −1) 2 n =1 n =1 ⎣ ⎦

∞

1

n→∞

1

n→∞

1 1 < = an n n 2 + 3 !

2 ( ) ( + 1)!

an + 1 =

for n ≥ 2

Converges by Theorem 9.14

18. ∑

(2n + 1)!

n=0

ln ⎡⎣( n + 1) + 1⎤⎦

an + 1 =

(−1)n

∞

20. ∑

n

n

n1 2 = lim n1 6 = ∞ n → ∞ n1 3 n→∞ lim

Diverges by the nth-Term Test

1 1 < = an n n + 1 ! ! ( ) = 0

Converges by Theorem 9.14

(−1)n +1 n! n = 1 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2 n − 1) (n + 1)! n! n +1 ⎛ n +1⎞ an + 1 = = ⋅ = an ⎜ ⎟ < an 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1)( 2n + 1) 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1) 2n + 1 ⎝ 2n + 1 ⎠ ∞

23. ∑

lim an = lim

n →∞

n!

n →∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅

(2n − 1)

= lim

1⋅ 2⋅3⋅⋅⋅ n n ⎤ 1 ⎡3 4 5 = lim 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 0 2n − 3 ⎥⎦ 2n − 1 (2n − 1) n →∞ ⎢⎣ 3 5 7

n →∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅

Converges by Theorem 9.14

(2n − 1) 1 ⋅ 4 ⋅ 7 3n − 2) ( n =1 1 ⋅ 3 ⋅ 5 ( 2n − 1)( 2n + 1) ⎛ 2n + 1 ⎞ an +1 = = an ⎜ ⎟ < an 1 ⋅ 4 ⋅ 7 (3n − 2)(3n + 1) ⎝ 3n + 1 ⎠ ∞

24. ∑ ( −1)

n +1 1 ⋅ 3 ⋅ 5

⎡5 7 9 lim an = lim 3⎢ ⋅ ⋅ n →∞ ⎣ 4 7 10

n →∞

Converges by Theorem 9.14

2n − 1 ⎤ 1 =0 3n − 5 ⎥⎦ 3n − 2

∞

25. ∑

(−1) (2) = ∞ (−1)

n =1

n +1

∑

en − e− n

Let f ( x) = f ′( x) =

n =1

n +1

( 2e n )

e2 n − 1

2e x . Then e −1 2x

−2e x (e 2 x + 1)

(e2 x − 1)

2

< 0.

So, f ( x) is decreasing. Therefore, an +1 < an , and 2e n 2e n 1 = lim 2 n = lim n = 0. n→∞ e n → ∞ 2e n→∞ e −1 lim

2n

The series converges by Theorem 9.14. © 2014 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 9.5 2( −1)

∞

26. ∑

n =1 e

n

n +1

+ e

( −1)

( 2e n )

(1 − e

(e

+ 1)

2x

2x 2

) < 0 for x > 0.

So, f ( x) is decreasing for x > 0 which implies an + 1 < an .

2e n 2e n 1 = lim 2 n = lim n = 0 2 n n→∞ e n → ∞ 2e n→∞ e +1 lim

The series converges by Theorem 9.14.

5

27. S6 = ∑

n!

6

4( −1)

6

(

)

≈ 2.7067 4 ≈ 1.9236 ln 8

R6 = S − S6 ≤ a7 = 0.7831 ≤ S ≤ 4.6303 6

29. S6 = ∑

(−1)

n +1

2

1

< 0.001

( N + 1)2

⇒ ( N + 1) > 1000.

2

By trial and error, this inequality is valid when N = 31(322 = 1024).

Use 31 terms. ∞

33. ∑

(−1)

n +1

3

−1

n =1 2n

1 2( N + 1) − 1 3

< 0.001

⇒ 2( N + 1) − 1 > 1000.

3

By trial and error, this inequality is valid when N = 7 ⎡⎣2(83 ) − 1 = 1024⎤⎦.

∞

34. ∑

n =1

(−1) n

n +1

5

By Theorem 9.15, RN ≤ aN + 1 =

n3

n =1

RN ≤ aN +1 =

Use 7 terms.

n +1

n = 1 ln n + 1

n2

RN ≤ aN + 1 =

5 1 = R6 = S − S6 ≤ a7 = 720 144 11 1 11 1 − ≤ S ≤ + 6 144 6 144 1.8264 ≤ S ≤ 1.8403 28. S6 = ∑

n =1

(−1)n +1

By Theorem 9.15,

(−1)n 5 = 11

n=0

∞

32. ∑

≈ 1.7996

1

( N + 1)

5

< 0.001

⇒ ( N + 1) > 1000.

5

R6 = S − S6 ≤ a7 =

2 ≈ 0.0058 73 1.7796 − 0.0058 ≤ S ≤ 1.7796 + 0.0058

By trial and error, this inequality is valid when N = 3( 45 = 1024).

1.7938 ≤ S ≤ 1.8054

Use 3 terms.

6

30. S6 = ∑

(−1)n +1 n ≈ 0.1852

n =1

3n

7 ≈ 0.0032 37 0.1852 − 0.0032 ≤ S ≤ 0.1852 + 0.0032 R6 = S − S6 ≤ a7 =

∞

n =1

(−1)

∞

(−1)

n=0

n!

35. ∑

n

By Theorem 9.15, RN ≤ aN + 1 =

1

( N + 1)!

< 0.001

⇒ ( N + 1)! > 1000.

0.1820 ≤ S ≤ 0.1884

31. ∑

901

By Theorem 9.15,

2e x . Then e +1 2x

2x

2e

n +1

e2 n + 1

n =1

Let f ( x) = f ′( x) =

∞

= ∑ −n

Alternating Series

By trial and error, this inequality is valid when N = 6(7! = 5040). Use 7 terms since the sum begins

n +1

n3

with n = 0.

By Theorem 9.15, RN ≤ aN + 1 =

1

( N + 1)

3

< 0.001

⇒ ( N + 1) > 1000 ⇒ N + 1 > 10.

3

Use 10 terms. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.