127
CHAPTER 9 Infinite Series Section 9.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Section 9.2 Series and Convergence . . . . . . . . . . . . . . . . . . . 247 Section 9.3 The Integral Test and p-Series . . . . . . . . . . . . . . . . 260 Section 9.4 Comparisons of Series . . . . . . . . . . . . . . . . . . . . 270 Section 9.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . 277 Section 9.6 The Ratio and Root Tests . . . . . . . . . . . . . . . . . . 286 Section 9.7 Taylor Polynomials and Approximations . . . . . . . . . . 298 Section 9.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 308 Section 9.9 Representation of Functions by Power Series . . . . . . . 321 Section 9.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . 329 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

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Page 1: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

C H A P T E R 9Infinite Series

Section 9.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Section 9.2 Series and Convergence . . . . . . . . . . . . . . . . . . . 247

Section 9.3 The Integral Test and p-Series . . . . . . . . . . . . . . . . 260

Section 9.4 Comparisons of Series . . . . . . . . . . . . . . . . . . . . 270

Section 9.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . 277

Section 9.6 The Ratio and Root Tests . . . . . . . . . . . . . . . . . . 286

Section 9.7 Taylor Polynomials and Approximations . . . . . . . . . . 298

Section 9.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 308

Section 9.9 Representation of Functions by Power Series . . . . . . . 321

Section 9.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . 329

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

Page 2: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

C H A P T E R 9Infinite Series

Section 9.1 Sequences

233

1.

a5 � 25 � 32

a4 � 24 � 16

a3 � 23 � 8

a2 � 22 � 4

a1 � 21 � 2

an � 2n 2.

a5 �35

5!�

243120

�8140

a4 �34

4!�

8124

�278

a3 �33

3!�

276

�92

a2 �32

2!�

92

a1 �31!

� 3

an �3n

n!3.

a5 � ��12�

5

� �132

a4 � ��12�

4

�1

16

a3 � ��12�

3

� �18

a2 � ��12�

2

�14

a1 � ��12�

1

� �12

an � ��12�

n

4.

a5 � �32243

a4 �1681

a3 � �827

a2 �49

a1 � �23

an � ��23�

n

5.

a5 � sin 5�

2� 1

a4 � sin 2� � 0

a3 � sin 3�

2� �1

a2 � sin � � 0

a1 � sin �

2� 1

an � sin n�

26.

a5 �108

�54

a4 �87

a3 �66

� 1

a2 �45

a1 �24

�12

an �2n

n � 3

7.

a5 ���1�15

52 � �125

a4 ���1�10

42 �1

16

a3 ���1�6

32 �19

a2 ���1�3

22 � �14

a1 ���1�1

12 � �1

an ���1�n�n�1��2

n2 8.

a5 �25

a4 � �24

� �12

a3 �23

a2 � �22

� �1

a1 �21

� 2

an � ��1�n�1�2n� 9.

a5 � 5 �15

�1

25�

12125

a4 � 5 �14

�1

16�

7716

a3 � 5 �13

�19

�439

a2 � 5 �12

�14

�194

a1 � 5 � 1 � 1 � 5

an � 5 �1n

�1n2

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234 Chapter 9 Infinite Series

10.

a5 � 10 �25

�6

25�

26625

a4 � 10 �12

�38

�878

a3 � 10 �23

�23

�343

a2 � 10 � 1 �32

�252

a1 � 10 � 2 � 6 � 18

an � 10 �2n

�6n2 11.

� 2�10 � 1� � 18

a5 � 2�a4 � 1�

� 2�6 � 1� � 10

a4 � 2�a3 � 1�

� 2�4 � 1� � 6

a3 � 2�a2 � 1�

� 2�3 � 1� � 4

a2 � 2�a1 � 1�

a1 � 3, ak�1 � 2�ak � 1� 12.

a5 � �4 � 12 �a4 � 30

a4 � �3 � 12 �a3 � 12

a3 � �2 � 12 �a2 � 6

a2 � �1 � 12 �a1 � 4

a1 � 4, ak�1 � �k � 12 �ak

13.

a5 �12

a4 �12

�4� � 2

a4 �12

a3 �12

�8� � 4

a3 �12

a2 �12

�16� � 8

a2 �12

a1 �12

�32� � 16

a1 � 32, ak�1 �12

ak 14.

a5 �13

a42 �

13

�768�2 � 196,608

a4 �13

a32 �

13

�482� � 768

a3 �13

a22 �

13

�122� � 48

a2 �13

a12 �

13

�62� � 12

a1 � 6, ak�1 �13

ak2 15.

decreases to 0; matches (f).

an �8

n � 1, a1 � 4, a2 �

83

,

16.

increases towards 8; matches (a).

an �8n

n � 1, a1 � 4, a2 �

163

, 17.decreases to 0; matches (e).an � 4�0.5�n�1, a1 � 4, a2 � 2, 18.

eventually approaches 0; matches (b).

an �4n

n!, a1 � 4, a2 � 8,

19. etc.; matches (d).a3 � �1,an��1�n, a1 � �1, a2 � 1, 20. etc.; matches (c).a3 � �13

,an ���1�n

n, a1 � �1, a2 �

12

,

21.

an �2

3n, n � 1, . . . , 10

−1

−1

12

8 22.

n � 1, . . . , 10an � 2 �4n

,

−1 12

−3

4 23.

an � 16��0.5�n�1, n � 1, . . . , 10

12−1

−10

18

24.

an �2n

n � 1, n � 1, 2, . . . , 10

12−1

−1

3 25.

Add 3 to preceding term.

a6 � 3�6� � 1 � 17

a5 � 3�5� � 1 � 14

an � 3n � 1 26.

a6 �6 � 6

2� 6

a5 �5 � 6

2�

112

an �n � 6

2

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Section 9.1 Sequences 235

27.

a6 � 2�80� � 160

a5 � 2�40� � 80

a1 � 5 an�1 � 2an, 28.

a5 �1

16, a6 � �

132

an � �12

an�1, a1 � 1 29.

Multiply the preceding term by �12.

a6 �3

��2�5 � �332

a5 �3

��2�4 �3

16

an �3

��2�n�1

30.

a5 �8116

, a6 � �24332

an � �32

an�1, a1 � 1 31.

� �9��10� � 90

10!8!

�8!�9��10�

8!32.

� �24��25� � 600

25!23!

�23!�24��25�

23!

33. �n � 1�!

n!�

n!�n � 1�n!

� n � 1 34.

� �n � 1��n � 2�

�n � 2�!

n!�

n!�n � 1��n � 2�n!

35.

�1

2n�2n � 1�

�2n � 1�!�2n � 1�! �

�2n � 1�!�2n � 1�!�2n��2n � 1�

36.

� �2n � 1��2n � 2�

�2n � 2�!

�2n�! ��2n�!�2n � 1��2n � 2�

�2n�!37. lim

n→�

5n2

n2 � 2� 5 38. lim

n→� �5 �

1n2� � 5 � 0 � 5

39. limn→�

2n

�n2 � 1� lim

n→�

2�1 � �1�n2�

�21

� 2 40. limn→�

5n

�n2 � 4� lim

n→�

5�1 � �4�n2�

�51

� 5

41. limn→�

sin�1n� � 0 42. lim

n→� cos�2

n� � 1

43.

The graph seems to indicate that the sequence convergesto 1. Analytically,

limn→�

an � limn→�

n � 1

n� lim

x→� x � 1

x� lim

x→� 1 � 1.

−1

−1

12

3 44.

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

limn→�

an � limn→�

1

n3�2 � limx→�

1

x3�2 � 0.

−1 12

−1

2

45.

The graph seems to indicate that the sequence diverges.Analytically, the sequence is

Hence, does not exist.limn→�

an

�an� � �0, �1, 0, 1, 0, �1, . . .�.

12−1

−2

2 46.

The graph seems to indicate that the sequence convergesto 3. Analytically,

limn→�

an � limn→�

�3 �12n� � 3 � 0 � 3.

−1 12

−1

4

Page 5: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

236 Chapter 9 Infinite Series

47.

does not exist (oscillates betweenand 1), diverges.�1

limn→�

��1�n� nn � 1� 48.

does not exist, (alternates between0 and 2), diverges.

limn→�

1 � ��1�n 49. convergeslimn→�

3n2 � n � 4

2n2 � 1�

32

,

50. convergeslimn→�

3�n

3�n � 1� 1, 51.

Thus, converges.limn→�

an � 0,

�1

2n�

32n

�52n

. . . 2n � 1

2n<

12n

an �1 � 3 � 5 . . . �2n � 1�

�2n�n

52. The sequence diverges. To prove this analytically, we use mathematical induction to show that Clearly, Assume that Then,

Since

we have which shows that for all n.an ≥ �32�n�1

ak�1 ≥ �32�k

,

2k � 1k � 1

≥32

,

� ak

2k � 1k � 1

≥ �32�

k�1�2k � 1k � 1 �. ak�1 �

1 � 3 � 5 . . . �2k � 1��2k � 1��k � 1�!

ak ≥ �32�k�1

.a1 � 1 ≥ �32�0

� 1.an ≥ �32�n�1

.

53. convergeslimn→�

1 � ��1�n

n� 0, 54. convergeslim

n→� 1 � ��1�n

n2 � 0,

55.

converges

(L’Hôpital’s Rule)

� limn→�

32 �

1n� � 0,

limn→�

ln�n3�

2n� lim

n→� 32

ln�n�

n56.

converges

(L’Hôpital’s Rule)

� limn→�

1

2n� 0,

limn→�

ln�n

n� lim

n→� 1�2 ln n

n

57. convergeslimn→�

�34�

n

� 0, 58. convergeslimn→�

�0.5�n � 0, 59.

� �, diverges

limn→�

�n � 1�!

n!� lim

n→� �n � 1�

60. convergeslimn→�

�n � 2�!

n!� lim

n→�

1n�n � 1� � 0, 61.

converges � limn→�

1 � 2nn2 � n

� 0,

limn→�

�n � 1n

�n

n � 1� � limn→�

�n � 1�2 � n2

n�n � 1�

62.

converges

limn→�

� n2

2n � 1�

n2

2n � 1� � limn→�

�2n2

4n2 � 1� �

12

, 63. converges

�p > 0, n ≥ 2�

limn→�

n p

en � 0,

Page 6: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.1 Sequences 237

64.

Let

(L’Hôpital’s Rule)

or,

Therefore limn→�

n sin 1n

� 1.limx→�

sin�1�x�

1�x� lim

y→0� sin� y�

y� 1.

limx→�

x sin 1x

� limx→�

sin�1�x�

1�x� lim

x→� ��1�x2� cos�1�x�

�1�x2 � limx→�

cos 1x

� cos 0 � 1

f �x� � x sin 1x.

an � n sin 1n

65.

where convergesu � k�n,

limn→�

�1 �kn�

n

� limu→0

�1 � u�1�uk � ek

an � �1 �kn�

n

66. convergeslimn→�

21�n � 20 � 1,

67.

converges (because isbounded)

�sin n�

limn→�

sin n

n� lim

n→� �sin n�1

n� 0, 68. convergeslim

n→� cos �n

n2 � 0, 69. an � 3n � 2

70. an � 4n � 1 71. an � n2 � 2 72. an ���1�n�1

n2 73. an �n � 1n � 2

74. an ���1�n�1

2n�275. an � 1 �

1n

�n � 1

n76.

�2n�1 � 1

2n

an � 1 �2n � 1

2n 77. an �n

�n � 1��n � 2�

81. an � �2n�!, n � 1, 2, 3, . . . 82. an � �2n � 1�!, n � 1, 2, 3, . . . 83.

monotonic; bounded�an� < 4,

an � 4 �1n

< 4 �1

n � 1� an�1,

84. Let Then

Thus, f is increasing which implies is increasing.

bounded�an� < 3,

�an�

f � �x� �6

�x � 2�2.f �x� �3x

x � 2.

78. an �1n! 79.

���1�n�1 2nn!

�2n�!

an ���1�n�1

1 � 3 � 5 . . . �2n � 1�80. an �

xn�1

�n � 1�!

Page 7: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

238 Chapter 9 Infinite Series

85.

Hence,

True; monotonic; bounded�an� ≤ 18,

an ≥ an�1.

n

2n�2 ≥ n � 1

2�n�1��2

2n�3n ≥ 2n�2�n � 1�

2n ≥ n � 1

n ≥ 1

n ≥ 1

2n ≥?

n � 1

2n�3n ≥?

2n�2�n � 1�

n

2n�2 ≥? n � 1

2�n�1��2 86.

Not monotonic; bounded�an� ≤ 0.7358,

a3 � 0.6694

a2 � 0.7358

a1 � 0.6065

an � ne�n�2

87.

Not monotonic; bounded�an� ≤ 1,

a3 � �13

a2 �12

a1 � �1

an � ��1�n �1n� 88.

Not monotonic; bounded�an� ≤ 23,

a3 � �827

a2 �49

a1 � �23

an � ��23�

n

89.

Monotonic; bounded�an� ≤ 23,

an � �23�

n

> �23�

n�1

� an�1

90.

Monotonic; notbounded

limn→�

an � �,

an � �32�

n

< �32�

n�1

� an�1 91.

Not monotonic; bounded�an� ≤ 1,

a4 � 0.8660

a3 � 1.000

a2 � 0.8660

a1 � 0.500

an � sin�n�

6 � 92.

Not monotonic; bounded�an� ≤ 1,

a3 � cos�3�

2 � � 0

a2 � cos � � �1

a1 � cos �

2� 0

an � cos�n�

2 �

93.

Not monotonic; bounded�an� ≤ 1,

a4 � �0.1634

a3 � �0.3230

a2 � �0.2081

a1 � 0.5403

an �cos n

n94.

Not monotonic, bounded�an� ≤ 1,

a6 �sin�6�

6� �0.0466a3 �

sin�3�3

� 0.0470

a5 �sin�5�

5� �0.1918a2 �

sin�2�2

� 0.4546

a4 �sin�4�

4� �0.1892a1 �

sin�1�1

� 0.8415

an �sin�n

n

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Section 9.1 Sequences 239

95. (a)

Therefore, converges.

(b)

limn→�

�5 �1n� � 5

−1

−1

12

7

�an�

� an�1 ⇒ �an�, monotonic

an � 5 �1n

> 5 �1

n � 1

�5 �1n� ≤ 6 ⇒ �an�, bounded

an � 5 �1n

96. (a)

Therefore, converges.

(b)

limn→�

�4 �3n� � 4

−10

12

8

�an�

an � 4 �3n

< 3 �4

n � 1� an�1 ⇒ monotonic

�4 �3n� < 4 ⇒ bounded

an � 4 �3n

97. (a)

Therefore, converges.

(b)

limn→�

13�1 �

13n�� �

13

−1

−0.1

12

0.4

�an�

� an�1 ⇒ �an�, monotonic

an �13�1 �

13n� <

13�1 �

13n�1�

�13�1 �13n�� <

13

⇒ �an�,bounded

an �13�1 �

13n� 98. (a)

Therefore, converges.

(b)

limn→�

�4 �12n� � 4

−1 12

−1

6

�an�

� an�1 ⇒ �an�, monotonic

an � 4 �12n > 4 �

12n�1

�4 �12n� ≤ 4.5 ⇒ �an�, bounded

an � 4 �12n

99. has a limit because it is a bounded, monotonicsequence. The limit is less than or equal to 4, and greaterthan or equal to 2.

2 ≤ limn→�

an ≤ 4

�an� 100. The sequence could converge or diverge. If is increasing, then it converges to a limit less than orequal to 1. If is decreasing, then it could converge(example: ) or diverge (example: ).an � �nan � 1�n

�an�

�an��an�

101.

(a) No, the sequence diverges,

The amount will grow arbitrarily large over time.

limn→�

An � �.�An�

An � P�1 �r

12�n

(b)

A10 � 9421.11A5 � 9208.15

A9 � 9378.13A4 � 9166.14

A8 � 9335.34A3 � 9124.32

A7 � 9292.75A2 � 9082.69

A6 � 9250.35A1 � 9041.25

P � 9000, r � 0.055, An � 9000�1 �0.055

12 �n

Page 9: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

240 Chapter 9 Infinite Series

102. (a)

(b) (c) A240 � 32,912.28A60 � 6480.83

A6 � 605.27

A5 � 503.76

A4 � 402.51

A3 � 301.50

A2 � 200.75

A1 � 100.25

A0 � 0

An � 100�401��1.0025n � 1� 103. (a) A sequence is a function whose domain is the set ofpositive integers.

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

(c) A sequence is monotonic if its terms are nondecreas-ing, or nonincreasing.

(d) A sequence is bounded if it is bounded belowfor some and bounded above

for some M�.�an ≤ MN��an ≥ N

104. The first sequence because everyother point is below the x-axis.

105. an � 10 �1n

106. Impossible. The sequenceconverges by Theorem 9.5.

107. an �3n

4n � 1108. Impossible. An unbounded sequence diverges.

109. (a)

(b)

(c) limn→�

�0.8�n�2.5� � 0

A4 � $1.024 billion

A3 � $1.28 billion

A2 � $1.6 billion

A1 � $2 billion

An � �0.8�n �2.5� billion 110.

P5 � $19,938.91

P4 � $19,080.30

P3 � $18,258.66

P2 � $17,472.40

P1 � $16,720.00

Pn � 16,000�1.045�n

111. (a)

(b) In 2008, and endangered species.a18 � 979n � 18

1500

13500

5

an � �6.60n2 � 151.7n � 387 112. (a)

(b) For 2008, and million.a18 � 7646n � 18

7000

130

2

an � 244.2n � 3250

113.

(a)

(b) Decreasing

(c) Factorials increase more rapidly than exponentials.

�1,562,500

567

�1,000,000,000

362,880

a9 � a10 �109

9!

an �10n

n!114.

limn→�

�1 �1n�

n

� e

a6 � 2.5216

a5 � 2.4883

a4 � 2.4414

a3 � 2.3704

a2 � 2.2500

a1 � 2.0000

an � �1 �1n�

n

Page 10: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.1 Sequences 241

115.

Let

Since we have Therefore,lim

n→� n�n � 1.

y � e0 � 1.ln y � 0,

ln y � limn→�

�1n

ln n� � limn→�

ln n

n� lim

n→� 1�nn

� 0

y � limn→�

n1�n.

a6 � 6�6 � 1.3480

a5 � 5�5 � 1.3797

a4 � 4�4 � 1.4142

a3 � 3�3 � 1.4422

a2 � �2 � 1.4142

a1 � 11�1 � 1

�an� � � n�n � � �n1�n� 116. Since

there exists for each

an integer N such that for every

Let and we have,

or

for each n > N.

0 < sn < 2L�L < sn � L < L,�sn � L� < L,

� � L > 0

n > N.�sn � L� < �

� > 0,

limn→�

sn � L > 0,

117. True 118. True 119. True 120. True

121.

(a)

(b)

b10 �8955

b5 �85

b9 �5534

b4 �53

b8 �3421

b3 �32

b7 �2113

b2 �21

� 2

b6 �138

b1 �11

� 1

bn �an�1

an

, n ≥ 1

a12 � 89 � 55 � 144 a6 � 5 � 3 � 8

a11 � 55 � 34 � 89 a5 � 3 � 2 � 5

a10 � 34 � 21 � 55 a4 � 2 � 1 � 3

a9 � 21 � 13 � 34 a3 � 1 � 1 � 2

a8 � 13 � 8 � 21 a2 � 1

a7 � 8 � 5 � 13 a1 � 1

an�2 � an � an�1

(c)

(d) If then

Since we have

Since and thus is positive, �1 ��5

2� 1.6180.bn,an,

�1 ± �1 � 4

2�

1 ± �52

0 � 2 � � 1

� 1 � 2

1 � �1�� � .

limn→�

bn � limn→�

bn�1,

limn→�

�1 �1

bn�1� � .lim

n→� bn � ,

�an � an�1

an

�an�1

an

� bn

� 1 �an�1

an

1 �1

bn�1� 1 �

1an�an�1

122.

The limit of the sequence appears to be In fact, this sequence is Newton’s Method applied to f �x� � x2 � 2.�2.

x10 � 1.414214 x5 � 1.414214

x9 � 1.414214 x4 � 1.414214

x8 � 1.414114 x3 � 1.414216

x7 � 1.414214 x2 � 1.41667

x6 � 1.414214 x1 � 1.5

x0 � 1, xn �12

xn�1 �1

xn�1, n � 1, 2, . . .

Page 11: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

242 Chapter 9 Infinite Series

123. (a) (b)

(c) We first use mathematical induction to show that clearly So assume Then

Now we show that is an increasing sequence. Since and

Since is a bounding increasing sequence, it converges to some number by Theorem 9.5.

⇒ �L � 2��L � 1� � 0 ⇒ L � 2 �L �1�

limn→�

an � L ⇒ �2 � L � L ⇒ 2 � L � L2 ⇒ L2 � L � 2 � 0

L,�an�

an ≤ an�1.

an ≤ �an � 2

an2 ≤ an � 2

an2 � an � 2 ≤ 0

�an � 2��an � 1� ≤ 0

an ≤ 2,an ≥ 0�an�

ak�1 ≤ 2.

�ak � 2 ≤ 2

ak � 2 ≤ 4

ak ≤ 2.a1 ≤ 2.an ≤ 2;

a5 � �2 � �2 � �2 � �2 � �2 � 1.9976

a4 � �2 � �2 � �2 � �2 � 1.9904

a3 � �2 � �2 � �2 � 1.9616

a2 � �2 � �2 � 1.8478

an � �2 � an�1, n ≥ 2, a1 � �2a1 � �2 � 1.4142

124. (a) (b)

(c) We first use mathematical induction to show that clearly So assume Then

Now we show that is an increasing sequence. Since and

Since is a bounded increasing sequence, it converges to some number Thus,

⇒ �L � 3��L � 2� � 0 ⇒ L � 3 �L �2�

�6 � L � L ⇒ 6 � L � L2 ⇒ L2 � L � 6 � 0

L: limn→�

an � L.�an�

an ≤ an�1.

an ≤ �an � 6

an2 ≤ an � 6

an2 � an � 6 ≤ 0

�an � 3��an � 2� ≤ 0

an ≤ 3,an ≥ 0�an�

ak�1 ≤ 3.

�6 � ak ≤ 3

6 � ak ≤ 9

ak ≤ 3.a1 ≤ 3.an ≤ 3;

a5 � �6 � �6 � �6 � �6 � �6 � 2.9996

a4 � �6 � �6 � �6 � �6 � 2.9974

a3 � �6 � �6 � �6 � 2.9844

a2 � �6 � �6 � 2.9068

an � �6 � an�1, n ≥ 2, a1 � �6a1 � �6 � 2.4495

Page 12: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.1 Sequences 243

125. (a) We use mathematical induction to show that

[Note that if then and if then ] Clearly,

Before proceeding to the induction step, note that

So assume Then

is increasing because

an ≤ an�1.

an ≤ �an � k

an2 ≤ an � k

an2 � an � k ≤ 0

�an �1 � �1 � 4k

2 ��an �1 � �1 � 4k

2 � ≤ 0

�an�

an�1 ≤1 � �1 � 4k

2.

�an � k ≤ �1 � �1 � 4k2

� k

an � k ≤1 � �1 � 4k

2� k

an ≤1 � �1 � 4k

2.

�1 � �1 � 4k2

� k �1 � �1 � 4k

2.

1 � �1 � 4k

2� k � 1 � �1 � 4k

2 �2

1 � �1 � 4k

2� k �

1 � 2�1 � 4k � 1 � 4k4

2 � 2�1 � 4k � 4k � 2 � 2�1 � 4k � 4k

a1 � �k ≤�1 � 4k

2≤

1 � �1 � 4k2

.

an ≤ 3.k � 6,an ≤ 3,k � 2,

an ≤1 � �1 � 4k

2.

(b) Since is bounded and increasing, it has a limit

(c) implies that

Since L > 0, L �1 � �1 � 4k

2.

⇒ L �1 ± �1 � 4k

2.

⇒ L2 � L � k � 0

L � �k � L ⇒ L2 � k � L

limn→�

an � L

L.�an�

126. (a)

The terms of are decreasing, and those of are increasing. They both seem to approach the same limit.

—CONTINUED—

�bn��an�

b5 � �a4b4 � 5.9777a5 �a4 � b4

2� 5.9777

b4 � �a3b3 � 5.9777a4 �a3 � b3

2� 5.9777

b3 � �a2b2 � 5.9777a3 �a2 � b2

2� 5.9777

b2 � �a1b1 � 5.9667a2 �a1 � b1

2� 5.9886

b1 � �a0b0 � �10�3� � 5.4772a1 �a0 � b0

2�

10 � 32

� 6.5

a0 � 10, b0 � 3

Page 13: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

244 Chapter 9 Infinite Series

126. —CONTINUED—

(b) For we need to show Since

Since

Since Thus, we have shown that

Now assume Since

Since

Since

Thus, we have shown that

(c) converges because it is decreasing and bounded below by 0. converges because it is increasing and bounded above by

(d) Let and Then A �A � B

2 ⇒ A � B.lim

n→� bn � B.lim

n→� an � A

a0.�bn��an�

ak�1 > ak�2 > bk�2 > bk�1.

ak�1bk�1 > bk�12 ⇒ �ak�1bk�1 > bk�1 ⇒ bk�2 > bk�1.ak�1 > bk�1,

⇒ ak�2 > bk�2.

⇒ ak�1 � bk�1 > 2�ak�1bk�1

⇒ �ak�1 � bk�1�2 > 4ak�1bk�1

ak�12 � 2ak�1bk�1 � bk�1

2 > 0 ⇒ ak�12 � 2ak�1bk�1 � bk�1

2 > 4ak�1bk�1

�ak�1 � bk�1�2 > 0,

2ak�1 > ak�1 � bk�1 ⇒ ak�1 >ak�1 � bk�1

2� ak�2.

ak�1 > bk�1,ak > ak�1 > bk�1 > bk.

a0 > a1 > b1 > b0.a0 > b0, a0b0 > b02 ⇒ �a0b0 > b0 ⇒ b1 > b0.

⇒ �a0 � b0�2 > 4a0b0 ⇒ a0 � b0 > 2�a0b0 ⇒ a1 > b1.

a02 � 2a0b0 � b0

2 > 0 ⇒ a02 � 2a0b0 � b0

2 > 4a0b0

�a0 � b0�2 > 0,

2a0 > a0 � b0 ⇒ a0 >a0 � b0

2� a1.a0 > b0,a0 > a1 > b1 > b0.n � 0,

127. (a)

(b)

� limn→�

an

� limn→�

nf �1n�

� limn→�

f �1�n��1�n�

� limh→0�

f �h�h

f ��0� � limh→0�

f �0 � h� � f �0�

h

limn→�

an � limn→�

n sin 1n

� limn→�

sin�1�n�

�1�n� � 1 � f ��0�

f ��x� � cos x, f ��0� � 1

f �x� � sin x, an � n sin 1n

128.

(a)

(b) diverges

(c) diverges

(d) If diverges. If

The sequence converges for �r� < 1.

→ 1

�r�n ln�r� ��rn

ln�r� → 0.

nrn �n

r�n “��

�r� < 1,�r� ≥ 1,

an � n�32�

2

,r �32

:

an � n,r � 1:

an � n�12�

n

�n2n → 0r �

12

:

an � nrn

Page 14: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.1 Sequences 245

129. (a)

� ln1 � 2 � 3 . . . n � ln�n!�

�n

1ln x dx < ln 2 � ln 3 � . . . � ln n

x

y

2 43

0.5

1.0

1.5

2.0 y = lnx

x

y

y = lnx

... n

(b)

� ln�n!�

�n�1

1ln x dx > ln 2 � ln 3 � . . . � ln n

x

y

2 3 4

0.5

1.0

1.5

2.0

n + 1...

y = lnx

n + 1...x

y

n + 1...

y = lnx

n + 1...

(c)

From part (a):

From part (b):

�n � 1�n�1

en > n!

eln�n�1�n�1�n > n!

ln�n � 1�n�1 � n > ln�n!�

�n�1

1 ln x dx � �n � 1� ln�n � 1� � �n � 1� � 1 � ln�n � 1�n�1 � n

nn

en�1 < n!

eln nn�n�1 < n!

ln nn � n � 1 < ln�n!�

�n

1ln x dx � n ln n � n � 1 � ln nn � n � 1

�ln x dx � x ln x � x � C

(d)

By the Squeeze Theorem, limn→�

n�n!n

�1e.

� �1�1e

�1e

limn→�

�n � 1�1��1�n�

ne� lim

n→� �n � 1�

n �n � 1�1�n

e

limn→�

1

e1��1�n� �1e

1

e1��1�n� <n�n!n

<�n � 1�1��1�n�

ne

n

e1��1�n� < n�n! <�n � 1��n�1��n

e

nn

en�1 < n! <�n � 1�n�1

en (e)

1e

� 0.3679

100�100!100

� 0.3799n � 100:

50�50!50

� 0.3897n � 50:

20�20!20

� 0.4152n � 20:

Page 15: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

246 Chapter 9 Infinite Series

130.

(a)

(b) limn→�

an � limn→�

1n

�n

k�1

11 � �k�n� � �1

0

11 � x

dx � ln�1 � x��1

0� ln 2

a5 �15

11 � �1�5� �

11 � �2�5� �

11 � �3�5� �

11 � �4�5� �

11 � 1� �

16272520

� 0.6456

a4 �14

11 � �1�4� �

11 � �2�4� �

11 � �3�4� �

11 � 1� �

533840

� 0.6345

a3 �13

11 � �1�3� �

11 � �2�3� �

11 � 1� �

3760

� 0.6167

a2 �12

11 � �1�2� �

11 � 1� �

12

23

�12� �

712

� 0.5833

a1 �1

1 � 1�

12

� 0.5

an �1n

�n

k�1

11 � �k�n�

131. For a given we must find such that

whenever That is,

or

So, let be given. Let be an integer satisfyingFor we have

Thus, limn→�

1n3 � 0.

� >1n3 ⇒ � 1

n3 � 0� < �.

n3 >1�

n > �1��

1�3

n > M,M > �1���1�3.M� > 0

n > �1��

1�3.n3 >

1�

n > M.

�an � L� � � 1n3� < �

M > 0� > 0, 132. For a given we must find such thatwhenever That is,

or

(since for ).

So, let be given. Let be an integer satisfying

For we have

Thus, for �1 < r < 1.limn→�

rn � 0

�rn � 0� < �.

�r�n < �

ln�r�n < ln���

n ln�r� < ln���

n >ln���ln�r�

n > M,

M >ln���ln�r� .

M� > 0

�r� < 1ln�r� < 0n >ln���ln�r�

n ln�r� < ln���n > M.�an � L� � �r n��

M > 0� > 0,

133. If is bounded, monotonic and nonincreasing,then Then

is a bounded, monotonic, nondecreasing sequence whichconverges by the first half of the theorem. Since converges, then so does �an�.

��an�

�a1 ≤ �a2 ≤ �a3 ≤ . . . ≤ �an ≤ . . .a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . .

�an� 134. Define

Hence, Thus,

xn�1 � an xn � xn�1 � axn � xn�1.

a1 � a2 � . . . � a.

an � an�1

xn�1 � xn�1

xn

�xn�2 � xn

xn�1

xn�1�xn�1 � xn�1� � xn�xn � xn�2�

xn�12 � xn xn�2 � 1 � xn

2 � xn�1xn�1 ⇒

an �xn�1 � xn�1

xn

, n ≥ 1.

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Section 9.2 Series and Convergence 247

135.

We use mathematical induction to verify the formula.

Assume Then

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

� �k � 1�! � 2k�1.

� �k2 � 5k � 4k � 4 � 4��k � 1�! � 8 � 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 � 5��k! � 2k� � 4�k � 1���k � 1�! � 2k�1� � �4k � 4���k � 2�! � 2k�2�

Tk�1 � �k � 1 � 4�Tk � 4�k � 1�Tk�1 � �4�k � 1� � 8�Tk�2

Tk � k! � 2k.

T2 � 2 � 4 � 6

T1 � 1 � 2 � 3

T0 � 1 � 1 � 2

Tn � n! � 2n

Section 9.2 Series and Convergence

1.

S5 � 1 �14 �

19 �

116 �

125 � 1.4636

S4 � 1 �14 �

19 �

116 � 1.4236

S3 � 1 �14 �

19 � 1.3611

S2 � 1 �14 � 1.2500

S1 � 1 2.

S5 �16 �

16 �

320 �

215 �

542 � 0.7357

S4 �16 �

16 �

320 �

215 � 0.6167

S3 �16 �

16 �

320 � 0.4833

S2 �16 �

16 � 0.3333

S1 �16 � 0.1667

3.

S5 � 3 �92 �

274 �

818 �

24316 � 10.3125

S4 � 3 �92 �

274 �

818 � �4.875

S3 � 3 �92 �

274 � 5.25

S2 � 3 �92 � �1.5

S1 � 3 4.

S5 � 1 �13 �

15 �

17 �

19 � 1.7873

S4 � 1 �13 �

15 �

17 � 1.6762

S3 � 1 �13 �

15 � 1.5333

S2 � 1 �13 � 1.3333

S1 � 1

5.

S5 � 3 �32 �

34 �

38 �

316 � 5.8125

S4 � 3 �32 �

34 �

38 � 5.625

S3 � 3 �32 �

34 � 5.250

S2 � 3 �32 � 4.5

S1 � 3 6.

S5 � 1 �12 �

16 �

124 �

1120 � 0.6333

S4 � 1 �12 �

16 �

124 � 0.6250

S3 � 1 �12 �

16 � 0.6667

S2 � 1 �12 � 0.5

S1 � 1

7.

Geometric series

Diverges by Theorem 9.6

r �32

> 1

��

n�0 3�3

2�n

8.

Geometric series

Diverges by Theorem 9.6

r �43

> 1

��

n�0�4

3�n

9.

Geometric series

Diverges by Theorem 9.6

r � 1.055 > 1

��

n�01000�1.055�n

Page 17: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

248 Chapter 9 Infinite Series

10.

Geometric series

Diverges by Theorem 9.6

r � 1.03 > 1

��

n�02��1.03�n 11.

Diverges by Theorem 9.9

limn→�

n

n � 1� 1 � 0

��

n�1

nn � 1

12.

Diverges by Theorem 9.9

limn→�

n

2n � 3�

12

� 0

��

n�1

n2n � 3

13.

Diverges by Theorem 9.9

limn→�

n2

n2 � 1� 1 � 0

��

n�1

n2

n2 � 114.

Diverges by Theorem 9.9

limn→�

n

n2 � 1� lim

n→�

11 � �1�n2�

� 1 � 0

��

n�1

nn2 � 1

15.

Diverges by Theorem 9.9

limn→�

2n � 12n�1 � lim

n→� 1 � 2�n

2�

12

� 0

��

n�1 2n � 12n�1

16.

Diverges by Theorem 9.9

limn→�

n!2n � �

��

n�1 n!2n

17.

Matches graph (c). Analytically, the series is geometric:

��

n�0 �9

4��14�

n

�9�4

1 � 1�4�

9�43�4

� 3

S0 �94

, S1 �94

�54

�4516

, S2 �94

�2116

� 2.95, . . .

��

n�0 94�

14�

n

�94�1 �

14

�1

16� . . . 18.

Matches graph (b). Analytically, the series is geometric:

��

n�0 �2

3�n

�1

1 � 2�3�

11�3

� 3

S0 � 1, S1 �53

, S2 � 2.11, . . . .

��

n�0�2

3�n

� 1 �23

�49

� . . .

19.

Matches graph (a). Analytically, the series is geometric:

��

n�0 154 ��

14�

n

�15�4

1 � ��1�4� �15�45�4

� 3

S0 �154

, S1 �4516

, S2 � 3.05, . . .

��

n�0 154 ��

14�

n

�154 �1 �

14

�1

16� . . . 20.

Matches (d). Analytically, the series is geometric:

��

n�0 173

��89�

n

�17�3

1 � ��8�9� �17�317�9

� 3

S0 �173

, S1 � 0.63, S3 � 5.1, . . .

��

n�0 173 ��

89�

n

�173

�1 �89

�6481

� . . .

21.

Matches (f). The series is geometric:

��

n�0 173 ��

12�

n�

173

1

1 � ��1�2� �173

23

�349

S0 �173

, S1 �176

, . . .

��

n�0 173 ��

12�

n�

173 �1 �

12

�14

� . . . 22.

Matches (e). The series is geometric:

��

n�0 �2

5�n

�1

1 � �2�5� �53

S0 � 1, S1 �75

, . . .

��

n�0 �2

5�n

� 1 �25

�4

25� . . .

23.

��

n�1

1n�n � 1� � lim

n→� Sn � lim

n→� �1 �

1n � 1� � 1

��

n�1

1n�n � 1� � �

n�1�1

n�

1n � 1� � �1 �

12� � �1

2�

13� � �1

3�

14� � �1

4�

15� � . . ., Sn � 1 �

1n � 1

Page 18: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 249

24.

��

n�1

1n�n � 2� � lim

n→� Sn � lim

n→� �1

2�

14

�1

2�n � 1� �1

2�n � 2� �12

�14

�34

� �12

�16� � �1

4�

18� � �1

6�

110� � �1

8�

112� � � 1

10�

114� � . . . �

n�1

1n�n � 2� � �

n�1� 1

2n�

12�n � 2��

25.

Geometric series with

Converges by Theorem 9.6

r �34 < 1

��

n�0 2�3

4�n

26.

Geometric series with

Converges by Theorem 9.6

r � �12 < 1

��

n�0 2��

12�

n

27.

Geometric series with

Converges by Theorem 9.6

r � 0.9 < 1

��

n�0�0.9�n 28.

Geometric series with

Converges by Theorem 9.6

r � �0.6 < 1

��

n�0��0.6�n

29. (a)

(b) (c)

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

0 110

5

� 2�1 �12

�13 �

113

� 3.667

�Sn � 2�1 �12

�13

� � 1n � 1

�1

n � 2�

1n � 3� �

� 2��1 �14� � �1

2�

15� � �1

3�

16� � �1

4�

17� � . . .

��

n�1

6

n�n � 3�� 2 �

n�1�1

n�

1

n � 3�

n 5 10 20 50 100

2.7976 3.1643 3.3936 3.5513 3.6078Sn

30. (a)

(b) (c)

(d) The terms of the series decrease in magnitude slowly. Thus,the sequence of partial sums approaches the sum slowly. 0

0 11

4

� 1 �12

�13

�14

�2512

� 2.0833

� �1 �15� � �1

2�

16� � �1

3�

17� � �1

4�

18� � �1

5�

19� � �1

6�

110� � . . .

��

n�1

4n�n � 4� � �

n�1�1

n�

1n � 4�

n 5 10 20 50 100

1.5377 1.7607 1.9051 2.0071 2.0443Sn

Page 19: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

250 Chapter 9 Infinite Series

31. (a) (c)

(b)

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

00

11

22��

n�1 2�0.9�n�1 � �

n�0 2�0.9�n �

21 � 0.9

� 20

n 5 10 20 50 100

8.1902 13.0264 17.5685 19.8969 19.9995Sn

32. (a) (Geometric series) (c)

(b)

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

00

11

24��

n�13�0.85�n�1 �

31 � 0.85

� 20

n 5 10 20 50 100

11.1259 16.0625 19.2248 19.9941 19.999998Sn

33. (a) (c)

(b)

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

1100

15��

n�110�0.25�n�1 �

101 � 0.25

�403

� 13.3333

n 5 10 20 50 100

13.3203 13.3333 13.3333 13.3333 13.3333Sn

34. (a) (c)

(b)

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

00 11

7��

n�1 5��

13�

n�1

�5

1 � ��1�3� �154

� 3.75

n 5 10 20 50 100

3.7654 3.7499 3.7500 3.7500 3.7500Sn

35.

�12�1 �

12� �

34

�12��1 �

13� � �1

2�

14� � �1

3�

15� � �1

4�

16� � . . .

��

n�2

1n2 � 1

� ��

n�2� 1�2

n � 1�

1�2n � 1� �

12

��

n�2� 1

n � 1�

1n � 1�

36. ��

n�1

4n�n � 2� � 2 �

n�1�1

n�

1n � 2� � 2��1 �

13� � �1

2�

14� � �1

3�

15� � . . . � 2�1 �

12� � 3

37. ��

n�1

8�n � 1��n � 2� � 8 �

n�1� 1

n � 1�

1n � 2� � 8��1

2�

13� � �1

3�

14� � �1

4�

15� � . . . � 8�1

2� � 4

38. ��

n�1

1�2n � 1��2n � 3� �

12

��

n�1� 1

2n � 1�

12n � 3� �

12��

13

�15� � �1

5�

17� � �1

7�

19� � . . . �

12�

13� �

16

Page 20: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 251

39. ��

n�0�1

2�n

�1

1 � �1�2� � 2 40.

(Geometric)

��

n�06�4

5�n

�6

1 � �4�5� � 30 41. ��

n�0��

12�

n

�1

1 � ��1�2� �23

42. ��

n�0 2��

23�

n

�2

1 � ��2�3� �65

43. ��

n�0� 1

10�n

�1

1 � �1�10� �109

44. ��

n�0 8�3

4�n

�8

1 � �3�4� � 32

45. ��

n�0 3��

13�

n

�3

1 � ��1�3� �94

46. ��

n�0 4��

12�

n

�4

1 � ��1�2� �83

47.

� 2 �32

�12

�1

1 � �1�2� �1

1 � �1�3�

��

n�0� 1

2n �13n� � �

n�0�1

2�n

� ��

n�0�1

3�n

48.

�103

� 10 � 2 �343

�1

1 � �7�10� �1

1 � �9�10� � 2

��

n�1 ��0.7�n � �0.9�n� � �

n�0� 7

10�n

� ��

n�0� 9

10�n

� 2

49. Note that The series is geometric with Thus,

��

n�1�sin�1��n � sin�1� �

n�0�sin�1��n �

sin�1�1 � sin�1� � 5.3080.

r � sin�1� < 1.��

n�1�sin�1��nsin�1� � 0.8415 < 1.

50.

limn→�

Sn � limn→�

13�

12

�1

3n � 2 �16

�13�

12

�1

3n � 2

�13��

12

�15� � �1

5�

18� � �1

8�

111� � . . . � � 1

3n � 1�

13n � 2�

� �n

k�1� 1

9k � 3�

19k � 6 �

13

�n

k�1� 1

3k � 1�

13k � 2

Sn � �n

k�1

19k2 � 3k � 2

� �n

k�1

1�3k � 1��3k � 2�

51. (a)

(b) Geometric series with and

S �a

1 � r�

4�101 � �1�10� �

49

r �110a �

410

0.4 � ��

n�0

410�

110�

n

52. (a)

(b) 0.9 �9

10

11 � �1�10� � 1

� ��

n�1 9� 1

10�n

�9

10 ��

n�0� 1

10�n

0.9 �9

10�

9100

�9

1000� . . .

53. (a)

(b) Geometric series with and

S �a

1 � r�

81�1001 � �1�100� �

8199

�9

11

r �1

100a �81100

0.81 � ��

n�0

81100

� 1100�

n

54. (a)

(b) 0.01 �1

100�

11 � �1�100� �

1100

�10099

�1

99

0.01 � ��

n�1� 1

100�n

�1

100 ��

n�0� 1

100�n

Page 21: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

252 Chapter 9 Infinite Series

55. (a)

(b) Geometric series with and

S �a

1 � r�

3�4099�100

�5

66

r �1

100a �340

0.075 � ��

n�0

340�

1100�

n

56. (a)

(b) Geometric series with and

S �15

�a

1 � r�

15

�3�200

99�100�

71330

r �1

100a �3

200

0.215 �15

� ��

n�0

3200�

1100�

n

57.

Diverges by Theorem 9.9

limn→�

n � 1010n � 1

�1

10� 0

��

n�1

n � 1010n � 1

58.

Diverges by Theorem 9.9

limn→�

n � 12n � 1

�12

� 0

��

n�1

n � 12n � 1

59. converges��

n�1�1

n�

1n � 2� � �1 �

13� � �1

2�

14� � �1

3�

15� � �1

4�

16� � . . . � 1 �

12

�32

,

60.

�13�1 �

12

�13� �

1118

, converges

�13��1 �

14� � �1

2�

15� � �1

3�

16� � �1

4�

17� � �1

5�

18� � �1

6�

19� � . . .

��

n�1

1n�n � 3� �

13

��

n�1�1

n�

1n � 3�

61.

Diverges by Theorem 9.9

limn→�

3n � 12n � 1

�32

� 0

��

n�1 3n � 12n � 1

62.

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

� limn→�

�ln 2�23n

6n� lim

n→�

�ln n�33n

6� �

limn→�

3n

n3 � limn→�

�ln 2�3n

3n2

��

n�1 3n

n3

63.

Geometric series with

Converges by Theorem 9.6

r �12

� �

n�0 42n � 4 �

n�0�1

2�n

64.

Geometric series with

Converges by Theorem 9.6

r �14

��

n�0 14n

65.

Geometric series with

Diverges by Theorem 9.6

r � 1.075

��

n�0�1.075�n 66.

Geometric series with

Diverges by Theorem 9.6

r � 2

��

n�1

2n

100

67. Since the terms

do not approach 0 as Hence, the series

diverges.��

n�2

nln�n�

n →�.

an �n

ln�n�n > ln�n�, 68.

Since diverges, diverges.��

n�1ln�1

n�limn→�

Sn

� 0 � ln 2 � ln 3 � . . . �ln�n�

Sn � �n

k�1ln�1

k� � �n

k�1�ln�k�

Page 22: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 253

72.

Diverges

� ln�n � 1� � ln�1� � ln�n � 1�

� �ln 2 � ln 1� � �ln 3 � ln 2� � . . . � �ln�n � 1� � ln n�

� ln�21� � ln�3

2� � . . . � ln�n � 1n �

Sn � �n

k�1ln�k � 1

k �

69. For

For

Hence, diverges.��

n�1�1 �

kn

n

limn→�

�1 � 0�n � 1 � 0.k � 0,

� ek � 0.

limn→�

�1 �kn�

n� lim

n→���1 �

kn�

n�k k

k � 0, 70. converges since

it is geometric with

r �1e

< 1.

��

n�1e�n � �

n�1�1

e�n

71.

Hence, diverges.��

n�1arctan n

limn→�

arctan n ��

2� 0

73. See definition on page 606.

74. means that the limit of the sequence is 5.

means that the limit of the

partial sums is 5.

��

n�1an � a1 � a2 � . . . � 5

�an�limn→�

an � 5 75. The series given by

is a geometric series with ratio r. When the series converges to The series diverges if

r ≥ 1.a��1 � r�.

0 < r < 1,

��

n�0arn � a � ar � ar2 � . . . � arn � . . . , a � 0

76. If then diverges.��

n�1anlim

n→� an � 0,

77.

diverges because limn→�

an � 0.��

n�1an

limn→�

an � limn→�

n � 1

n� 1 ��an� converges�

an �n � 1

n78. (a)

(b)

These are the same. The third series is different,unless is constant.

(c) ��

n�1ak � ak � ak � . . .

a1 � a2 � . . . � a

��

k�1ak � a1 � a2 � a3 � . . .

��

n�1an � a1 � a2 � a3 � . . .

79.

Geometric series: converges for or

�x2

1

1 � �x�2� �x2

2

2 � x�

x2 � x

, x < 2

f �x� �x2

��

n�0�x

2�n

x < 2x2 < 1

��

n�1 xn

2n � ��

n�1�x

2�n

�x2

��

n�0�x

2�n

80.

Geometric series: converges for

f �x� � �3x� ��

n�0�3x�n � �3x� 1

1 � 3x�

3x1 � 3x

, x <13

3x < 1 ⇒ x <13

��

n�1�3x�n � �3x� �

n�0�3x�n

Page 23: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

254 Chapter 9 Infinite Series

81.

Geometric series: converges for

� �x � 1� 11 � �x � 1� �

x � 12 � x

, 0 < x < 2

f �x� � �x � 1� ��

n�0�x � 1�n

x � 1 < 1 ⇒ 0 < x < 2

��

n�1�x � 1�n � �x � 1� �

n�0�x � 1�n 82.

Geometric series: converges for

�4

�4 � x � 3��4�

167 � x

, �1 < x < 7

f �x� � ��

n�04�x � 3

4 �n�

41 � ��x � 3��4�

x � 34 < 1 ⇒ x � 3 < 4 ⇒ �1 < x < 7

��

n�04�x � 3

4 �n

83.

Geometric series: converges for

f �x� � ��

n�0��x�n �

11 � x

, �1 < x < 1

�x < 1 ⇒ x < 1 ⇒ �1 < x < 1

��

n�0��1�nxn � �

n�0��x�n 84.

Geometric series: converges for

f �x� � ��

n�0��x2�n �

11 � ��x2� �

11 � x2, �1 < x < 1

�x2 < 1 ⇒ �1 < x < 1

��

n�0��1�nx2n � �

n�0��x2�n

85.

Geometric series: converges if

or

x > 1 or x < �1f �x� � ��

n�0�1

x�n

�1

1 � �1�x� �x

x � 1,

x > 1⇒ x > 1 ⇒ x < �1

1x < 1

��

n�0�1

x�n

86.

Geometric series: converges for

f �x� �x2

x2 � 4�

1

1 �x2

x2 � 4

�x2

x2 � 4 x2 � 4

4�

x2

4

⇒ x2 < x2 � 4 ⇒ converges for all x

x2

x2 � 4 < 1

��

n�1� x2

x2 � 4�n

�x2

x2 � 4 �

n�0� x2

x2 � 4�n

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

(b) Yes, the new series will converge.

88. Neither statement is true. The formula

holds for �1 < x < 1.

1x � 1

� 1 � x � x2 � . . .

89. (a) is the common ratio.

(b) 1 � x � x2 � . . . � ��

n�0xn �

11 � x

, x < 1

x (c)

y3 � S5 � 1 � x � x2 � x3 � x 4

y2 � S3 � 1 � x � x2

3

1.5−1.50

f

S3

S5

y1 �1

1 � x

90. (a) is the common ratio.

(b)

�2

2 � x, x < 2

�1

1 � ��x�2�

1 �x2

�x2

4�

x3

8� . . . � �

n�0��

x2�

n

��x2� (c)

y3 � S5 � 1 �x2

�x2

4�

x3

8�

x 4

16

y2 � S3 � 1 �x2

�x2

4

f

S3

S5

5

5

−5

−5

y1 �2

2 � x

Page 24: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 255

91.

Horizontal asymptote:

The horizontal asymptote is the sum of the series. is the partial sum.nthf �n�

S �3

1 � �1�2� � 6

��

n�0 3�1

2�n

y � 6

−2 10

−1

7y = 6f �x� � 3�1 � 0.5x

1 � 0.5 92.

Horizontal asymptote:

The horizontal asymptote is the sum of the series. is the partial sum.nthf �n�

S �2

1 � �4�5� � 10

��

n�0 2�4

5�n

y � 10

−2 16

−2

y = 1012

f �x� � 2�1 � 0.8x

1 � 0.8

93.

Choosing the positive value for n we have The first term that is less than 0.0001 is

This inequality is true when This series convergesat a faster rate.

n � 5.

10,000 < 8n

�18�

n

< 0.0001

n � 100.n � 99.5012.

n ��1 ± 12 � 4�1���10,000�

2

0 < n2 � n � 10,000

10,000 < n2 � n

1

n�n � 1� < 0.0001 94.

This inequality is true when

This inequality is true when This series convergesat a faster rate.

n � 5.

10,000 < 10n

�0.01�n < 0.0001

n � 14.

10,000 < 2n

12n < 0.0001

95.

� 80,000�1 � 0.9n�, n > 0

�n�1

i�0 8000�0.9�i �

8000�1 � �0.9��n�1��1�1 � 0.9

96.

V�5� � �0.7�5�225,000� � $37,815.75

V�t� � 225,000�1 � 0.3�n � �0.7�n�225,000�

97.

Sum � 400 million dollars

� 400�1 � 0.75n� million dollars

�n�1

i�0 100�0.75�i �

100�1 � 0.75�n�1��1�1 � 0.75

98.

million dollars

Sum � 250 million dollars

� 250�1 � 0.6n�

�n�1

i�0100�0.60�i �

100�1 � 0.6n�1 � 0.6

99.

� 152.42 feet

� �16 � ��

n�0 32�0.81�n � �16 �

321 � 0.81

D � 16 � 32�0.81� � 32�0.81�2 � . . .

D3 � 16�0.81�2 � 16�0.81�2 � 32�0.81�2

D2 � 0.81�16� � 0.81�16� � 32�0.81�

D1 � 16

up down

� �

100. The ball in Exercise 99 takes the following times for each fall.

Beginning with 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

� �1 �2

1 � 0.9� 19 seconds.

t � 1 � 2 ��

n�1�0.9�n � �1 � 2 �

n�0�0.9�n

s2,

sn � 0 if t � �0.9�n�1 sn � �16t 2 � 16�0.81�n�1

��

s3 � 0 if t � �0.9�2s3 � �16t 2 � 16�0.81�2

s2 � 0 if t � 0.9 s2 � �16t 2 � 16�0.81�

s1 � 0 if t � 1 s1 � �16t 2 � 16

Page 25: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

256 Chapter 9 Infinite Series

101.

��

n�0 12�

12�

n

�1�2

1 � �1�2� � 1

P�2� �12�

12�

2

�18

P�n� �12�

12�

n

102.

��

n�0 13�

23�

n

�1�3

1 � �2�3� � 1

P�2� �13�

23�

2

�4

27

P�n� �13�

23�

n

103. (a)

(b) No, the series is not geometric.

(c) ��

n�1n�1

2�n

� 2

�12

1

�1 � �1�2�� � 1

��

n�1�1

2�n

� ��

n�0

12

�12�

n

104. Person 1:

Person 2:

Person 3:

Sum:47

�27

�17

� 1

123 �

126 �

129 � . . . �

18

��

n�0�1

8�n

�18

1

1 � �1�8� �17

122 �

125 �

128 � . . . �

14

��

n�0�1

8�n

�14

1

1 � �1�8� �27

12

�124 �

127 � . . . �

12

��

n�0�1

8�n

�12

1

1 � �1�8� �47

105. (a)

(b)

Note: This is one-half of the area of the original square!

��

n�0 64�1

2�n

�64

1 � �1�2� � 128 in.216 in.

16 in.

64 � 32 � 16 � 8 � 4 � 2 � 126 in.2

106. (a) (b) If and then

Total: z sin � � z sin2 � � z sin3 � � . . . � zsin �

1 � sin �

sin � �x1y2x1y1 ⇒ x1y2 � x1y1 sin � � z sin3 �

sin � �x1y1Yy1 ⇒ x1y1 � Yy1 sin � � z sin2 �

total �1�2

1 � �1�2� � 1.� ��

6,z � 1sin � �

Yy1z

⇒ Yy1 � z sin �

107.

� 573,496.06

�50,0001.06 �1 � 1.06�20

1 � 1.06�1 , �n � 20, r � 1.06�1�

�20

n�150,000� 1

1.06�n

�50,0001.06

�19

i�0� 1

1.06�i

108. Surface area � 4� �1�2 � 9�4� �13�2� � 92 � 4� �1

9�2� . . . � 4�� � � � . . .� � �

109.

(a) When

(b) When

(c) When n � 31: w � $21,474,836.47

n � 30: w � $10,737,418.23

n � 29: w � $5,368,709.11

w � �n�1

i�0 0.01�2�i �

0.01�1 � 2n�1 � 2

� 0.01�2n � 1�

Page 26: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 257

110.

�12t�1

n�0 P�er�12�n �

P�1 � �er�12�12t�1 � er�12 �

P�ert � 1�er�12 � 1

� P�12r ���1 �

r12�

12t

� 1 � P��12r ���1 � �1 �

r12�

12t

�12t�1

n�0 P�1 �

r12�

n

P�1 � �1 �r

12�12t

1 � �1 �

r12�

113.

(a)

(b) A �100�e0.04�40� � 1�

e0.04�12 � 1� $118,393.43

A � 100� 120.04���1 �

0.0412 �

12�40�� 1 � $118,196.13

P � 100, r � 0.04, t � 40 114.

(a)

(b) A �20�e0.06�50� � 1�

e0.06�12 � 1� $76,151.45

A � 20� 120.06���1 �

0.0612 �

12�50�� 1 � $75,743.82

P � 20, r � 0.06, t � 50

115. (a)

6500

3500132

an � 3484.1363�1.0502�n � 3484.1363e0.04897n (b) Adding the ten values you obtain

or $50,809,000,000.

(c) or $50,803,000,000

(Answers will vary.)

�12

n�33484.1363e0.04897n � 50,803,

�12

n�3an � 50,809,

a3, a4, . . . , a12

116.

� 40,000�1 � 1.0440

1 � 1.04 � � $3,801,020 � �39

n�040,000�1.04�n

T � 40,000 � 40,000�1.04� � . . . � 40,000�1.04�39 117. False. but diverges.��

n�1 1n

limn→�

1n

� 0,

111.

(a)

(b) A �50�e0.03�20� � 1�

e0.03�12 � 1� $16,421.83

A � 50� 120.03���1 �

0.0312 �

12�20�� 1 � $16,415.10

P � 50, r � 0.03, t � 20 112.

(a)

(b) A �75�e0.05�25� � 1�

e0.05�12 � 1� $44,732.85

A � 75 � 120.05���1 �

0.0512 �

12�25�� 1 � $44,663.23

P � 75, r � 0.05, t � 25

118. True 119. False;

The formula requires that the geo-metric series begins with n � 0.

��

n�1 arn � � a

1 � r� � a 120. True

limn→�

n

1000�n � 1� �1

1000� 0

121. True

� 0.74 �1

100� 0.75

� 0.74 �9

103 �109

� 0.74 �9

103 �1

1 � �1�10�

� 0.74 �9

103 ��

n�0� 1

10�n

0.74999 . . . � 0.74 �9

103 �9

104 � . . .

122. True

Page 27: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

258 Chapter 9 Infinite Series

123. By letting we have

Thus,

� ��

n�1��c � Sn�1� � �c � Sn��.

� ��

n�1�Sn � Sn�1�c � c�

��

n�1 an � �

n�1�Sn � Sn�1�

an � �n

k�1 ak � �

n�1

k�1 ak � Sn � Sn�1.

S0 � 0, 124. Let be the sequence of partial sums for theconvergent series

Then and since

we have

� limn→�

L � limn→�

Sn � L � L � 0.

limn→�

Rn � limn→�

�L � Sn�

Rn � ��

k�n�1ak � L � Sn,

limn→�

Sn � L��

n�1 an � L.

�Sn�

125. Let and

Both are divergent series.

��an � bn� � ��

n�0�1 � ��1�� � �

n�0�1 � 1� � 0

� bn � ��

n�0 ��1�.� an � �

n�0 1 126. If converged, then

would converge, which is a contradiction. Thus,diverges.� �an � bn�

� bn

� �an � bn� � � an �� �an � bn�

127. Suppose, on the contrary, that converges. Because

converges. This is a contradiction since diverged.Hence, diverges.� can

� can

��1c�can � � an

c � 0,� can 128. If converges, then Hence,

which implies that diverges.� 1�an

limn→�

1an

� 0,

limn→�

an � 0.� an

129. (a)

(b)

��

n�0

1an�1an�3

� limn→�

Sn � limn→�

�1 �1

an�2an�3 � 1

� 1 �1

an�2an�3

�1

a1a2�

1an�2an�3

� � 1a1a2

�1

a2a3 � � 1

a2a3�

1a3a4

� . . . � � 1an�1an�2

�1

an�2an�3

� �n

k�0� 1

ak�1ak�2�

1ak�2ak�3

Sn � �n

k�0

1ak�1ak�3

�1

an�1an�3

�an�2

an�1an�2an�3

1

an�1an�2�

1an�2an�3

�an�3 � an�1

an�1an�2an�3

Page 28: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.2 Series and Convergence 259

130.

As the two geometric series converge for

limn→�

S2n �1

1 � x2 � 2x� 11 � x2� �

1 � 2x1 � x2 , x < 1

x < 1:n →�,

� �n

k�0�x2�k � 2x �

n

k�0�x2�k

� �1 � x2 � x 4 � . . . � x2n� � �2x � 2x3 � . . . � 2x2n�1�

S2n � 1 � 2x � x2 � 2x3 � . . . � x2n � 2x2n�1

131.

This is a geometric series which converges if

1r < 1 ⇔ r > 1.

�since 1r < 1� �1

r � 1

�1�r

1 � �1�r�

1r

�1r 2 �

1r 3 � . . . � �

n�0 1r�

1r�

n

132. The entire rectangle has area 2 because the height is 1and the base is The squares alllie inside the rectangle, and the sum of their areas is

Thus, ��

n�1 1n2 < 2.

1 �122 �

132 �

142 � . . . .

1 �12 �

14 � . . . � 2.

133. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug,administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by where One time interval after the last dose is administered is given by Two time intervals after the last dose is administered is given by and so on. Since as where s is an integer.s →�,k < 0, Tn�s→ 0

Tn�2 � Pe2kt � Pe3kt � Pe4kt � . . . � Pe�n�1�kt

Tn�1 � Pekt � Pe2kt � Pe3kt � . . . � Penkt.k � ��ln 2��H.Tn � P � Pekt � Pe2kt � . . . � Pe�n�1�kt

134. The series is telescoping:

limn→�

Sn � 3 � 1 � 2

� 3 �3n�1

3n�1 � 2n�1

� �n

k�1� 3k

3k � 2k �3k�1

3k�1 � 2k�1

Sn � �n

k�1

6k

�3k�1 � 2k�1��3k � 2k�

135.

In general:

(See below for a proof of this.)

and are either both odd or both even. If both even, then

If both odd,

Proof by induction that the formula for is correct. It is true for Assume that the formula is valid for

If is even, then and

The argument is similar if is odd.k

�k2 � 2k

4�

�k � 1�2 � 14

.

f �k � 1� � f �k� �k2

�k2

4�

k2

f �k� � k2�4kk.n � 1.

f �n�

� xy.

f �x � y� � f �x � y� ��x � y�2 � 1

4�

�x � y�2 � 14

f �x � y� � f �x � y� ��x � y�2

4�

�x � y�2

4� xy.

x � yx � y

f �n� � �n2�4,

�n2 � 1��4, n even

n odd.

f �1� � 0, f �2� � 1, f �3� � 2, f �4� � 4, . . .

Page 29: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.3 The Integral Test and p-Series

260 Chapter 9 Infinite Series

1.

Let

f is positive, continuous, and decreasing for

Diverges by Theorem 9.10

��

1

1x � 1

dx � �ln�x � 1��1

�� �

x ≥ 1.

f �x� �1

x � 1.

��

n�1

1n � 1

2.

Let

f is positive, continuous, and decreasing for

Diverges by Theorem 9.10

��

1

23x � 5

dx � �23

ln�3x � 5���

1� �

x ≥ 1.

f �x� �2

3x � 5.

��

n�1

2

3n � 5

3.

Let

f is positive, continuous, and decreasing for

Converges by Theorem 9.10

��

1 e�x dx � ��e�x�

1�

1e

x ≥ 1.

f �x� � e�x.

��

n�1e�n 4.

Let

f is positive, continuous, and decreasing for since

for

Converges by Theorem 9.10

��

3xe�x�2 dx � ��2�x � 2�e�x�2�

3� 10e�3�2

x ≥ 3.f � �x� �2 � x2e x�2 < 0

x ≥ 3

f �x� � xe�x�2.

��

n�1ne�n�2

5.

Let

is positive, continuous, and decreasing for

Converges by Theorem 9.10

��

1

1x2 � 1

dx � �arctan x��

1�

4

x ≥ 1.f

f �x� �1

x2 � 1.

��

n�1

1n2 � 1

6.

Let

f is positive, continuous, and decreasing for

Diverges by Theorem 9.10

��

1

12x � 1

dx � �ln �2x � 1��

1� �

x ≥ 1.

f �x� �1

2x � 1.

��

n�1

12n � 1

7.

Let

f is positive, continuous, and decreasing for since

Diverges by Theorem 9.10

��

1 ln�x � 1�

x � 1 dx � �ln�x � 1� 2

2 ��

1� �

f��x� �1 � ln�x � 1�

�x � 1�2 < 0 for x ≥ 2.

x ≥ 2

f �x� �ln�x � 1�

x � 1.

��

n�1 ln�n � 1�

n � 18.

Let

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

2 ln x�x

dx � �2�x�ln x � 2���

2� �,

x > e2 � 7.4.f

f �x� �ln x�x

, f ��x� �2 � ln x

2x3�2 .

��

n�2 ln n�n

Page 30: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.3 The Integral Test and p-Series 261

9.

Let

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

1

1�x��x � 1� dx � �2 ln��x � 1��

1� �,

x ≥ 1.f

f �x� �1

�x��x � 1�, f ��x� � �1 � 2�x

2x3�2��x � 1�2 < 0.

��

n�1

1�n��n � 1�

10.

Let

is positive, continuous, and decreasing for since

Diverges by Theorem 9.10

��

1

xx2 � 3

dx � �ln�x2 � 3��

1� �

f��x� �3 � x2

�x2 � 3� < 0 for x ≥ 2.

x ≥ 2f �x�

f �x� �x

x2 � 3.

��

n�1

nn2 � 3

11.

Let

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

1

1�x � 1

dx � �2�x � 1��

1� �,

x ≥ 1.f

f �x� �1

�x � 1, f ��x� �

�12�x � 1�3�2 < 0.

��

n�1

1�n � 1

12.

Let

is positive, continuous, and decreasing for

Hence, the series converges by Theorem 9.10.

�2 ln 2 � 1

16, converges

��

2 ln xx2 dx � ��

�2 ln x � 1�4x 4 �

2

x > 2.f

f �x� �ln xx3 , f ��x� �

1 � 3 ln xx4 .

��

n�2

ln nn3

13.

Let

is positive, continuous, and decreasing for

converges

Hence, the series converges by Theorem 9.10.

��

1 ln xx2 dx � ���ln x � 1�

x ��

1� 1,

x > e1�2 � 1.6.f

f �x� �ln xx2 , f ��x� �

1 � 2 ln xx3 .

��

n�1 ln nn2 14.

Let

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

2

1

x�ln x dx � �2�ln x�

2� �,

x ≥ 2.f

f �x� �1

x�ln x, f ��x� � �

2 ln x � 12x2�ln x�3�2.

��

n�2

1

n�ln n

15.

Let for

is positive, continuous, and decreasing for

converges

Hence, the series converges by Theorem 9.10.

��

1 arctan xx2 � 1

dx � ��arctan x�2

2 ��

1�

3� 2

32,

x ≥ 1.f

x ≥ 1.f �x� �arctan xx2 � 1

, f ��x� �1 � 2x arctan x

�x2 � 1�2 < 0

��

n�1 arctan nn2 � 1

16.

Let

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

3

1x ln x ln�ln x� dx � �ln�ln�ln x���

3� �,

x ≥ 3.f

f ��x� ���ln x � 1� ln�ln x� � 1

x2�ln x�2�ln�ln x��2 .

f �x� �1

x ln x ln�ln x�,

��

n�3

1n ln n ln�ln n�

Page 31: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

262 Chapter 9 Infinite Series

17.

Let for

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

1

2xx2 � 1

dx � �ln�x2 � 1���

1� �,

x > 1.f

x > 1.f �x� �2x

x2 � 1, f ��x� � 2

1 � x2

�x2 � 1�2 < 0

��

n�1

2nn2 � 1

18.

Let for

is positive, continuous, and decreasing for

converges

Hence, the series converges by Theorem 9.10.

��

1

xx4 � 1

dx � �12

arctan�x2���

1�

8,

x > 1.f

x > 1.f �x� �x

x 4 � 1, f ��x� �

1 � 3x 4

�x 4 � 1�2 < 0

��

n�1

nn4 � 1

19.

Let

is positive, continuous, and decreasing forsince

Diverges by Theorem 9.10

��

1

xk�1

xk � c dx � �1

k ln�xk � c��

1� �

for x > k�c�k � 1�.

f � �x� �xk�2c�k � 1� � xk

�xk � c�2 < 0

x > k�c�k � 1�f

f �x� �xk�1

xk � c.

��

n�1

nk�1

nk � c20.

Let

f is positive, continuous, and decreasing for since

for

We use integration by parts.

Converges by Theorem 9.10

�1e

�ke

�k�k � 1�

e� . . . �

k!e

��

1 x ke�x dx � ��x ke�x�

1� k ��

1 x k�1e�x dx

x > k.f � �x� �x k�1�k � x�

ex < 0

x > k

f �x� �xk

ex .

��

n�1 n ke�n

21. Let

The function is not positive for x ≥ 1.f

f �x� ���1�x

x, f �n� � an.

22. Let

The function is not positive for x ≥ 1.f

f �x� � e�x cos x, f �n� � an.

23. Let

The function is not decreasing for x ≥ 1.f

f �x� �2 � sin x

x, f �n� � an. 24. Let

The function is not decreasing for x ≥ 1.f

f �x� � �sin xx 2

, f �n� � an.

25.

Let

f is positive, continuous, and decreasing for

Converges by Theorem 9.10

��

1

1x3 dx � ��

12x2�

1�

12

x ≥ 1.

f �x� �1x3.

��

n�1 1n3 26.

Let

is positive, continuous, and decreasing for

Diverges by Theorem 9.10

��

1

1x1�3 dx � �3

2 x2�3�

1� �

x ≥ 1.f

f �x� �1

x1�3.

��

n�1

1n1�3

Page 32: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.3 The Integral Test and p-Series 263

27.

Let for

is positive, continuous, and decreasing for

diverges

Hence, the series diverges by Theorem 9.10.

��

1

1�x

dx � �2�x��

1� �,

x ≥ 1.f

x ≥ 1.f �x� �1�x

, f ��x� ��1

2x3�2 < 0

��

n�1

1�n

28.

Let for

is positive, continuous, and decreasing for

converges

Hence, the series converges by Theorem 9.10.

��

1

1x2 dx � ��

1x�

1� 1,

x ≥ 1.f

x ≥ 1.f �x� �1x2, f ��x� �

�2x3 < 0

��

n�1 1n2

29.

Divergent p-series with p �15 < 1

��

n�1

15�n

� ��

n�1

1n1�5 30.

Convergent p-series with p �53 > 1

��

n�1

3n5�3 31.

Divergent p-series with p �12 < 1

��

n�1

1n1�2

32.

Convergent p-series with p � 2 > 1

��

n�1 1n2 33.

Convergent p-series with p �32 > 1

��

n�1

1n3�2 34.

Divergent p-series with p �23 < 1

��

n�1

1n2�3

35.

Convergent p-series withp � 1.04 > 1

��

n�1

1n1.04 36.

Convergent p-series withp � � > 1

��

n�1

1n� 37.

Matches (c), diverges

S4 � 4.7740

S3 � 4.0666

S2 � 3.1892

S1 � 2

��

n�1

2n3�4

38.

Matches (f), diverges

S4 � 4.1667

S3 � 3.6667

S2 � 3

S1 � 2

��

n�1 2n

39.

Matches (b), converges

Note: The partial sums for 39 and41 are very similar because��2 � 3�2.

S4 � 3.2560

S3 � 3.0293

S2 � 2.6732

S1 � 2

��

n�1

2�n�

� ��

n�1

2n��2 40.

Matches (a), diverges

S3 � 4.8045

S2 � 3.5157

S1 � 2

��

n�1

2n2�5

41.

Matches (d), converges

Note: The partial sums for 39 and 41 are very similarbecause ��2 � 3�2.

S4 � 3.3420

S3 � 3.0920

S2 � 2.7071

S1 � 2

��

n�1

2

n�n� �

n�1

2n3�2 42.

Matches (e), converges

S3 � 2.7222

S2 � 2.5

S1 � 2

��

n�1 2n2

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49. The area under the rectangles is greater than the area under the graph of

Since diverges, the series diverges.��

n�1

1�n

��

1

1�x

dx � �2�x��

1� �

��

n�1

1�n

> ��

1

1�x

dx

y � 1��x, x ≥ 1.

x

y

1 2 3 4 5 n

0.5

1.0

264 Chapter 9 Infinite Series

43. (a)

The partial sums approach the sum 3.75 very rapidly.

(b)

The partial sums approach the sum slower than the series in part (a).

� 2�6 � 1.6449 00

11

5

00

11

8

n 5 10 20 50 100

3.7488 3.75 3.75 3.75 3.75Sn

n 5 10 20 50 100

1.4636 1.5498 1.5962 1.6251 1.635Sn

44.

(a)

�N

n�1 1n

� 1 �12

�13

�14

� . . . �1N

> M

(b) No. Since the terms are decreasing (approaching zero),more and more terms are required to increase the partialsum by 2.

M 2 4 6 8

N 4 31 227 1674

45. Let f be positive, continuous, and decreasing for and Then,

and

either both converge or both diverge (Theorem 9.10). See Example 1, page 618.

��

1f �x� dx�

n�1an

an � f �n�.x ≥ 1 46. A series of the form is a p-series,

The p-series converges if and diverges if0 < p ≤ 1.

p > 1

p > 0.��

n�1

1np

47. Your friend is not correct. The series

is the harmonic series, starting with the term,and hence diverges.

10,000th

��

n�10,000 1n

�1

10,000�

110,001

� . . .

48. No. Theorem 9.9 says that if the series converges, then the terms tend to zero. Some of the series in Exercises37–42 converge because the terms tend to 0 very rapidly.

an

51.

If then the series diverges by the Integral Test. If

Converges for �p � 1 < 0 or p > 1

��

2

1x�ln x� p dx � ��

2�ln x��p

1x dx � ��ln x��p�1

�p � 1 ��

2.

p � 1,p � 1,

��

n�2

1n�ln n� p50.

1 2 3 4 5 6 7

1

x

y

�6

n�1an ≥ �7

1f �x� dx ≥ �

7

n�2an

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Section 9.3 The Integral Test and p-Series 265

52.

If then the series diverges by the Integral Test. If

(Use integration by parts.)

Converges for �p � 1 < 0 or p > 1

��

2 ln xx p dx � ��

2x�p ln x dx � � x�p�1

��p � 1�2 �1 � ��p � 1� ln x��

2.

p � 1,p � 1,

��

n�2 ln nn p

53.

If diverges (see Example 1). Let

For a fixed is eventually negative. is positive, continuous, and eventually decreasing.

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

��

1

x�1 � x2� p dx � � 1

�x2 � 1� p�1�2 � 2p���

1

fp > 0, p � 1, f ��x�

f ��x� �1 � �2p � 1�x2

�1 � x2� p�1 .

f �x� �x

�1 � x2� p , p � 1

��

n�1

n1 � n2p � 1,

��

n�1

n�1 � n2� p

54.

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

��

n�1n�1 � n2� p 55. converges for

Hence, diverges.��

n�2

1n ln n

,

p > 1.��

n�2

1n�ln n� p

56. converges for

Hence, diverges.��

n�2

1

n 3��ln n�2� �

n�2

1n�ln n�2�3,

p > 1.��

n�2

1n�ln n� p 57. converges for

Hence, converges.��

n�2

1n�ln n�2,

p > 1.��

n�2

1n�ln n� p

58. converges for

�12

��

n�2

1n ln�n�, diverges

Hence, ��

n�2

1n ln�n2� � �

n�2

12n ln n

p > 1.��

n�2

1n�ln n� p 59. Since f is positive, continuous, and decreasing for

and we have,

Also,

Thus, 0 ≤ RN ≤ ��

N

f �x� dx.

≤ ��

N

f �x� dx.

RN � S � SN � ��

n�N�1

an ≤ aN�1 � ��

N�1 f �x� dx

RN � S � SN � ��

n�1 an � �

N

n�1 an � �

n�N�1 an > 0.

an � f �n�,x ≥ 1

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266 Chapter 9 Infinite Series

60. From Exercise 59, we have:

�N

n�1 an ≤ S ≤ �

N

n�1 an � ��

N

f �x� dx

SN ≤ S ≤ SN � ��

N

f �x� dx

0 ≤ S � SN ≤ ��

N

f �x� dx

61.

1.0811 ≤ ��

n�1 1n4 ≤ 1.0811 � 0.0015 � 1.0826

R6 ≤ ��

6 1x4 dx � ��

13x3�

6� 0.0015

S6 � 1 �124 �

134 �

144 �

154 �

164 � 1.0811

62.

1.0363 ≤ ��

n�1 1n5 ≤ 1.0363 � 0.0010 � 1.0373

R4 ≤ ��

4 1x5 dx � ��

14x 4�

4� 0.0010

S4 � 1 �125 �

135 �

145 � 1.0363

63.

0.9818 ≤ ��

n�1

1n2 � 1

≤ 0.9818 � 0.0997 � 1.0815

R10 ≤ ��

10

1x2 � 1

dx � �arctan x��

10�

2� arctan 10 � 0.0997

� 1

50�

165

�1

82�

1101

� 0.9818 S10 �12

�15

�1

10�

117

�1

26�

137

64.

1.9821 ≤ ��

n�1

1�n � 1�ln�n � 1�3 ≤ 1.9821 � 0.0870 � 2.0691

R10 ≤ ��

10

1�x � 1�ln�x � 1�3 dx � ��

12ln�x � 1�2�

10�

12�ln 11�3 � 0.0870

S10 �1

2�ln 2�3 �1

3�ln 3�3 �1

4�ln 4�3 � . . . �1

11�ln 11�3 � 1.9821

65.

0.4049 ≤ ��

n�1 ne�n2

≤ 0.4049 � 5.6 � 10�8

R4 ≤ ��

4 xe�x2

dx � ��12

e�x2��

4�

e�16

2� 5.6 � 10�8

S4 �1e

�2e4 �

3e9 �

4e16 � 0.4049 66.

0.5713 ≤ ��

n�0 e�n ≤ 0.5713 � 0.0183 � 0.5896

R4 ≤ ��

4 e�x dx � ��e�x�

4� 0.0183

S4 �1e

�1e2 �

1e3 �

1e4 � 0.5713

67.

N ≥ 7

N > 6.93

N 3 > 333.33

1

N 3 < 0.003

0 ≤ RN ≤ ��

N

1x4 dx � ��

13x3�

N�

13N 3 < 0.001 68.

N ≥ 4,000,000

�N > 2000

N�1�2 < 0.0005

0 ≤ RN ≤ ��

N

1

x3�2 dx � ��2

x1�2��

N�

2�N

< 0.001

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Section 9.3 The Integral Test and p-Series 267

69.

N ≥ 2

N > 1.0597

N > ln 200

5

5N > ln 200

e5N > 200

1

e5N < 0.005

RN ≤ ��

N

e�5x dx � ��15

e�5x��

N �

e�5N

5 < 0.001 70.

N ≥ 16

N > 2 ln 2000 � 15.2

N2

> ln 2000

eN�2 > 2000

2

eN�2 < 0.001

RN ≤ ��

N

e�x�2 dx � ��2e�x�2��

N�

2eN�2 < 0.001

71.

N ≥ 1004

N > tan 1.5698

arctan N > 1.5698

�arctan N < �1.5698

��

2� arctan N < 0.001

RN ≤ ��

N

1

x2 � 1 dx � �arctan x�

N72.

N ≥ 2004

N�5

> tan 1.56968

1.56968 < arctan� N�5

2� arctan� N

�5 < 0.001118

�2�5��

2� arctan� N

�5 < 0.001

Rn ≤ ��

N

2x2 � 5

dx � 2� 1�5

arctan� x�5 �

N

73. (a) This is a convergent p-series with a divergent series. Use the Integral Test.

is positive, continuous, and decreasing for

(b)

For the terms of the convergent series seem to be larger than those of the divergent series!

(c)

This inequality holds when Or, Then ln e40 � 40 < �e40�0.1 � e4 � 55.n > e40. n ≥ 3.5 � 1015.

ln n < n0.1

n ln n < n1.1

1

n1.1 < 1

n ln n

n ≥ 4,

�6

n�2

1n ln n

�1

2 ln 2�

13 ln 3

�1

4 ln 4�

15 ln 5

�1

6 ln 6� 0.7213 � 0.3034 � 0.1803 � 0.1243 � 0.0930

�6

n�2

1n1.1 �

121.1 �

131.1 �

141.1 �

151.1 �

161.1 � 0.4665 � 0.2987 � 0.2176 � 0.1703 � 0.1393

��

2

1x ln x

dx � �ln�ln x���

2� �

x ≥ 2.f �x� �1

x ln x

��

n�2

1n ln n

isp � 1.1 > 1.��

n�2

1n1.1 .

74. (a)

(c) The horizontal asymptote is As n increases, theerror decreases.

y � 0.

��

10 1x p dx � � x�p�1

�p � 1��

10�

1�p � 1�10 p�1, p > 1 (b)

≤ Area under the graph of f over the interval 10, ��

R10� p� � ��

n�11

1n p

f �x� �1x p

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268 Chapter 9 Infinite Series

75. (a) Let f is positive, continuous, and decreasing on

Hence, Similarly,

Thus,

(b) Since we have Also, since is an increasing function,for Thus, and the sequence is bounded.

(c)

Thus, and the sequence is decreasing.

(d) Since the sequence is bounded and monotonic, it converges to a limit,

(e) (Actually ) � 0.577216.a100 � S100 � ln 100 � 0.5822

.

an ≥ an�1

an � an�1 � Sn � ln n � Sn�1 � ln�n � 1� � �n�1

n

1x dx �

1n � 1

≥ 0

�an�0 ≤ Sn � ln n ≤ 1n ≥ 1.ln�n � 1� � ln n > 0ln xln�n � 1� � ln n ≤ Sn � ln n ≤ 1.ln�n � 1� ≤ Sn ≤ 1 � ln n,

ln�n � 1� ≤ Sn ≤ 1 � ln n.

Sn ≥ �n�1

1 1x dx � ln�n � 1�.

Sn ≤ 1 � ln n.

Sn � 1 ≤ ln n

Sn � 1 ≤ �n

1 1x dx

1 2 3 ...n−1

n n+1x

1

12

y1, ��.f �x� � 1�x.

76.

� �ln 7 � ln 5 � 2 ln 6� � �ln 8 � ln 6 � 2 ln 7� � �ln 9 � ln 7 � 2 ln 8� � . . . � �ln 2

� 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�

��

n�2 ln�1 �

1n2 � �

n�2 ln�n2 � 1

n2 � ��

n�2 ln

�n � 1�(n � 1�n2 � �

n�2 ln�n � 1� � ln�n � 1� � 2 ln n

77.

(a) diverges

(b) diverges��

n�2�1

e ln n

� ��

n�2e�ln n � �

n�2 1n

,x �1e

:

��

n�21ln n � �

n�21,x � 1:

��

n�2xln n

(c) Let be given, Put

This series converges for p > 1 ⇒ x <1e.

��

n�2xln n � �

n�2e�p ln n � �

n�2n�p � �

n�2

1n p

x � e�p ⇔ ln x � �p.x > 0.x

78.

Converges for by Theorem 9.11x > 1

�x� � ��

n�1 n�x � �

n�1 1nx

79.

Let

f is positive, continuous, and decreasing for

Diverges by Theorem 9.10

��

1

12x � 1

dx � �ln �2x � 1��

1� �

x ≥ 1.

f �x� �1

2x � 1.

��

n�1

12n � 1

80.

Let

f is positive, continuous, and decreasing for

Converges by Theorem 9.10

��

2

1

x�x2 � 1 dx � �arcsec x�

2�

2�

3

x ≥ 2.

f �x� �1

x�x2 � 1.

��

n�2

1

n�n2 � 1

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Section 9.3 The Integral Test and p-Series 269

83.

Geometric series with

Converges by Theorem 9.6

r �23

��

n�0 �2

3 n

84.

Geometric series with

Diverges by Theorem 9.6

r � 1.075

��

n�0 �1.075�n

89.

Let

f is positive, continuous and decreasing for

Converges by Theorem 9.10. See Exercise 51.

��

2

1x�ln x�3 dx � ��

2 �ln x��3

1x dx � ��ln x��2

�2 ��

2� ��

12�ln x�2��

2�

12�ln 2�2

x ≥ 2.

f �x� �1

x�ln x�3.

��

n�2

1n�ln n�3

90.

Let

f is positive, continuous, and decreasing for since for

(Use integration by parts.)

Converges by Theorem 9.10. See Exercise 30.

��

2 ln xx3 dx � ��

ln x2x2��

2�

12

��

2 1x3 dx �

ln 28

� ��1

4x2��

2�

ln 28

�1

16

x ≥ 2.f��x� �1 � 3 ln x

x4 < 0x ≥ 2

f �x� �ln xx3 .

��

n�2 ln nn3

81.

p-series with

Converges by Theorem 9.11

p �54

��

n�1

1

n 4�n� �

n�1

1n5�4 82.

p-series with

Diverges by Theorem 9.11

p � 0.95

3 ��

n�1

1n0.95

85.

Diverges by Theorem 9.9

limn→�

n

�n2 � 1� lim

n→�

1�1 � �1�n2�

� 1 � 0

��

n�1

n�n2 � 1

86.

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

��

n�1 � 1

n2 �1n3 � �

n�1 1n2 � �

n�1 1n3

87.

Fails nth-Term Test

Diverges by Theorem 9.9

limn→�

�1 �1n

n

� e � 0

��

n�1 �1 �

1n

n

88.

Diverges by Theorem 9.9

limn→�

ln�n� � �

��

n�2 ln�n�

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Section 9.4 Comparisons of Series

270 Chapter 9 Infinite Series

1. (a)

(b) The first series is a p-series. It converges

(c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge.

(d) The smaller the magnitude of the terms, the smaller the magnitude of the termsof the sequence of partial sums.

n

2

2

4

4

6

6 8

8

10

10

12

6k3/2Σ

k = 1

n

Σ 6k3/2 + 3k = 1

n

Σ 6

k2+ 0.5kk = 1

n

Sn

� p �32 > 1�.

��

n�1

6

n�n2 � 0.5�

6

1�1.5�

62�4.5

� . . . ; S1 �6

�1.5� 4.9

��

n�1

6n3�2 � 3

�64

�6

23�2 � 3� . . . ; S1 �

32

n

2

1

2

4

3

4

6

5

6 8 10

6n3/2

an =

6n3/2 + 3

an =

6

n2+ 0.5an =

n

an��

n�1

6n3�2 �

61

�6

23�2 � . . . ; S1 � 6

2. (a)

(b) The first series is a p-series. It diverges

(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge.

(d) The larger the magnitude of the terms, the larger the magnitude of the terms of thesequence of partial sums.

n4

4

8

8

10

12

16

20

62

n + 0.54

2n

n − 0.52Σ

Σ

Σ

Sn

� p �12 < 1�.

��

n�1

4�n � 0.5

�4

�1.5�

4�2.5

� . . . S1 � 3.3

��

n�1

2�n � 0.5

�2

0.5�

2�2 � 0.5

� . . . S1 � 4

n

2

2

4

4 6 8 10

1

3an =

n + 0.54

an =n − 0.5

2

an = 2n

an��

n�1

2�n

� 2 �2�2

� . . . S1 � 2

3.

Therefore,

converges by comparison with theconvergent p-series

��

n�1 1n2.

��

n�1

1n2 � 1

0 <1

n2 � 1 <

1n2 4.

Therefore,

converges by comparison with theconvergent p-series

13

��

n�1 1n2.

��

n�1

13n2 � 2

13n2 � 2

< 1

3n2 5. for

Therefore,

diverges by comparison with thedivergent p-series

��

n�2 1n

.

��

n�2

1n � 1

n ≥ 21

n � 1 >

1n

> 0

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Section 9.4 Comparisons of Series 271

6. for

Therefore,

diverges by comparison with thedivergent p-series

��

n�2

1�n

.

��

n�2

1�n � 1

n ≥ 21

�n � 1 >

1�n

7.

Therefore,

converges by comparison with theconvergent geometric series

��

n�0 �1

3�n

.

��

n�0

13n � 1

0 <1

3n � 1 <

13n 8.

Therefore,

converges by comparison with theconvergent geometric series

��

n�0�3

4�n

.

��

n�0

3n

4n � 5

3n

4n � 5< �3

4�n

9. For

Therefore,

diverges by comparison with thedivergent series

Note: diverges by the

Integral Test.

��

n�1

1n � 1

��

n�1

1n � 1

.

��

n�1

ln nn � 1

ln nn � 1

> 1

n � 1> 0.n ≥ 3, 10.

Therefore,

converges by comparison with theconvergent p-series

��

n�1

1n3�2.

��

n�1

1�n3 � 1

1�n3 � 1

< 1

n3�2 11. For

Therefore,

converges by comparison with theconvergent p-series

��

n�1 1n2.

��

n�0 1n!

1n2 >

1n!

> 0.n > 3,

12.

Therefore,

diverges by comparison with thedivergent p-series

14

��

n�1

14�n

.

��

n�1

1

4 3�n � 1

1

4 3�n � 1>

1

4 4�n13.

Therefore,

converges by comparison with theconvergent geometric series

��

n�0 �1

e�n

.

��

n�0

1en2

0 <1

en2 ≤ 1en 14.

Therefore,

diverges by comparison with thedivergent geometric series

��

n�1 �4

3�n

.

��

n�1

4n

3n � 1

4n

3n � 1 >

4n

3n

15.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1

nn2 � 1

limn→�

n��n2 � 1�

1�n� lim

n→�

n2

n2 � 1� 1 16.

Therefore,

converges by a limit comparison with the convergentgeometric series

��

n�1�1

3�n

.

��

n�1

23n � 5

limn→�

2��3n � 5�

1�3n � limn→�

2 � 3n

3n � 5� 2

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272 Chapter 9 Infinite Series

17.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�0

1�n2 � 1

limn→�

1��n2 � 1

1�n� lim

n→�

n�n2 � 1

� 1 18.

Therefore,

diverges by a limit comparison with the divergentharmonic series

��

n�3 1n

.

��

n�3

3�n2 � 4

limn→�

3��n2 � 4

1�n� lim

n→�

3n�n2 � 4

� 3

19.

Therefore,

converges by a limit comparison with the convergent p-series

��

n�1 1n3.

��

n�1

2n2 � 13n5 � 2n � 1

limn→�

2n2 � 13n5 � 2n � 1

1�n3 � limn→�

2n5 � n3

3n5 � 2n � 1�

23

20.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1

5n � 3n2 � 2n � 5

limn→�

5n � 3n2 � 2n � 5

1�n� lim

n→�

5n2 � 3nn2 � 2n � 5

� 5

21.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1

n � 3n�n � 2�

limn→�

n � 3n�n � 2�

1�n� lim

n→� n2 � 3nn2 � 2n

� 1 22.

Therefore,

converges by a limit comparison with the convergent p-series

��

n�1 1n3.

��

n�1

1n�n2 � 1�

limn→�

1n�n2 � 1�

1�n3 � limn→�

n3

n3 � n� 1

23.

Therefore,

converges by a limit comparison with the convergent p-series

��

n�1 1n2.

��

n�1

1

n�n2 � 1

limn→�

1��n�n2 � 1�

1�n2 � limn→�

n2

n�n2 � 1� 1 24.

Therefore,

converges by a limit comparison with the convergentgeometric series

��

n�1�1

2�n�1

.

��

n�1

n�n � 1�2n�1

limn→�

n��n � 1�2n�1

1��2n�1� � limn→�

n

n � 1� 1

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Section 9.4 Comparisons of Series 273

25.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1

nk�1

nk � 1

limn→�

�nk�1���nk � 1�

1�n� lim

n→�

nk

nk � 1� 1 26.

Therefore,

diverges by a limit comparison with the divergentharmonic series

��

n�1 1n

.

��

n�1

5

n � �n2 � 4

limn→�

5��n � �n2 � 4�

1�n� lim

n→�

5n

n � �n2 � 4�

52

27.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1 sin�1

n�

� limn→�

cos�1n� � 1

limn→�

sin�1�n�

1�n� lim

n→� ��1�n2� cos�1�n�

�1�n2 28.

Therefore,

diverges by a limit comparison with the divergent p-series

��

n�1 1n

.

��

n�1 tan�1

n�

� limn→�

sec2�1n� � 1

limn→�

tan�1�n�

1�n� lim

n→� ��1�n2� sec2�1�n�

�1�n2

29.

Diverges

p-series with p �12

��

n�1 �nn

� ��

n�1

1�n

30.

Converges

Geometric series with r � �15

��

n�0 5��

15�

n

31.

Converges

Direct comparison with ��

n�1�1

3�n

��

n�1

13n � 2

32.

Converges

Limit comparison with ��

n�1 1n2

��

n�4

13n2 � 2n � 15

33.

Diverges; nth-Term Test

limn→�

n

2n � 3�

12

� 0

��

n�1

n2n � 3

34.

Converges; telescoping series

��

n�1� 1

n � 1�

1n � 2� � �1

2�

13� � �1

3�

14� � �1

4�

15� � . . . �

12

35.

Converges; Integral Test

��

n�1

n�n2 � 1�2 36.

Converges; telescoping series

��

n�1 �1

n�

1n � 3�

��

n�1

3n�n � 3�

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274 Chapter 9 Infinite Series

37. By given conditions

is finite and nonzero. Therefore,

diverges by a limit comparison with the p-series

��

n�1 1n

.

��

n�1an

limn→�

nanlimn→�

an

1�n� lim

n→� nan. 38. If then The p-series with

converges and since

the series

converges by the Limit Comparison Test. Similarly, ifthen which implies that

diverges by the Limit Comparison Test.

��

n�1 P�n�Q�n�

k � j ≤ 1j ≥ k � 1,

��

n�1 P�n�Q�n�lim

n→� P�n��Q�n�

1�nk� j � L > 0,

p � k � jk � j > 1.j < k � 1,

39.

which diverges since the degree of the numerator is onlyone less than the degree of the denominator.

12

�25

�3

10�

417

�5

26� . . . � �

n�1

nn2 � 1

, 40.

which converges since the degree of the numerator is twoless than the degree of the denominator.

13

�18

�1

15�

124

�1

35� . . . � �

n�2

1n2 � 1

,

41.

converges since the degree of the numerator is three lessthan the degree of the denominator.

��

n�1

1n3 � 1

42.

diverges since the degree of the numerator is only one lessthan the degree of the denominator.

��

n�1

n2

n3 � 1

43.

Therefore, diverges.��

n�1

n3

5n4 � 3

limn→�

n� n3

5n4 � 3� � limn→�

n4

5n4 � 3�

15

� 0 44.

Therefore, diverges.��

n�2

1ln n

limn→�

n

ln n� lim

n→�

11�n

� limn→�

n � � � 0

45.

diverges, (harmonic)

1200

�1

400�

1600

� . . . � ��

n�1

1200n

46.

diverges

1200

�1

210�

1220

� . . . � ��

n�0

1200 � 10n

47.

converges

1201

�1

204�

1209

�1

216� �

n�1

1200 � n2

48.

converges

1201

�1

208�

1227

�1

264� . . . � �

n�1

1200 � n3

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

50. See Theorem 9.12, page 624. One example is

converges because and

converges ( p-series).��

n�1 1n2

1n2 � 1

<1n2�

n�1

1n2 � 1

51. See Theorem 9.13, page 626. One example is

diverges because and

diverges ( p-series).��

n�2

1�n

limn→�

1��n � 1

1��n� 1�

n�2

1�n � 1

52. This is not correct. The beginning terms do not affect theconvergence or divergence of a series. In fact,

diverges (harmonic)

and converges ( p-series).1 �14

�19

� . . . � ��

n�1 1n2

11000

�1

1001� . . . � �

n�1000 1n

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Section 9.4 Comparisons of Series 275

53.

For Hence, the lower terms are those of � an2.0 < an < 1, 0 < an

2 < an < 1.

0.2

0.4

0.6

0.8

1.0

n4 8 12 16 20

Σan∞

n = 1

Σan∞

n = 1

2

Terms of

Terms of

54. (a)

converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.)

(c) ��

n�3

1�2n � 1�2 �

� 2

8� S2 � 0.1226

��

n�1

1�2n � 1�2 � �

n�1

14n2 � 4n � 1

(b)

(d) ��

n�10

1�2n � 1�2 �

� 2

8� S9 � 0.0277

n 5 10 20 50 100

1.1839 1.02087 1.2212 1.2287 1.2312Sn

55. False. Let and and both and converge.��

n�1 1n2�

n�1 1n30 < an ≤ bnbn �

1n2.an �

1n3

58. False. Let Then,

but converges.��

n�1cnan ≤ bn � cn,

an � 1�n, bn � 1�n, cn � 1�n2. 59. True

56. True 57. True

60. False. could converge or diverge.

For example, let which diverges.

and diverges, but

and converges.��

n�1 1n20 <

1n2 <

1�n

��

n�1 1n

0 <1n

<1�n

��

n�1bn � �

n�1

1�n

,

��

n�1an 61. Since converges, There exists N such

that for Thus, for and

converges by comparison to the convergent

series ��

i�1 an .

��

n�1 an bn

n > Nanbn < ann > N.bn < 1

limn→�

bn � 0.��

n�1 bn

62. Since converges, then

converges by Exercise 61.

��

n�1an an � �

n�1 an

2��

n�1 an 63. and both converge, and hence so does

�� 1n2�� 1

n3� � � 1n5.

� 1n3�

1n2

64. converge, and hence so does �� 1n2�2

� � 1n4.�

1n2

65. Suppose and converges.

From the definition of limit of a sequence, there existssuch that

whenever Hence, for From theComparison Test, converges.� an

n > M.an < bnn > M.

�an

bn

� 0� < 1

M > 0

� bnlimn→�

an

bn

� 0

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276 Chapter 9 Infinite Series

66. Suppose and diverges. From the definition of limit of a sequence, there exists such that

for Thus, for By the Comparison Test, diverges.� ann > M.an > bnn > M.

an

bn

> 1

M > 0� bnlimn→�

an

bn

� �

67. (a) Let and converges.

By Exercise 65, converges.��

n�1

1�n � 1�3

limn→�

an

bn

� limn→�

1���n � 1�3�

1��n2� � limn→�

n2

�n � 1�3 � 0

� bn � � 1n2,� an � �

1�n � 1�3, (b) Let and converges.

By Exercise 65, converges.��

n�1

1�n� n

limn→�

an

bn

� limn→�

1���n� n�

1��� n� � limn→�

1�n

� 0

� bn � � 1

� n,� an � � 1

�n� n,

68. (a) Let and diverges.

By Exercise 66, diverges.��

n�1 ln n

n

limn→�

an

bn

� limn→�

�ln n��n

1�n� lim

n→� ln n � �

� bn � � 1n

,� an � � ln n

n, (b) Let and diverges.

By Exercise 66, diverges.� 1

ln n

limn→�

an

bn

� limn→�

n

ln n� �

� bn � � 1n

,� an � � 1

ln n,

69. Since the terms of are positive for

sufficiently large Since

and

converges, so does � sin�an�.

� anlimn→�

sin�an�

an

� 1

n.

� sin�an�limn→�

an � 0, 70.

Since converges, and

converges. � 1

1 � 2 � . . . � n

limn→�

2��n�n � 1��

1��n2� � limn→�

2n2

n�n � 1� � 2,

� 1�n2

� ��

n�1

2n�n � 1�

��

n�1

11 � 2 � . . . � n

� ��

n�1

1�n�n � 1���2

71. The series diverges. For

Since diverges, so does � 1

n�n�1��n.� 12n

1

n�n�1��n >12n

1

n1�n >12

n1�n < 2

n < 2n

n > 1, 72. Consider two cases:

If then and

If then and

combining,

Since converges, so does by

the Comparison Test.

��

n�1an

n��n�1���

n�1�2an �

12n

ann��n�1� ≤ 2an �

12n.

ann��n�1� ≤ � 1

2n�1n��n�1��

12n,an ≤

12n�1,

ann��n�1� �

an

an1��n�1� ≤ 2an.

an1��n�1� ≥ � 1

2n�11��n�1�

�12

,an ≥1

2n�1,

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Section 9.5 Alternating Series

Section 9.5 Alternating Series 277

1.

Matches (d).

S3 � 8.1667

S2 � 7.5

S1 � 6

��

n�1 6n2 2.

Matches (f ).

S3 � 5.1667

S2 � 4.4

S1 � 6

��

n�1

��1�n�16n2 3.

Matches (a).

S3 � 5.0

S2 � 4.5

S1 � 3

��

n�1 3n!

4.

Matches (b).

S3 � 2.0

S2 � 1.5

S1 � 3

��

n�1 ��1�n�13

n!5.

Matches (e).

S2 � 6.25

S1 � 5

��

n�1 10n2n 6.

Matches (c).

S2 � 3.75

S1 � 5

��

n�1 ��1�n10

n2n

7.

(a)

(b)

(c) The points alternate sides of the horizontal line 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 of the series.

0.60 11

1.1

��

n�1 ��1�n�1

2n � 1�

4� 0.7854

n 1 2 3 4 5 6 7 8 9 10

1 0.6667 0.8667 0.7238 0.8349 0.7440 0.8209 0.7543 0.8131 0.7605Sn

8.

(a)

(b)

(c) The points alternate sides of the horizontal line 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.

0 110

2

��

n�1 ��1�n�1

�n � 1�! �1e

� 0.3679

n 1 2 3 4 5 6 7 8 9 10

1 0 0.5 0.3333 0.375 0.3667 0.3681 0.3679 0.3679 0.3679Sn

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n 1 2 3 4 5 6 7 8 9 10

1 0.75 0.8611 0.7986 0.8386 0.8108 0.8312 0.8156 0.8280 0.8180Sn

278 Chapter 9 Infinite Series

9.

(a)

(b)

(c) The points alternate sides of the horizontal line 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.

0 110.6

1.1

��

n�1 ��1�n�1

n2 ��2

12� 0.8225

10.

(a)

(b)

(c) The points alternate sides of the horizontal line 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.

0 110

2

��

n�1

��1�n�1

�2n � 1�! � sin�1� � 0.8415

n 1 2 3 4 5 6 7 8 9 10

1 0.8333 0.8417 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415Sn

11.

Converges by Theorem 9.14

limn→�

1n

� 0

an�1 �1

n � 1 <

1n

� an

��

n�1 ��1�n�1

n12.

Diverges by the nth-Term Test

limn→�

n

2n � 1�

12

��

n�1 ��1�n�1 n

2n � 1

13.

Converges by Theorem 9.14

limn→�

1

2n � 1� 0

an�1 �1

2�n � 1� � 1 <

12n � 1

� an

��

n�1 ��1�n�1

2n � 114.

Converges by Theorem 9.14

limn→�

1

ln�n � 1� � 0

an�1 �1

ln�n � 2� <1

ln�n � 1� � an

��

n�1

��1�n

ln�n � 1�

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Section 9.5 Alternating Series 279

15.

Thus,

Diverges by the nth-Term Test

limn→�

an � 0.limn→�

n2

n2 � 1� 1.

��

n�1 ��1�n n2

n2 � 116.

Converges by Theorem 9.14

limn→�

n

n2 � 1� 0

an�1 �n � 1

�n � 1�2 � 1 <

nn2 � 1

� an

��

n�1 ��1�n�1n

n2 � 1

17.

Converges by Theorem 9.14

limn→�

1�n

� 0

an�1 �1

�n � 1 <

1�n

� an

��

n�1 ��1�n

�n18.

Diverges by nth-Term Test

limn→�

n2

n2 � 5� 1

��

n�1

��1�n�1n2

n2 � 5

19.

Diverges by the nth-Term Test

� limn→�

�n � 1� � �limn→�

n � 1

ln�n � 1� � limn→�

1

1��n � 1�

��

n�1 ��1�n�1�n � 1�

ln�n � 1� 20.

Converges by Theorem 9.14

limn→�

ln�n � 1�

n � 1� lim

n→� 1��n � 1�

1� 0

an�1 �ln��n � 1� � 1�

�n � 1� � 1 <

ln�n � 1�n � 1

for n ≥ 2

��

n�1 ��1�n�1 ln�n � 1�

n � 1

21.

Diverges by the nth-Term Test

��

n�1 sin�2n � 1��

2 � ��

n�1 ��1�n�1 22.

Converges; (See Exercise 11.)

��

n�1 1n

sin�2n � 1��2 � �

n�1 ��1�n�1

n

23.

Diverges by the nth-Term Test

��

n�1 cos n� � �

n�1 ��1�n 24.

Converges; (See Exercise 11.)

��

n�1 1n

cos n� � ��

n�1 ��1�n

n25.

Converges by Theorem 9.14

limn→�

1n!

� 0

an�1 �1

�n � 1�! < 1n!

� an

��

n�0 ��1�n

n!

26.

Converges by Theorem 9.14

limn→�

1

�2n � 1�! � 0

an�1 �1

�2n � 3�! <1

�2n � 1�! � an

��

n�0

��1�n

�2n � 1�! 27.

Converges by Theorem 9.14

limn→�

�n

n � 2� 0

<�n

n � 2 for n ≥ 2

an�1 ��n � 1

�n � 1� � 2

��

n�1 ��1�n�1�n

n � 228.

Diverges by the nth-Term Test

limn→�

n1�2

n1�3 � limn→�

n1�6 � �

��

n�1 ��1�n�1�n

3�n

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280 Chapter 9 Infinite Series

29.

Converges by Theorem 9.14

� limn→�

233

�45

�57

. . . n

2n � 3 �1

2n � 1� 0

� limn→�

1 � 2 � 3 . . . n

1 � 3 � 5 . . . �2n � 1�

limn→�

an � limn→�

n!

1 � 3 � 5 . . . �2n � 1�

an�1 ��n � 1�!

1 � 3 � 5 . . . �2n � 1��2n � 1� �n!

1 � 3 � 5 . . . �2n � 1� �n � 12n � 1

� an� n � 12n � 1� < an

��

n�1

��1�n�1n!1 � 3 � 5 . . . �2n � 1�

30.

Converges by Theorem 9.14

limn→�

an � limn→�

354

�77

�910

. . . 2n � 13n � 5

13n � 2

� 0

an�1 �1 � 3 � 5 . . . �2n � 1��2n � 1�1 � 4 � 7 . . . �3n � 2��3n � 1� � an�2n � 1

3n � 1� < an

��

n�1��1�n�1

1 � 3 � 5 . . . �2n � 1�1 � 4 � 7 . . . �3n � 2�

31.

Let Then

Thus, is decreasing. Therefore, and

The series converges by Theorem 9.14.

limn→�

2en

e2n � 1� lim

n→� 2en

2e2n � limn→�

1en � 0.

an�1 < an ,f �x�

f��x� ��2ex �e2x � 1�

�e2x � 1�2 < 0.

f �x� �2ex

e2x � 1.

��

n�1 ��1�n�1�2�

en � e�n � ��

n�1 ��1�n�1�2en�

e2n � 132.

Let Then

for

Thus, is decreasing for which implies

The series converges by Theorem 9.14.

limn→�

2en

e2n � 1� lim

n→� 2en

2e2n � limn→�

1en � 0

an�1 < an.x > 0f �x�

x > 0.f��x� �2e2x �1 � e2x�

�e2x � 1�2 < 0

f �x� �2ex

e2x � 1.

��

n�1 2��1�n�1

en � e�n � ��

n�1 ��1�n�1�2en�

e2n � 1

33.

2.3713 ≤ S ≤ 2.4937

R6 � S � S6 ≤ a7 �3

49� 0.0612

2.4325 � 0.0612 ≤ S ≤ 2.4325 � 0.0612

S6 � �6

n�1

3��1�n�1

n2� 2.4325 34.

0.7831 ≤ S ≤ 4.6303

R6 � S � S6 ≤ a7 �4

ln 8� 1.9236

S6 � �6

n�1

4��1�n�1

ln�n � 1� � 2.7067

35.

0.7305 ≤ S ≤ 0.7361

R6 � S � S6 ≤ a7 �26!

� 0.002778

0.7333 � 0.002778 ≤ S ≤ 0.7333 � 0.002778

S6 � �5

n�0

2��1�n

n!� 0.7333 36.

0.1328 ≤ S ≤ 0.2422

R6 � S � S6 ≤ a7 �727 � 0.05469

S6 � �6

n�1

��1�n�1n2n � 0.1875

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Section 9.5 Alternating Series 281

37.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(7 terms. Note that the sum begins with )n � 0.

� 0.368.

�6

n�0 ��1�n

n!� 1 � 1 �

12

�16

�1

24�

1120

�1

720

N � 6.

RN ≤ aN�1 �1

�N � 1�! < 0.001.

��

n�0 ��1�n

n!38.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(5 terms. Note that the sum begins with )n � 0.

�4

n�0 ��1�n

2n n!� 1 �

12

�18

�1

48�

1348

� 0.607.

N � 4.

Rn ≤ aN�1 �1

2N�1�N � 1�! < 0.001.

��

n�0 ��1�n

2n n!

39.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(3 terms. Note that the sum begins with )n � 0.

�2

n�0

��1�n

�2n � 1�! � 1 �16

�1

120� 0.842.

N � 2.

RN ≤ aN�1 �1

�2�N � 1� � 1�! < 0.001.

��

n�0

��1�n

�2n � 1�! 40.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(4 terms. Note that the sum begins with )n � 0.

�3

n�0 ��1�n

�2n�! � 1 �12

�1

24�

1720

� 0.540.

N � 3.

RN ≤ aN�1 �1

�2N � 2�! < 0.001.

��

n�0 ��1�n

�2n�!

41.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(1000 terms)

� 0.693.

�1000

n�1 ��1�n�1

n� 1 �

12

�13

�14

� . . . �1

1000

N � 1000.

RN ≤ aN�1 �1

N � 1 < 0.001.

��

n�1 ��1�n�1

n42.

(a) By Theorem 9.15,

This inequality is valid when

(b) We may approximate the series by

(3 terms)

�3

n�1 ��1�n�1

4n n�

14

�1

32�

1192

� 0.224.

N � 3.

RN ≤ aN�1 �1

4N�1�N � 1� < 0.001.

��

n�1 ��1�n�1

4n n

43.

By Theorem 9.15,

Use 10 terms.

⇒ �N � 1�3 > 1000 ⇒ N � 1 > 10.

RN ≤ aN�1 �1

�N � 1�3 < 0.001

��

n�1 ��1�n�1

n3 44.

By Theorem 9.15,

Use 31 terms.

⇒ �N � 1�2 > 1000 ⇒ N � 31.

RN ≤ aN�1 �1

�N � 1�2 < 0.001

��

n�1 ��1�n�1

n2

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282 Chapter 9 Infinite Series

45.

By Theorem 9.15,

This inequality is valid when Use 7 terms.N � 7.

RN ≤ aN�1 �1

2�N � 1�3 � 1 < 0.001.

��

n�1 ��1�n�1

2n3 � 146.

By Theorem 9.15,

This inequality is valid when N � 5.

RN ≤ aN�1 �1

�N � 1�4 < 0.001.

��

n�1 ��1�n�1

n4

47.

converges by comparison to the p-series

Therefore, the given series converges absolutely.

��

n�1 1n2.

��

n�1

1�n � 1�2

��

n�1 ��1�n�1

�n � 1�2 48.

The given series converges by the Alternating Series Test,but does not converge absolutely since the series

diverges by the Integral Test. Therefore, the seriesconverges conditionally.

��

n�1

1n � 1

��

n�1 ��1�n�1

n � 1

49.

The given series converges by the Alternating Series Test,but does not converge absolutely since

is a divergent p-series. Therefore, the series convergesconditionally.

��

n�1

1�n

��

n�1 ��1�n�1

�n50.

which is a convergent p-series.

Therefore, the given series converges absolutely.

��

n�1

1

n�n� �

n�1

1n3�2

��

n�1 ��1�n�1

n�n

51.

Therefore, the series diverges by the nth-Term Test.

limn→�

n2

�n � 1�2 � 1

��

n�1 ��1�n�1 n2

�n � 1�2 52.

Therefore, the series diverges by the nth-Term Test.

limn→�

2n � 3n � 10

� 2

��

n�1 ��1�n�1�2n � 3�

n � 10

53.

The given series converges by the Alternating Series Testbut does not converge absolutely since the series

diverges by comparison to the harmonic series

Therefore, the series converges conditionally.��

n�1 1n

.

��

n �2

1ln n

��

n�2 ��1�n

ln n54.

converges by a comparison to

the convergent geometric series Therefore, the

given series converges absolutely.

��

n�0 �1

e�n

.

��

n�0

1e n2

��

n�0 ��1�n

en2

55.

converges by a limit comparison to

the convergent p-series Therefore, the given series

converges absolutely.

��

n�2 1n2.

��

n�2

nn3 � 1

��

n�2 ��1�n nn3 � 1

56.

is a convergent p-series.

Therefore, the given series converges absolutely.

��

n�1

1n1.5

��

n�1 ��1�n�1

n1.5

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Section 9.5 Alternating Series 283

57.

is convergent by comparison to the convergent geometricseries

since

Therefore, the given series converges absolutely.

1�2n � 1�! <

12n for n > 0.

��

n�0 �1

2�n

��

n�0

1�2n � 1�!

��

n�0

��1�n

�2n � 1�! 58.

The given series converges by the Alternating Series Test,but

diverges by a limit comparison to the divergent p-series

Therefore, the given series converges conditionally.

��

n�1

1�n

.

��

n�0

1�n � 4

��

n�0

��1�n

�n � 4

59.

The given series converges by the Alternating Series Test,but

diverges by a limit comparison to the divergent harmonicseries,

therefore the series

converges conditionally.

limn→�

cos n� ��n � 1�1�n

� 1,

��

n�1 1n

.

��

n�0 cos n�

n � 1� �

n�0

1n � 1

��

n�0 cos n�

n � 1� �

n�0 ��1�n

n � 160.

Therefore, the series diverges by the nth-Term Test.

limn→�

arctan n ��

2� 0

��

n�1 ��1�n�1 arctan n

61.

is a convergent p-series.

Therefore, the given series converges absolutely.

��

n�1 1n2

��

n�1 cos n�

n2 � ��

n�1 ��1�n

n2 62.

The given series converges by the Alternating Series Test,but

is a divergent p-series. Therefore, the series convergesconditionally.

��

n�1 sin��2n � 1���2�

n � ��

n�1 1n

��

n�1 sin��2n � 1���2�

n� �

n�1 ��1�n�1

n

63. An alternating series is a series whose terms alternate insign. See Theorem 9.14.

64. (Theorem 9.15) S � Sn � Rn ≤ an�1

65. is absolutely convergent if converges. is conditionally convergent if diverges, but converges.

� an� an � an� an � an 66. (b). The partial sums alternate above and below the

horizontal line representing the sum.

67. False. Let

Then converges and converges.

But, diverges.� an � � 1n

� ��an�� an

an ���1�n

n. 68. True. If converged, then so would by

Theorem 9.16.� an� an

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284 Chapter 9 Infinite Series

69. True.

Since the next term is negative, is anoverestimate of the sum.

S100�1

101

S100 � �1 �12

�13

� . . . �1

10070. False. Let

Then both converge by the Alternating Series Test. But,

which diverges.�anbn � � 1n

,

�an � �bn � � ��1�n

�n.

71.

If then diverges.

If then diverges.

If then and

Therefore, the series converges for p > 0.

an�1 �1

�n � 1� p <1n p � an.

limn→�

1n p � 0p > 0,

��

n�1��1�nn�pp < 0,

��

n�1��1�np � 0,

��

n�1��1�n

1n p 72.

Assume that so that aredefined for all For all

Therefore, the series converges for all p.

an�1 �1

n � 1 � p<

1n � p

< an.

limn→�

an � limn→�

1

n � p� 0

p,n.an � 1��n � p�n � p � 0

��

n�1��1�n 1

n � p

73. Since

converges we have Thus, there must exist an such that for all

and it follows that for all Hence, by the Comparison Test,

converges. Let to see that the converse is false.an � 1�n

��

n�1 an

2

n > N.an2 ≤ �an�

n > N�aN� < 1N > 0limn→�

�an� � 0.

��

n�1�an�

74. converges, but diverges.��

n�1 1n�

n�1 ��1�n�1

n75. converges, hence so does �

n�1 1n4.�

n�1 1n2

76. (a)

converges absolutely (by comparison) for since

and

is a convergent geometric series for

(b) When we have the convergent alternating series

When we have the divergent harmonic series Therefore,

converges conditionally for x � �1.��

n�1 xn

n

1�n.x � 1,

��

n�1 ��1�n

n.

x � �1,

�1 < x < 1.

� xn�xn

n � < �xn�

�1 < x < 1,

��

n�1 xn

n77. (a) No, the series does not satisfy for all For

example,

(b) Yes, the series converges.

As

S2n → 11 � �1�2� �

11 � �1�3� � 2 �

32

�12

.

n →�,

� �1 �12

� . . . �12n� � �1 �

13

� . . . �13n�

� �12

� . . . �12n� � �1

3� . . . �

13n�

S2n �12

�13

� . . . �12n �

13n

19 < 1

8.n.an�1 ≤ an

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Section 9.5 Alternating Series 285

78. (a) No, the series does not satisfy

and

18

<1�3

.

��

n�1��1�n�1an � 1 �

18

�1�3

�1

64� . . .

an�1 ≤ an: (b) No, the series diverges because diverges.� 1�n

88. Diverges by comparison toDivergent Harmonic Series:

for n ≥ 3ln n

n>

1n

89. The first term of the series is zero,not one. You cannot regroup seriesterms arbitrarily.

90. Rearranging the terms of an alternating series can change the sum.

79.

convergent p-series

��

n�1

10n3�2 � 10 �

n�1

1n3�2, 80.

converges by limit comparison toconvergent p-series

� 1n2.

��

n�1

3n2 � 5

81. Diverges by nth-Term Test

limn→�

an � �

82. Converges by limit comparison to

convergent geometric series � 12n.

83. Convergent geometric series

�r �78

< 1�84. Diverges by nth-Term Test

limn→�

an �32

85. Convergent geometric seriesor Integral Test�r � 1��e �

86. Converges (conditionally) byAlternating Series Test

87. Converges (absolutely) byAlternating Series Test

91.

(i)

Adding:

(ii) (In fact, )

since

Thus, S � s.

S � limn→�

S3n � s4n �12

s2n � s �12

s �32

s

s >12

.s � 0

s � ln 2.limn→�

sn � s

s4n �12

s2n � 1 �13

�12

�15

�17

�14

� . . . �1

4n � 3�

14n � 1

�1

2n� S3n

12

s2n �12

�14

�16

�18

�1

10� . . . �

14n � 2

�1

4n

s4n � 1 �12

�13

�14

�15

�16

� . . . �1

4n � 1�

14n

S � 1 �13

�12

�15

�17

�14

�19

�1

11�

16

� . . .

s � 1 �12

�13

�14

� . . .

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Section 9.6 The Ratio and Root Tests

286 Chapter 9 Infinite Series

1.

� �n � 1��n��n � 1�

�n � 1�!�n � 2�! �

�n � 1��n��n � 1��n � 2�!�n � 2�!

2.

�1

�2k��2k � 1�

�2k � 2�!

�2k�! ��2k � 2�!

�2k��2k � 1��2k � 2�!

3. Use the Principle of Mathematical Induction. When the formula is valid since Assume that

and show that

To do this, note that:

The formula is valid for all n ≥ 1.

��2n � 2�!

2n�1�n � 1�!

��2n�!�2n � 1��2n � 2�

2n�1n!�n � 1�

��2n�!�2n � 1�

2n n!�

�2n � 2�2�n � 1�

��2n�!2n n!

� �2n � 1� �Induction hypothesis�

1 � 3 � 5 . . . �2n � 1��2n � 1� � �1 � 3 � 5 . . . �2n � 1���2n � 1�

1 � 3 � 5 . . . �2n � 1��2n � 1� ��2n � 2�!

2n�1�n � 1�!.

1 � 3 � 5 . . . �2n � 1� ��2n�!2n n!

1 ��2�1��!21 � 1!

.k � 1,

4. Use the Principle of Mathematical Induction. When the formula is valid since Assume that

and show that

To do this, note that:

The formula is valid for all n ≥ 3.

�2n�1�n � 1�!�2n � 1��2n � 1�

�2n � 2�!

�2n �2��n � 1�n!�2n � 1��2n � 1�

�2n�!�2n � 1��2n � 2�

�2n n!�2n � 1�

�2n�! ��2n � 1��2n � 2��2n � 1��2n � 2�

�2n n!�2n � 3��2n � 1�

�2n�! �1

�2n � 3�

1

1 � 3 � 5 . . . �2n � 5��2n � 3� �1

1 � 3 � 5 . . . �2n � 5� �1

�2n � 3�

11 � 3 � 5 . . . �2n � 5��2n � 3� �

2n�1�n � 1�!�2n � 1��2n � 1��2n � 2�! .

11 � 3 � 5 . . . �2n � 5� �

2n n!�2n � 3��2n � 1��2n�!

11

�233!�3��5�

6!� 1.k � 3,

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Section 9.6 The Ratio and Root Tests 287

5.

Matches (d).

S1 �34

, S2 � 1.875

��

n�1 n�3

4�n

� 1�34� � 2� 9

16� � . . . 6.

Matches (c).

S1 �34

, S2 � 1.03

��

n�1 �3

4�n

� 1n!� �

34

�9

16 �1

2� � . . . 7.

Matches (f).

S1 � 9

��

n�1 ��3�n�1

n!� 9 �

33

2� . . .

8.

Matches (b).

S1 � 2

��

n�1 ��1�n�1 4

(2n�! �42

�4

24� . . . 9.

Matches (a).

S1 � 2, S2 � 3.31

��

n�1� 4n

5n � 3�n

�4

2� �8

7�2

� . . . 10.

Matches (e).

S1 � 4

��

n�0

4e�n � 4 �4

e� . . .

11. (a) Ratio Test: Converges

(b)

(c)

(d) The sum is approximately 19.26.

(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sumof the series.

00

12

20

� limn→�

�n � 1n �

2

58

�58

< 1.limn→�

an�1

an � limn→�

�n � 1�2�58�n�1

n2�58�n

n 5 10 15 20 25

9.2104 16.7598 18.8016 19.1878 19.2491Sn

12. (a) Ratio Test: Converges

(b)

(c)

0 110

10

� limn→�

�n2 � 2n � 2n2 � 1 �� 1

n � 1� � 0 < 1.limn→�

an�1

an � limn→�

�n � 1�2 � 1�n � 1�!n2 � 1

n!

n 5 10 15 20 25

7.0917 7.1548 7.1548 7.1548 7.1548Sn

(d) The sum is approximately 7.15485

(e) The more rapidly the terms of the series approach 0, the more rapidly thesequence of the partial sums approaches the sum of the series.

13.

Therefore, by the Ratio Test, the series diverges.

� limn→�

n � 1

3� �

limn→�

an�1

an � limn→�

�n � 1�!3n�1 �

3n

n!��

n�0 n!3n 14.

Therefore, by the Ratio Test, the series converges.

� limn→�

3

n � 1� 0

limn→�

an�1

an � limn→�

3n�1

�n � 1�! �n!3n

��

n�0 3n

n!

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288 Chapter 9 Infinite Series

15.

Therefore, by the Ratio Test, the series converges.

� limn→�

3�n � 1�4n �

34

limn→�

an�1

an � limn→�

�n � 1��34�n�1

n�34�n ��

n�1n�3

4�n

16.

Therefore, by the Ratio Test, the series diverges.

� limn→�

3�n � 1�

2n�

32

limn→�

an�1

an � limn→�

�n � 1�3n�1

2n�1 �2n

n3n��

n�1 n�3

2�n

17.

Therefore, by the Ratio Test, the series converges.

� limn→�

n � 1

2n�

12

limn→�

an�1

an � limn→�

n � 12n�1 �

2n

n ��

n�1 n2n

19.

Therefore, by the Ratio Test, the series diverges.

� limn→�

2n2

�n � 1�2 � 2

limn→�

an�1

an � limn→�

2n�1

�n � 1�2 �n2

2n��

n�1 2n

n2

18.

Therefore, by the Ratio Test, the series converges.

� limn→�

�n � 1�3

2n3 �12

limn→�

an � 1

an � limn→�

�n � 1�32n�1

n32n ��

n�1

n3

2n

20.

Therefore, by Theorem 9.14, the series converges.

Note: The Ratio Test is inconclusive since

The series converges conditionally.

limn→� an�1

an � 1.

limn→�

n � 2

n�n � 1� � 0

an�1 �n � 3

�n � 1��n � 2� ≤ n � 2

n�n � 1� � an

��

n�1 ��1�n�1�n � 2�

n�n � 1�

21.

Therefore, by the Ratio Test, the series converges.

� limn→�

2

n � 1� 0

limn→�

an�1

an � limn→�

2n�1

�n � 1�! �n!2n

��

n�0 ��1�n 2n

n!

23.

Therefore, by the Ratio Test, the series diverges.

� limn→�

n3

� �

limn→�

an�1

an � limn→�

�n � 1�!�n � 1�3n�1 �

n3n

n! ��

n�1

n!n3n

22.

Therefore, by the Ratio Test, the series diverges.

� limn→�

3n2

2�n2 � 2n � 1� �32

> 1

limn→�

an�1

an � limn→�

�32�n�1

n2 � 2n � 1�

n2

�32�n��

n�1 ��1�n�1�32�n

n2

24.

Therefore, by the Ratio Test, the series diverges.

� limn→�

�2n � 2��2n � 1�n5

�n � 1�5 � �

limn→�

an�1

an � limn→�

�2n � 2�!�n � 1�5 �

n5

�2n�!��

n�1 �2n�!

n5

Page 58: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.6 The Ratio and Root Tests 289

25.

Therefore, by the Ratio Test, the series converges.

� limn→�

4

n � 1� 0

limn→�

an�1

an � limn→�

4n�1

�n � 1�! �n!4n

��

n�0 4n

n!26.

Therefore, by the Ratio Test, the series diverges.

� limn→�

�n � 1n �

n

� e > 1

� limn→�

�n � 1��n � 1�n n!

�n � 1�n!nn

limn→�

an�1

an � limn→�

�n � 1�n�1

�n � 1�! �n!nn

��

n�1 nn

n!

27.

To find let Then,

by L’Hôpital’s Rule

Therefore, by the Ratio Test, the series converges.

y � e�1 �1e.

ln y � limn→�

��1��n � 1�� � ��1��n � 2��

��1n2� � �1

ln y � limn→�

n ln�n � 1n � 2� � lim

n→� ln��n � 1��n � 2��

1n�

00

y � limn→�

�n � 1n � 2�

n

.limn→�

�n � 1n � 2�

n

,

limn→�

an�1

an � limn→�

3n�1

�n � 2�n�1 ��n � 1�n

3n � limn→�

3�n � 1�n

�n � 2�n�1 � limn→�

3

n � 2 �n � 1

n � 2�n

� �0��1e� � 0

��

n�0

3n

�n � 1�n

28.

Therefore, by the Ratio Test, the series converges.

� limn→�

�n � 1�2

�3n � 3��3n � 2��3n � 1� � 0

limn→�

an�1

an � limn→�

��n � 1�!�2

�3n � 3�! ��3n�!�n!�2

��

n�0 �n!�2

�3n�! 29.

Therefore, by the Ratio Test, the series diverges.

� limn→�

4�1 � 13n�3 � 13n �

43

� limn→�

4�3n � 1�3n�1 � 1

limn→�an�1

an � limn→� 4n�1

3n�1 � 1�

3n � 14n

��

n�0

4n

3n � 1

30.

Therefore, by the Ratio Test, the series converges.

limn→�

an�1

an � limn→�

24n�4

�2n � 3�! ��2n � 1�!

24n � limn→�

24

�2n � 3��2n � 2� � 0

��

n�0 ��1�n 24n

�2n � 1�!

31.

Therefore, by the Ratio Test, the series converges.

Note: The first few terms of this series are �1 �1

1 � 3�

2!1 � 3 � 5

�3!

1 � 3 � 5 � 7� . . ..

limn→�

an�1

an � limn→�

�n � 1�!1 � 3 � 5 . . . �2n � 1��2n � 3� �

1 � 3 � 5 . . . �2n � 1�n! � lim

n→�

n � 12n � 3

�12

��

n�0

��1�n�1n!1 � 3 � 5 . . . �2n � 1�

Page 59: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

290 Chapter 9 Infinite Series

32.

Therefore, by the Ratio Test, the series converges.

Note: The first few terms of this series are �22

�2 � 42 � 5

�2 � 4 � 62 � 5 � 8

� . . ..

limn→�

an�1

an � limn→�

2 � 4 . . . 2n�2n � 2�2 � 5 . . . �3n � 1��3n � 2� �

2 � 5 . . . �3n � 1�2 � 4 . . . 2n � lim

n→� 2n � 23n � 2

�23

��

n�1 ��1�n 2 � 4 � 6 . . . 2n2 � 5 � 8 . . . �3n � 1�

33.

Ratio Test is inconclusive.

� limn→� �

nn � 1�

32

� 1

limn→�

an�1

an � limn→� 1

�n � 1�32 �n32

1 ��

n�1

1n32 34.

Ratio Test is inconclusive.

� limn→�

� nn � 1�

12

� 1

limn→�

an�1

an � limn→�

1�n � 1�12 �

n12

1 ��

n�1

1n12

35.

limn→�

an�1

an � limn→�

1�n � 1�4 �

n4

1 � limn→�

� nn � 1�

4

� 1

��

n�1 1n4 36.

limn→�

an�1

an � limn→�

1�n � 1� p �

n p

1 � limn→�

� nn � 1�

p

� 1

��

n�1 1np

37.

Therefore, by the Root Test, the series converges.

� limn→�

n

2n � 1�

12

limn→�

n�an � limn→�

n�� n2n � 1�

n

��

n�1 � n

2n � 1�n

38.

Therefore, by the Root Test, the series diverges.

� limn→�

2n

n � 1� 2

limn→�

n�an � limn→�

n�� 2nn � 1�

n

��

n�1 � 2n

n � 1�n

39.

Since the series diverges.2 > 1,

� limn→�

�2n � 1n � 1 � � 2

limn→�

n�an � limn→�

n��2n � 1n � 1 �

n

��

n�2 �2n � 1

n � 1 �n

40.

Since the series diverges.2 > 1,

� limn→�

4n � 32n � 1

� 2

limn→�

n�an � limn→�

n��4n � 32n � 1�

n

��

n�1 �4n � 3

2n � 1�n

41.

Therefore, by the Root Test, the series converges.

� limn→�

1

ln n � 0

limn→�

n�an � limn→�

n���1�n

�ln n�n��

n�2 ��1�n

�ln n�n 42.

Therefore, by the Root Test, the series diverges.

� limn→�

� 3n2n � 1�

3

� �32�

3

�278

limn→�

n�an � limn→�

n�� �3n2n � 1�3n

��

n�1� �3n

2n � 1�3n

Page 60: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.6 The Ratio and Root Tests 291

43.

To find let Then

Thus, so and Therefore, by the Root Test, the series diverges.limn→�

�2 n�n � 1� � 2�1� � 1 � 3.y � e0 � 1ln y � 0,

ln y � limn→�

�ln n�n � � limn→�

1n

ln n � limn→�

ln n

n� lim

n→� 1n1

� 0.

y � limn→�

n�n .limn→�

n�n,

limn→�

n�an � limn→�

n��2 n�n � 1�n � limn→�

�2 n�n � 1�

��

n�1 �2 n�n � 1�n

44.

Therefore, by the Root Test, the series converges.

limn→�

n�an � limn→�

n� 1en �

1e

��

n�0 e�n 45.

Since the series converges.

Note: You can use L’Hôpital’s Rule to show

Let

Hence, ln y → 0 ⇒ y � n1n → 1.

limn→�

ln nn

� limn→�

1n1

� 0.

y � n1n ⇒ ln y �ln n

n

limn→�

n1n � 1.

14 < 1,

� limn→�

n1n

4�

14

limn→�

n�an � limn→�

n� n4n

��

n�1 n4n

46.

The series diverges.

� limn→�

� n500� � �

limn→�

n�an � lim

n→� n�� n

500�n

��

n�1 � n

500�n

47.

Hence, the series converges.

� limn→�

�1n

�1n2� � 0 � 0 � 0 < 1

limn→�

n�an � limn→�

n��1n

�1n2�

n

��

n�1 �1

n�

1n2�

n

48.

By the Root Test, the series converges.

� limn→�

ln n

n� 0 < 1

limn→�

n�an � limn→�

n��ln nn �

n

��

n�1 �ln n

n �n

49.

Since the series converges by the Root Test.0 < 1,

� limn→�

n1n

ln n� 0

limn→�

n�an � limn→�

n� n�ln n�n

��

n�2

n�ln n�n

50.

The series diverges.

� limn→�

n!n2 � �

limn→�

n�an � limn→�

n��n!��n2�n

��

n�1 �n!�n

�nn�2 � ��

n�1 �n!�n

�n2�n 51.

Therefore, by the Alternating Series Test, the seriesconverges (conditional convergence).

limn→�

5n

� 0

an�1 �5

n � 1 <

5n

� an

��

n�1 ��1�n�1 5

n

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292 Chapter 9 Infinite Series

52.

This is the divergent harmonic series.

��

n�1 5n

� 5 ��

n�1 1n

53.

This is convergent p-series.

��

n�1

3

n�n� 3 �

n�1

1n32

55.

This diverges by the nth-Term Test for Divergence.

limn→�

2n

n � 1� 2 � 0

��

n�1

2nn � 1

59.

Therefore, the series converges by a Limit ComparisonTest with the geometric series

��

n�0 �1

2�n

.

limn→�

�10n � 3�n2n

12n � limn→�

10n � 3

n� 10

��

n�1 10n � 3

n2n

57.

Since this is a divergent geometric series.r �32 > 1,

� ��

n�1 19��

32�

n

��

n�1 ��1�n 3n�2

2n � ��

n�1 ��1�n 3n 3�2

2n56.

This series diverges by limit comparison to the divergentharmonic series

��

n�1 1n

.

limn→�

n�2n2 � 1�

1n� lim

n→�

n2

2n2 � 1�

12

> 0

��

n�1

n2n2 � 1

58.

Therefore, the series converges by a Limit ComparisonTest with the p-series

��

n�1

1n32.

limn→�

103n32

1n32 �103

��

n�1

10

3�n3

60.

Therefore, the series diverges by the nth-Term Test forDivergence.

� limn→�

�ln 2�22n

8� � lim

n→�

2n

4n2 � 1� lim

n→� �ln 2�2n

8n

��

n�1

2n

4n2 � 1

62.

Therefore, by the Alternating Series Test, the seriesconverges.

limn→�

1

n ln�n� � 0

an�1 �1

�n � 1� ln�n � 1� ≤ 1

n ln�n� � an

��

n�2 ��1�n

n ln n

54.

Since this is convergent geometric series.�4 < 1,

��

n�1��

4�n

63.

Therefore, by the Ratio Test, the series converges.

� limn→�

7n

� 0limn→�

an�1

an � limn→�

�n � 1�7n�1

�n � 1�! �n!

n7n��

n�1 n7n

n!

61.

Therefore, the series

converges by comparison with the geometric series

��

n�0 �1

2�n

.

��

n�1 cos n

2n cos n

2n ≤ 12n

��

n�1 cos n

2n

Page 62: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.6 The Ratio and Root Tests 293

64.

Therefore, the series converges by comparison with the p-series

��

n�1

1n32.

ln�n�n2 ≤

1n32

��

n�1 ln�n�

n265.

Therefore, by the Ratio Test, the series converges.

(Absolutely)

limn→�

an�1

an � limn→�

3n

�n � 1�! �n!

3n�1 � limn→�

3

n � 1� 0

��

n�1 ��1�n 3n�1

n!

66.

Therefore, by the Ratio Test, the series diverges.

limn→�

an�1

an � limn→�

3n�1

�n � 1�2n�1 �n2n

3n � limn→�

3n

2�n � 1� �32

��

n�1 ��1�n 3n

n2n

68.

Therefore, by the Ratio Test, the series converge.

limn→�

an�1

an � limn→�

3 � 5 � 7 . . . �2n � 1��2n � 3�18n�1�2n � 1��2n � 1�n!

�18n �2n � 1�n!

3 � 5 � 7 . . . �2n � 1� � limn→�

�2n � 3��2n � 1�

18�2n � 1��2n � 1� �1

18

��

n�1 3 � 5 � 7 . . . �2n � 1�

18n �2n � 1�n!

67.

Therefore, by the Ratio Test, the series converges.

limn→�

an�1

an � limn→�

��3�n�1

3 � 5 � 7 . . . �2n � 1��2n � 3� �3 � 5 � 7 . . . �2n � 1�

��3�n � limn→�

3

2n � 3� 0

��

n�1

��3�n

3 � 5 � 7 . . . �2n � 1�

69. (a) and (c)

� 5 ��2��5�2

2!�

�3��5�3

3!�

�4��5�4

4!� . . .

��

n�1 n5n

n!� �

n�0 �n � 1�5n�1

�n � 1�!

71. (a) and (b) are the same.

70. (b) and (c)

� 1 � 2�34� � 3�3

4�2

� 4�34�

3

� . . .

��

n�0 �n � 1��3

4�n

� ��

n�1 n �3

4�n�1

72. (a) and (b) are the same.

��

n�1 ��1�n�1

n2n �12

�1

2 � 22 �1

3 � 23 � . . .

��

n�2

��1�n

�n � 1�2n�1 �12

�1

2 � 22 �1

3 � 23 � . . .

73. Replace n with

��

n�1 n4n � �

n�0 n � 14n�1

n � 1. 75. Since

use 9 terms.

�9

k�1 ��3�k

2k k!� �0.7769

310

210 10!� 1.59 � 10�5,

74. Replace n with

��

n�2

2n

�n � 2�! � ��

n�0 2n�2

n!

n � 2.

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294 Chapter 9 Infinite Series

76.

(See Exercise 3 and use 10 terms, .)k � 9

� 0.40967

� ��

k�0

��6�k k!�2k � 1�!

��

k�0

��3�k

1 � 3 � 5 . . . �2k � 1� � ��

k�0

��3�k 2k k!�2k�!�2k � 1�

77.

The series diverges by the Ratio Test.

� limn→�

4n � 13n � 2

�43

> 1

limn→�an�1

an � limn→��4n � 1��3n � 2�an

an

78.

The series converges by the Ratio Test.

� limn→�

2n � 15n � 4

�25

< 1

limn→�an�1

an � limn→��2n � 1��5n � 4�an

an 79.

The series converges by the Ratio Test.

� limn→�

sin n � 1�n

� 0 < 1

limn→�an�1

an � limn→��sin n � 1���n�an

an

80.

The series converges by the Ratio Test.

� limn→�

cos n � 1n

� 0 < 1

limn→�an�1

an � limn→��cos n � 1��n�an

an 81.

The Ratio Test is inconclusive.

But, so the series diverges.limn→�

an � 0,

� limn→�

�1 �1n� � 1

limn→�an�1

an � limn→��1 � �1��n��an

an

82. The series diverges because

In general, an�1 > an > 0.

a3 � �12�

13

� 0.7937

a2 � �14�

12

�12

a1 �14

limn→�

an � 0. 83.

The series converges by the Ratio Test.

� limn→�

n � 12n � 1

�12

< 1

limn→�an�1

an � limn→� 1 � 2 . . . n�n � 1�

1 � 3 . . . �2n � 1��2n � 1�

1 � 2 . . . n1 � 3 . . . �2n � 1�

84.

Let

Since so

Therefore, by the Root Test, the series converges.

limn→�

�n � 1

3�

13

.

ln y � 0, y � e0 � 1,

� limn→�

ln�n � 1�

n�

1n � 1

� 0.

� limn→�

1n

ln�n � 1�

ln y � limn→�

�ln n�n � 1 � y � lim

n→� n�n � 1

limn→�

n�an � limn→�

n�n � 13n � lim

n→�

n�n � 13

��

n�0 n � 1

3n 85.

Therefore, by the Root Test, the series converges.

limn→�

n�an � limn→�

n� 1�ln n�n � lim

n→�

1ln n

� 0

��

n�3

1�ln n�n

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Section 9.6 The Ratio and Root Tests 295

93. See Theorem 9.17, page 597.

86.

The series converges by the Ratio Test.

� limn→�

2n � 1

�2n��2n � 1� � 0 < 1

limn→�an�1

an � limn→� 1 � 3 � 5 . . . �2n � 1��2n � 1�

1 � 2 � 3 . . . �2n � 1��2n��2n � 1�

1 � 3 � 5 . . . �2n � 1�1 � 2 � 3 . . . �2n � 1� 87.

For the series to converge:

For the series diverges.

For the series diverges.

Answer: �3 < x < 3

x � �3,

x � 3,

x3 < 1 ⇒ �3 < x < 3.

� limn→�

x3 � x3 limn→�an�1

an � limn→�2�x3�n�1

2�x3�n

88.

For the series to converge,

For diverges.

For diverges.

Answer: �5 < x < 3

x � �5, ��

n�0��1�n,

x � 3, ��

n�0�1�n,

⇒ �5 < x < 3.

x � 14 < 1 ⇒ �4 < x � 1 < 4

� limn→�

x � 14 � x � 1

4 limn→�

n�an � limn→�

n�x � 14 n

89.

For the series to converge,

For converges.

For diverges.

Answer: �2 < x ≤ 0

x � �2, ��

n�1

��1�n��1�n

n� �

n�1 1n

,

x � 0, ��

n�1

��1�n

n,

⇒ �2 < x < 0.

x � 1 < 1 ⇒ �1 < x � 1 < 1

� limn→�

nn � 1

�x � 1� � x � 1

limn→�

an�1

an � limn→��x � 1�n�1�n � 1�

xnn

90.

For the series to converge,

For diverges.

For diverges.

Answer: 0 < x < 2

x � 0, ��

n�02��1�n,

x � 2, ��

n�02�1�n,

⇒ 0 < x < 2.

x � 1 < 1 ⇒ �1 < x � 1 < 1

� limn→�

x � 1 � x � 1

limn→�

an�1

an � limn→�2x � 1n�1

2x � 1n 91.

The series converges only at x � 0.

� limn→�

�n � 1�x2 � �

limn→�

an�1

an � limn→�

�n � 1�!x2n�1

n!x2n

92.

The series converges for all x.

� limn→�

x � 1n � 1

� 0 limn→�

an�1

an � limn→�

x � 1n�1

�n � 1�! x � 1n

n!

94. See Theorem 9.18.

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296 Chapter 9 Infinite Series

95. No. Let

The series diverges.��

n�1

1n � 10,000

an �1

n � 10,000. 96. One example is �

n�1��100 �

1n�.

97. The series converges absolutely. See Theorem 9.17.

98. Assume that

or that

Then there exists such that for all Therefore,

⇒ limn→�

an � 0 ⇒ � an diverges. an�1 > an, n > N

n > N.an�1an > 1N > 0

limn→�

an�1an � �.limn→�

an�1an � L > 1

99. First, let

and choose R such that There must

exist some such that for all

Thus, for and since the geometric

series

converges, we can apply the Comparison Test to conclude that

converges which in turn implies that converges.��

n�1 an

��

n�1 an

��

n�0 Rn

an < Rnn > N

n > N.n�an < RN > 0

0 ≤ r < R < 1.

limn→�

n�an � r < 1

Second, let

Then there must exist some such that for infinitely many Thus, for infinitely many we have which implies that which in turn implies that

diverges.��

n�1 an

limn→�

an � 0an > Rn > 1n > M,n > M.

n�an > RM > 0

limn→�

n�an � r > R > 1.

100. , p-series

Hence, the Root Test is inconclusive.

Note: because if then

and

Hence y → 1 as n →�.

pn

ln n → 0 as n →�.ln y �pn

ln n

y � npn,limn→�

npn � 1

limn→�

n�an � limn→�

n� 1np � lim

n→�

1npn � 1

��

n�1

1 np 101. Ratio Test:

inconclusive.

Root Test:

Furthermore, let

Thus, inconclusive.limn→�

1

n1n�ln n�pn � 1,

� limn→�

p

ln�n��1n� � 0 ⇒ limn→�

�ln n�pn � 1.

limn→�

ln y � limn→�

p ln �ln n�n

ln y �pn

ln �ln n�.

y � �ln n�pn ⇒ limn→�

n1n � 1.

limn→�

n�an � limn→�

n� 1n�ln n�p � lim

n→�

1n1n�ln n�pn

limn→�an�1

an � limn→�

n�ln n�p

�n � 1��ln�n � 1��p � 1,

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Section 9.6 The Ratio and Root Tests 297

102. positive integer

(a) diverges

(b) converges by the Ratio Test:

(c) converges by the Ratio Test:

(d) Use the Ratio Test:

The cases were solved above. For the limit is 0. Hence, the series converges for all integers x ≥ 2.x > 3,x � 1, 2, 3

limn→�

��n � 1�!�2

�x�n � 1��! �n!�2

�xn�! � limn→�

�n � 1�2 �xn�!

�xn � x�!

limn→�

��n � 1�!�2

�3n � 3�! �n!�2

�3n�! � limn→�

�n � 1�2

�3n � 3��3n � 2��3n � 1� � 0 < 1

x � 3: � �n!�2

�3n�!

limn→�

��n � 1�!�2

�2n � 2�! �n!�2

�2n�! � limn→�

�n � 1�2

�2n � 2��2n � 1� �14

< 1

x � 2: � �n!�2

�2n�!

x � 1: � �n!�2

n!� � n!,

x��

n�1 �n!�2

�xn�!,

103. For

Taking limits as � ��

n�1an ≤ �

n�1an ≤ �

n�1an ⇒ ��n�1

an ≤ ��

n�1an.k →�,

n � 1, 2, 3, . . . ,�an ≤ an ≤ an ⇒ � �k

n�1an ≤ �

k

n�1an ≤ �

k

n�1an.

104. The differentiation test states that if

is an infinite series with real terms and is a real function such that for all positive integers n and exists at then

converges absolutely if and diverges otherwise. Below are some examples.

� sin 1n

, f �x� � sin x��1 � cos 1n�, f �x� � 1 � cos x

� 1n

, f �x� � x� 1n3, f �x� � x3

Divergent SeriesConvergent Series

f �0� � f�0� � 0

��

n�1Un

x � 0,d 2 fdx2f �1n� � Unf �x�

��

n�1Un

105. Using the Ratio Test,

Hence, the series converges.

�197

�1e

< 1

� limn→�

� 1�1 � �1n��n �19

7 �

� limn→�

�n � nn�1

�n � 1�n �197 �

limn→�an�1

an � limn→�

� n!�n � 1�n �19

7 �n�n � 1�!nn�1 �19

7 �n�1

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298 Chapter 9 Infinite Series

106. We first prove Abel’s Summation Theorem:

If the partial sums of are bounded and if decreases to zero, then converges.

Let Let be a bound for

The series is absolutely convergent because and

converges to

Also, because bounded and Thus, converges.

Now let to finish the problem.bn �1n

��

n�1anbn � lim

n→� �

n

i�1aibibn → 0.�Sn�lim

n→�Snbn � 0

b1.��

i�1�bi � bi�1�

�Si�bi � bi�1�� ≤ M�bi � bi�1���

i�1Si�bi � bi�1�

� �n�1

i�1Si�bi � bi�1� � Snbn

� S1�b1 � b2� � S2�b2 � b3� � . . . � Sn�1�bn�1 � bn� � Snbn

� S1b1 � �S2 � S1�b2 � . . . � �Sn � Sn�1�bna1b1 � a2b2 � . . . � anbn

��Sk��.MSk � �k

i�1ai.

�anbn�bn��an

Section 9.7 Taylor Polynomials and Approximations

1.

Parabola

Matches (d)

y � �12 x2 � 1 2.

y-axis symmetry

Three relative extrema

Matches (c)

y �18 x 4 �

12 x2 � 1

3.

Linear

Matches (a)

y � e�1�2��x � 1� � 1 4.

Cubic

Matches (b)

y � e�1�2�13 �x � 1�3 � �x � 1� � 1

5.

is called the first degree Taylor polynomial for f at c.P1

−2 6

−2

10

P1

f

(1, 4)

P1�x� � �2x � 6

� 4 � ��2��x � 1�

P1�x� � f �1� � f��1��x � 1�

f��x� � �2x�3�2 f� �1� � �2

f �x� �4

x� 4x�1�2 f �1� � 4 6.

is called the first degree Taylor polynomial for f at c.P1

(8, 2)

−4

−4

14

8

P1�x� � �112

x �83

� 2 � ��1

12��x � 8�

P1�x� � f �8� � f��8��x � 8�

f��x� � �43

x�4�3 f��8� � �1

12

f �x� �43x

� 4x�1�3 f �8� � 2

Page 68: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.7 Taylor Polynomials and Approximations 299

7.

is called the first degree Taylor polynomial for f at c.P1

−1

5

P1

f

( , 2)�4

�2

P1�x� � 2 � 2�x ��

4�

P1�x� � f ��

4� � f���

4��x ��

4�

f��x� � sec x tan x f���

4� � 2

f �x� � sec x f ��

4� � 2 8.

is called the first degree Taylor polynomial for f at c.P1

�2

�2

−3

3

P1�x� � 2x � 1 ��

2

� 1 � 2�x ��

4�

P1 � f ��

4� � f���

4��x ��

4�

f��x� � sec2 x f���

4� � 2

f �x� � tan x f ��

4� � 1

9.

� 4 � 2�x � 1� �32

�x � 1�2

P2 � f �1� � f��1��x � 1� �f��1�

2�x � 1�2

f��x� � 3x�5�2 f��1� � 3

f��x� � �2x�3�2 f��1� � �2

−2 6

−2

10

P2

f

(1, 4)

f �x� �4x

� 4x�1�2 f �1� � 4

x 0 0.8 0.9 1.0 1.1 1.2 2

Error 4.4721 4.2164 4.0 3.8139 3.6515 2.8284

7.5 4.46 4.215 4.0 3.815 3.66 3.5P2�x�

f �x�

10.

P2�x� � 2 � 2�x ��

4� �322�x �

4�2

P2�x� � f ��

4� � f���

4��x ��

4� �f� ���4�

2 �x ��

4�2

f��x� � sec3 x � sec x tan2 x f���

4� � 32

f��x� � sec x tan x f���

4� � 2

−30

3

4 f �x� � sec x f ��

4� � 2

x 0.585 0.685 0.885 0.985 1.785

1.1995 1.2913 1.4142 1.5791 1.8088

15.5414 1.2160 1.2936 1.4142 1.5761 1.7810 4.9475P2�x�

�4.7043�1.8270f �x�

��4�2.15

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300 Chapter 9 Infinite Series

11.

(a)

3−3

−2

2

P6

P2

P4

f

P6�x� � 1 �12x2 �

124x 4 �

1720x6

P4�x� � 1 �12x2 �

124x 4

P2�x� � 1 �12x2

f �x� � cos x (b)

(c) In general, for all n.f �n��0� � Pn�n��0�

f �6��0� � �1 � P6�6��0�

f �6��x� � �cos x P�6��x� � �1

f �5��x� � �sin x P6�5��x� � �x

f �4��0� � 1 � P4�4��0�

f �4��x� � cos x P4�4��x� � 1

f����x� � sin x P4����x� � x

f ��0� � P2��0� � �1

f ��x� � �cos x P2��x� � �1

f��x� � �sin x P2��x� � �x

12.

(a)

P4�x� � x2 � x3 �12x 4

4!� x2 � x3 �

x4

2

P3�x� � x2 �6x3

3!� x2 � x3

P2�x� �2x2

2!� x2

f �4��x� � �x2 � 8x � 12�ex f �4��0� � 12

f���x� � �x2 � 6x � 6�ex f���0� � 6

f ��x� � �x2 � 4x � 2�ex f ��0� � 2

f��x� � �x2 � 2x�ex f��0� � 0

f �x� � x2ex, f �0� � 0

(b)

(c)

(d) f �n��0� � Pn�n��0�

f �4��0� � 12 � P4�4��0�

f���0� � 6 � P3���0�

f ��0� � 2 � P2��0�

−1

−3 3

f

P3

P2P4

2

13.

� 1 � x �x2

2�

x3

6

P3�x� � f �0� � f��0�x �f ��0�

2! x2 �

f����0�3!

x3

f����x� � �e�x f����0� � �1

f ��x� � e�x f ��0� � 1

f��x� � �e�x f��0� � �1

f �x� � e�x f �0� � 1 14.

�f �5��0�

5! x5 � 1 � x �

x2

2�

x3

6�

x 4

24�

x5

120

P5�x� � f �0� � f��0�x �f��0�2!

x2 �f���0�

3! x3 �

f �4��0�4!

x4

f �5��x� � �e�x f �5��0� � �1

f �4��x� � e�x f �4��0� � 1

f ���x� � �e�x f���0� � �1

f ��x� � e�x f ��0� � 1

f��x� � �e�x f��0� � �1

f �x� � e�x f �0� � 1

15.

� 1 � 2x � 2x2 �43

x3 �23

x 4

P4�x� � 1 � 2x �42!

x2 �83!

x3 �164!

x 4

f �4��x� � 162x f �4��0� � 16

f����x� � 8e2x f����0� � 8

f ��x� � 4e2x f ��0� � 4

f��x� � 2e2x f��0� � 2

f �x� � e2x f �0� � 1 16.

� 1 � 3x �92

x2 �92

x3 �278

x 4

P4�x� � 1 � 3x �92!

x2 �273!

x3 �814!

x 4

f �4��x� � 81e3x f �4��0� � 81

f���x� � 27e3x f���0� � 27

f ��x� � 9e3x f � �0� � 9

f��x� � 3e3x f��0� � 3

f �x� � e3x f �0� � 1

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Section 9.7 Taylor Polynomials and Approximations 301

17.

� x �16

x3 �1

120 x5

P5�x� � 0 � �1�x �02!

x2 ��13!

x3 �04!

x 4 �15!

x5

f �5��x� � cos x f �5��0� � 1

f �4��x� � sin x f �4��0� � 0

f����x� � �cos x f����0� � �1

f ��x� � �sin x f ��0� � 0

f��x� � cos x f��0� � 1

f �x� � sin x f �0� � 0

19.

� x � x2 �12

x3 �16

x 4

P4�x� � 0 � x �22!

x2 �33!

x3 �44!

x 4

f �4��x� � xex � 4ex f �4��0� � 4

f����x� � xex � 3ex f����0� � 3

f ��x� � xex � 2ex f ��0� � 2

f��x� � xex � ex f��0� � 1

f �x� � xex f �0� � 0

18.

� � x ��3

6 x3 P3�x� � 0 � �x �

02!

x2 ��� 3

3! x3

f���x� � ��3 cos � x f���0� � ��3

f ��x� � ��2 sin � x f ��0� � 0

f��x� � � cos � x f��0� � �

f �x� � sin � x f �0� � 0

20.

� x2 � x3 �12

x 4

P4�x� � 0 � 0x �22!

x2 ��63!

x3 �124!

x 4

f �4��x� � 12e�x � 8xe�x � x2e�x f �4��0� � 12

f���x� � �6e�x � 6xe�x � x2e�x f���0� � �6

f ��x� � 2e�x � 4xe�x � x2e�x f ��0� � 2

f��x� � 2xe�x � x2e�x f��0� � 0

f �x� � x2e�x f �0� � 0

21.

� 1 � x � x2 � x3 � x 4

P4�x� � 1 � x �22!

x2 ��63!

x3 �244!

x 4

f �4��x� �24

�x � 1�5 f �4��0� � 24

f����x� ��6

�x � 1�4 f����0� � �6

f ��x� �2

�x � 1�3 f ��0� � 2

f��x� � �1

�x � 1�2 f��0� � �1

f �x� �1

x � 1 f �0� � 1

23.

P2�x� � 1 � 0x �12!

x2 � 1 �12

x2

f ��x� � sec3 x � sec x tan2 x f ��0� � 1

f��x� � sec x tan x f��0� � 0

f �x� � sec x f �0� � 1

22.

� x � x2 � x3 � x4

P4�x� � 0 � 1�x� �22

x2 �66

x3 �2424

x4

f �4��x� � �24�x � 1��5 f �4��0� � �24

f����x� � 6�x � 1��4 f����0� � 6

f��x� � �2�x � 1��3 f��0� � �2

f��x� � �x � 1��2 f��0� � 1

� 1 � �x � 1��1

f �x� �x

x � 1�

x � 1 � 1x � 1

f �0� � 0

24.

� x �13

x3 P3�x� � 0 � 1�x� � 0 �26

x3

f����x� � 4 sec2 x tan2 x � 2 sec4 x f����0� � 2

f��x� � 2 sec2 x tan x f��0� � 0

f��x� � sec2 x f��0� � 1

f �x� � tan x f �0� � 0

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302 Chapter 9 Infinite Series

25.

� 1 � �x � 1� � �x � 1�2 � �x � 1�3 � �x � 1�4

�244!

�x � 1�4

P4�x� � 1 � �x � 1� �22!

�x � 1�2 ��63!

�x � 1�3

f �4��x� �24x5 f �4��1� � 24

f����x� � �6x4 f����1� � �6

f ��x� �2x3 f ��1� � 2

f��x� � �1x2 f��1� � �1

f �x� �1x f �1� � 1 26.

�5

32�x � 2�4

P4�x� �12

�12

�x � 2� �38

�x � 2�2 �14

�x � 2�3

f �4��x� � 240x�6 f �4��x� �154

f����x� � �48x�5 f����x� � �32

f��x� � 12x�4 f��2� �34

f��x� � �4x�3 f��2� � �12

f �x� � 2x�2 f �2� �12

27.

�1

16�x � 1�3 �

5128

�x � 1�4

P4�x� � 1 �12

�x � 1� �18

�x � 1�2

f �4��x� � �15

16x3x f �4��1� � �

1516

f����x� �3

8x2x f����1� �

38

f ��x� � �1

4xx f ��1� � �

14

f��x� �1

2x f��1� �

12

f �x� � x f �1� � 1

29.

�13

�x � 1�3 �14

�x � 1�4

P4�x� � 0 � �x � 1� �12

�x � 1�2

f �4��x� � �6x 4

f �4��1� � �6

f����x� �2x3 f����1� � 2

f ��x� � �1x2 f ��1� � �1

f��x� �1x f��1� � 1

f �x� � ln x f �1� � 0

28.

P3�x� � 2 �1

12�x � 8� �

1288

�x � 8�2 �5

20,736�x � 8�3

f����x� �1027

x�8�3 f����8� �1027

�128 �

53456

f� �x� � �29

x�5�3 f� �8� � �1

144

f��x� �13

x�2�3 f��8� �1

12

f �x� � x1�3 f �8� � 2

30.

P2�x� � ��2 � 2��x � �� ���2 � 2�

2�x � ��2

f ��x� � 2 cos x � 4x sin x � x2 cos x f ���� � �2 � � 2

f��x� � cos x � x 2 sin x f���� � �2�

f �x� � x2 cos x f ��� � �� 2

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Section 9.7 Taylor Polynomials and Approximations 303

31.

f

P3

Q3

P5

−6

6

−�2

�2

f �5��x� � 16 sec2 x tan4 x � 88 sec4 x tan2 x � 16 sec6 x

f �4��x� � 8 sec2 x tan3 x � 16 sec4 x tan x

f����x� � 4 sec2 x tan2 x � 2 sec4 x

f ��x� � 2 sec2 x tan x

f��x� � sec2 x

f �x� � tan x (a)

(b)

� 1 � 2�x ��

4� � 2�x ��

4�2

�83�x �

4�3

Q3�x� � 1 � 2�x ��

4� �42!�x �

4�2

�163!�x �

4�3

n � 3, c ��

4

P3�x� � 0 � x �02!

x2 �23!

x3 � x �13

x3

n � 3, c � 0

32.

(a)

(b)

�12

�12

�x � 1� �14

�x � 1�2 �18

�x � 1�4 Q4�x� �12

� ��12��x � 1� �

1�22!

�x � 1�2 �03!

�x � 1�3 ��34!

�x � 1�4

n � 4, c � 1

P4�x� � 1 � 0x ��22!

x2 �03!

x3 �244!

x 4 � 1 � x2 � x 4

n � 4, c � 0

f �4��x� �24�5x 4 � 10x2 � 1�

�x2 � 1�5

f����x� �24x�1 � x2�

�x2 � 1�4

f ��x� �2�3x2 � 1��x2 � 1�3

f��x� ��2x

�x2 � 1�2−3 3

−2

f

Q

P4

4P2

2 f �x� �1

x2 � 1

33.

(a)

(c) As the distance increases, the accuracy decreases.

P5�x� � x �16 x3 �

1120 x5

P3�x� � x �16 x3

P1�x� � x

f �x� � sin x

x 0.00 0.25 0.50 0.75 1.00

0.0000 0.2474 0.4794 0.6816 0.8415

0.0000 0.2500 0.5000 0.7500 1.0000

0.0000 0.2474 0.4792 0.6797 0.8333

0.0000 0.2474 0.4794 0.6817 0.8417P5�x�

P3�x�

P1�x�

sin x

(b)

−3

−2

P1P3

P5

f

3

� 2�

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304 Chapter 9 Infinite Series

34.

(a)

(c) As the degree increases, the accuracy increases. As the distance from to 1 increases, the accuracy decreases.x

P4�x� � �x � 1� �12�x � 1�2 �

13�x � 1�3 �

14�x � 1�4

P2�x� � �x � 1� �12�x � 1�2

P1�x� � x � 1

f �x� � ln x, c � 1

x 1.00 1.25 1.50 1.75 2.00

0 0.2231 0.4055 0.5596 0.6931

0 0.2500 0.5000 0.7500 1.0000

0 0.2188 0.375 0.4688 0.5000

0 0.2230 0.4010 0.5303 0.5833P4�x�

P2�x�

P1�x�

ln x

(b)

5

−2

−1

P1

P4P2

2

f

35.

(a)

(b)

P3�x� � x �x3

6

f �x� � arcsin x

x 0 0.25 0.50 0.75

0 0.253 0.524 0.848

0 0.253 0.521 0.820�0.253�0.521�0.820P3�x�

�0.253�0.524�0.848f �x�

�0.25�0.50�0.75

(c)

1

2

−1

f

x

3P

π

y

36. (a)

(b)

P3�x� � x �x3

3

f �x� � arctan x (c)

x

f

1−1 12

π

π

4

4

π2

π2

P3

y

x 0 0.25 0.50 0.75

0 0.2450 0.4636 0.6435

0 0.2448 0.4583 0.6094�0.2448�0.4583�0.6094P3�x�

�0.2450�0.4636�0.6435f �x�

�0.25�0.50�0.75

37.6

6 8

4

2

−4

−6

−6x

P6 P2

P8 P4y

f (x) = cos x

f �x� � cos x

39.3

2

1

−2 2 3 4−3−4

−3

x

P6P2

P8P4

y

f (x) = ln (x2 + 1)

f �x) � ln�x2 � 1�

38.

−2

2

1

x

f x x( ) = arctan

1 3−3 −2

PP33PP11PP13P9

yPP7 PP1P5f �x� � arctan x

40.

x

2

4

−4 4

yy = 4xe(−x2/4)f �x� � 4xe�x2�4

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Section 9.7 Taylor Polynomials and Approximations 305

41.

f �12� 0.6042

f �x� � e�x 1 � x �x2

2�

x3

642.

f �15� 0.0328

f �x� � x2e�x x2 � x3 �12

x4

43.

f �1.2� 0.1823

f �x� � ln x �x � 1� �12 �x � 1�2 �

13 �x � 1�3 �

14 �x � 1�4

45.

Note: you could use

Exact error: 0.000001 � 1.0 10 �6

R5�x� ≤16!

�0.3�6 � 1.0125 10�6

R5�x�: f �6��x� � �cos x, max on �0, 0.3 is 1.

R4�x� ≤15!

�0.3�5 � 2.025 10�5

f �x� � cos x; f �5��x� � �sin x ⇒ Max on �0, 0.3 is 1.

44.

f�7�

8 � �6.7954

f �x� � x2 cos x ��2 � 2��x � �� � �� 2 � 22 ��x � ��2

46.

R5�x� ≤ e1

6! �1�6 0.00378 � 3.78 10�3

f �x� � ex; f �6��x� � ex ⇒ Max on �0, 1 is e1.

47.

The exact error is Note: You could use R4.�8.5 10�4.R3�x� ≤ 7.3340

4!�0.4�4 0.00782 � 7.82 10�3.

f �x� � arcsin x; f �4��x� �x�6x2 � 9��1 � x2�7�2 ⇒ Max on �0, 0.4 is f �4��0.4� 7.3340.

48.

R3�x� ≤ 22.3672

4! �0.4�4 0.0239

⇒ Max on �0, 0.4 is f �4��0.4� 22.3672.

f �x� � arctan x; f �4��x� �24x�x2 � 1�

�1 � x2�4 49.

for all x.

By trial and error, n � 3.

Rn�x� ≤ 1

�n � 1�!�0.3�n�1 < 0.001

�g�n�1��x�� ≤ 1

g�x� � sin x

50.

for all and all

By trial and error, n � 2.

≤�0.1�n�1

�n � 1�! < 0.001

�Rn�x�� � � f �n�1��z�xn�1

�n � 1�! �n.x� f �n�1��x�� ≤ 1

f �x� � cos x 51.

Max on is

By trial and error, n � 5.

Rn ≤1.8221

�n � 1�! �0.6�n�1 < 0.001

e0.6 1.8221.�0, 0.6

f �n�1��x� � ex

f �x� � ex

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306 Chapter 9 Infinite Series

52.

The maximum value on is

By trial and error, .n � 3

�Rn� ≤1.3499

�n � 1�! �0.3�n�1 < 0.001

e0.3 1.3499.�0, 0.3

f �n�1��x� � e x

f �x� � e x 53.

By trial and error, (See Example 9.) Using 9 terms,ln�1.5� 0.4055.

n � 9.

Rn ≤ n!

�n � 1�! �0.5�n�1 ��0.5�n�1

n � 1 < 0.0001

f �n�1��x� ���1�n n!

�x � 1�n�1 ⇒ Max on �0, 0.5 is n!.

f �x� � ln�x � 1�

55.

By trial and error, n � 16.

�Rn� ≤ ���n�1

�n � 1�! �1.3�n�1 < 0.0001

f �n�1��x� � ����n�1e��x ≤ �����n�1� on �0, 1.3

f��x� � ����e��x

f �x� � e��x, f (1.3�54.

Since this is an alternating series,

By trial and error, Using 4 terms f �0.6� 0.4257.n � 4.

Rn ≤ an�1 ��2n

�2n�! �0.6�4n < 0.0001.

f �0.6� � 1 �� 2

2! �0.6�4 �

�4

4! �0.6�8 �

� 6

6! �0.6�12 � . . .

� 1 �� 2x 4

2!�

�4x8

4!�

�6x12

6!� . . .

� 1 ���x2�2

2!�

��x2�4

4!�

��x2�6

6!� . . .

f �x� � g��x2�

g�x� � cos x � 1 �x2

2!�

x 4

4!�

x6

6!� . . .

f �x� � cos��x2�

58.

�0.3936 < x < 0.3936

�x� < 0.3936

x 4 < 0.024

�R3�x�� � �sin z4!

x4� ≤ �x4�

4! < 0.001

f �x� � sin x x �x3

3!59. fifth degree polynomial

for all and all

Note: Use a graphing utility to graphin the viewing window

to verify theanswer.��0.9467, 0.9467 ��0.001, 0.001y � cos x � �1 � x2�2 � x 4�24�

�0.9467 < x < 0.9467

�x� < 0.9467

�x�6 < 0.72

�R5�x�� ≤16!

�x�6 < 0.001

n.x� f �n�1��x�� ≤ 1

f �x� � cos x 1 �x2

2!�

x 4

4!,

56.

By trial and error, n � 7.

�Rn� ≤ 1

�n � 1�! 1n�1 ≤ 0.0001

f �n�1��x� � ��1�n�1e�x ≤ 1 on �0, 1

f��x� � �e�x

f �x� � e�x 57.

�0.3936 < x < 0

�x� < 0.3936

ez�4 < 0.3936, z < 0

�xez�4� < 0.3936

ez x4 < 0.024

R3�x� �ez

4! x 4 < 0.001

f �x� � ex 1 � x �x2

2�

x3

6, x < 0

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Section 9.7 Taylor Polynomials and Approximations 307

60.

Thus

In fact, by graphing and you can verify that on ��0.19294, 0.20068�

� f �x� � y� < 0.001y � 1 � 2x � 2x2 �43 x

3,f �x� � e�2x

0 < x < 0.1970.

x < �0.0015e�2z �

1�4

0.1970e2z < 0.1970, for z < 0.

e�2zx 4 < 0.0015

R3 �x� �f 4�z�

4!�x � 0�4 �

16e�2z

24 x 4 �

23

e�2z x 4 < 0.001

f��x� � �2e�2x, f��x� � 4e�2x, f���x� � �8e�2x, f �4��x� � 16e�2x

f �x� � e�2x 1 � 2x � 2x2 �43

x3

61. The graph of the approximating polynomial P and theelementary function f both pass through the point and the slopes of P and f agree at Depending onthe degree of P, the nth derivatives of P and f agree at�c, f �c��.

�c, f �c��.�c, f �c��

62. f �c� � P2�c�, f��c� � P2��c�, and f��c� � P2��c�

63. See definition on page 650. 64. See Theorem 9.19, page 654.

65. The accuracy increases as the degree increases (for valueswithin the interval of convergence).

66.

x20−20 10

2

−2

−4

4

6

8

10 f

y

P1

P3

P2

67. (a)

(b)

(c) g�x� �sin x

x�

1x P5�x� � 1 �

x2

3!�

x 4

5!

Q6�x� � x P5�x� � x2 �x 4

3!�

x6

5!

g�x� � x sin x

P5�x� � x �x3

3!�

x5

5!

f �x� � sin x

Q5�x� � x P4�x�

Q5�x� � x � x2 �12

x3 �16

x 4 �1

24 x5

g�x� � xex

P4�x� � 1 � x �12

x2 �16

x3 �1

24 x 4

f �x� � ex 68. (a) for

This is the Maclaurin polynomial of degree 4 for

(b) for

(c)

The first four terms are the same!

R��x� � 1 � x �x2

2!�

x3

3!

R�x� � 1 � x �x2

2!�

x3

3!�

x 4

4!

Q6��x� � �x �x3

3!�

x5

5!� �P5�x�

cos xQ6�x� � 1 �x2

2�

x 4

4!�

x6

6!

g�x� � cos x.

P5��x� � 1 �x2

2!�

x 4

4!

f �x� � sin xP5�x� � x �x3

3!�

x5

5!

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308 Chapter 9 Infinite Series308 Chapter 9 Infinite Series

69. (a) Q2�x� � �1 ��2�x � 2�2

32(b) R2�x� � �1 �

�2�x � 6�2

32(c) No. The polynomial will be linear.

Translations are possible atx � �2 � 8n.

70. Let f be an odd function and be the nth Maclaurin polynomial for f. Since f is odd, is even:

Similarly, is odd, is even, etc. Therefore, f, etc. are all odd functions, which implies that Hence, in the formula

all the coefficients of the even power of x are zero.Pn�x� � f �0� � f��0�x �f ��0�x2

2!� . . .

f �0� � f ��0� � . . . � 0.f �4�,f�,f��f�

f���x� � limh→0

f ��x � h� � f ��x�

h� lim

h→0

�f �x � h� � f �x�h

� limh→0

f �x � ��h�� � f �x�

�h� f��x�.

f�Pn

71. Let f be an even function and be the nth Maclaurin polynomial for f. Since f is even, is odd, is even, is odd, etc. All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. will only haveterms with even powers of x.

Pn

f���f�f�Pn

72. Let

For Pn�k��c� � an k! � �f �k��c�

k! �k! � f �k��c�.1 ≤ k ≤ n,

Pn�c� � a0 � f �c�

Pn �x� � a0 � a1�x � c� � a2�x � c�2 � . . . � an �x � c�n where ai �f �i��c�

i!.

73. As you move away from the Taylor Polynomial becomes less and less accurate.x � c,

Section 9.8 Power Series

1. Centered at 0 2. Centered at 0 3. Centered at 2 4. Centered at �

5.

�x� < 1 ⇒ R � 1

� limn→�

�n � 1n � 2��x� � �x�

L � limn→�

�un�1

un � � limn→�

���1�n�1xn�1

n � 2�

n � 1��1�n xn�

��

n�0��1�n

xn

n � 1

7.

2�x� < 1 ⇒ R �12

� limn→�

� 2n2x�n � 1�2� � 2�x�

L � limn→�

�un�1

un � � limn→�

� �2x�n�1

�n � 1�2 �n2

�2x�n���

n�1 �2x�n

n2

6.

2�x� < 1 ⇒ R �12

L � limn→�

�un � 1un � � lim

n→� ��2x�n�1

�2x�n � � 2�x�

��

n�0�2x�n

8.

12

�x� < 1 ⇒ R � 2

�12�x�

L � limn→�

�un�1

un � � limn→�

���1�n�1xn�1

2n�1 �2n

��1�n xn���

n�0 ��1�n xn

2n

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Section 9.8 Power Series 309

9.

Thus, the series converges for all x. R � �.

� limn→�

� �2x�2

�2n � 2��2n � 1�� � 0

L � limn→�

�un�1

un � � limn→�

��2x�2n�2��2n � 2�!�2x�2n��2n�! �

��

n�0

�2x�2n

�2n�!

11.

Since the series is geometric, it converges only ifor �2 < x < 2.�x�2� < 1

��

n�0�x

2�n

10.

The series only converges at R � 0.x � 0.

� limn→�

��2n � 2��2n � 1�x2

�n � 1� � � �

L � limn→�

�un � 1un � � lim

n→� ��2n � 2�!x2n�2��n � 1�!

�2n�!x2n�n! ���

n�0

�2n�!x2n

n!

12.

is geometric, so it converges if

�x�5� < 1 ⇒ �x� < 5 ⇒ �5 < x < 5.

��

n�0�x

5�n

13.

Interval:

When the alternating series converges.

When the p-series diverges.

Therefore, the interval of convergence is �1 < x ≤ 1.

��

n�1 1n

x � �1,

��

n�1 ��1�n

n x � 1,

�1 < x < 1

� limn→�

� nxn � 1� � �x�

limn→�

�un�1

un � � limn→�

���1�n�1xn�1

n � 1�

n��1�n xn�

��

n�1 ��1�n xn

n

15.

The series converges for all x. Therefore, the interval of convergence is �� < x < �.

� limn→�

� xn � 1� � 0

limn→�

�un�1

un � � limn→�

� xn�1

�n � 1�! �n!xn�

��

n�0 xn

n!

14.

Interval:

When the series diverges.

When the series diverges.

Therefore, the interval of convergence is �1 < x < 1.

��

n�0��n � 1�x � �1,

��

n�0��1�n�1�n � 1�x � 1,

�1 < x < 1

� limn→�

��n � 2�xn � 1 � � �x�

limn→�

�un � 1un � � lim

n→� ���1�n�2�n � 2�xn�1

��1�n�n � 1�xn ���

n�0��1�n�1�n � 1�xn

16.

Therefore, the interval of convergence is �� < x < �.

� limn→�

� 3x�2n � 2��2n � 1�� � 0

limn→�

�un�1

un � � limn→�

� �3x�n�1

�2n � 1�! ��2n�!�3x�n�

��

n�0 �3x�n

�2n�!

17.

Therefore, the series converges only for x � 0.

� limn→�

��2n � 2��2n � 1� x2 � � � lim

n→� �un�1

un � � limn→�

��2n � 2�!xn�1

2n�1 �2n

�2n�!xn� ��

n�0�2n�!�x

2�n

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310 Chapter 9 Infinite Series

19.

Since the series is geometric, it converges only if or �4 < x < 4.�x�4� < 1

��

n�1 ��1�n�1xn

4n

18.

Interval:

When the alternating series converges.

When the series converges by limit comparison to

Therefore, the interval of convergence is �1 ≤ x ≤ 1.

��

n�1 1n2.�

n�0

1�n � 1��n � 2�x � �1,

��

n�0

��1�n

�n � 1��n � 2�x � 1,

�1 < x < 1

� limn→�

��n � 1�xn � 3 � � �x� lim

n→� �un�1

un � � limn→�

� ��1�n�1xn�1

�n � 2��n � 3� ��n � 1��n � 2�

��1�n xn ���

n�0

��1�n xn

�n � 1��n � 2�

20.

Center:

Therefore, the series converges only for x � 4.

x � 4

R � 0

� limn→�

��n � 1��x � 4�3 � � � lim

n→� �un�1

un � � limn→�

���1�n�1�n � 1�!�x � 4�n�1

3n�1 �3n

��1�n n!�x � 4�n���

n�0 ��1�n n!�x � 4�n

3n

21.

Center:

Interval: or

When the p-series diverges. When the alternating series converges.

Therefore, the interval of convergence is 0 < x ≤ 10.

��

n�1

��1�n�1

nx � 10,�

n�1 �1n

x � 0,

0 < x < 10�5 < x � 5 < 5

x � 5

R � 5

� limn→�

�n�x � 5�5�n � 1�� �

15�x � 5� lim

n→� �un�1

un � � limn→�

���1�n�2�x � 5�n�1

�n � 1�5n�1 �n5n

��1�n�1�x � 5�n���

n�1 ��1�n�1�x � 5�n

n5n

22.

Center:

Interval: or

When the alternating series converges.

When the series diverges.

Therefore, the interval of convergence is �2 ≤ x < 6.

��

n�0

1n � 1

x � 6,

��

n�0

��1�n�1

�n � 1�x � �2,

�2 < x < 6�4 < x � 2 < 4

x � 2

R � 4

� limn→�

��x � 2��n � 1�4�n � 2� � �

14�x � 2� lim

n→� �un�1

un � � limn→�

� �x � 2�n�2

�n � 2�4n�2 ��n � 1�4n�1

�x � 2�n�1 ���

n�0

�x � 2�n�1

�n � 1�4n�1

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Section 9.8 Power Series 311

23.

Center:

Interval: or

When the series diverges by the integral test.

When the alternating series converges.

Therefore, the interval of convergence is 0 < x ≤ 2.

��

n�0 ��1�n�1

n � 1x � 2,

��

n�0

1n � 1

x � 0,

0 < x < 2�1 < x � 1 < 1

x � 1

R � 1

� limn→�

��n � 1��x � 1�n � 2 � � �x � 1�lim

n→� �un�1

un � � limn→�

���1�n�2�x � 1�n�2

n � 2�

n � 1��1�n�1�x � 1�n�1�

��

n�0 ��1�n�1�x � 1�n�1

n � 1

24.

At

diverges.

At

converges.

Interval of convergence: 0 < x ≤ 4

��

n�1

��1�n�1 2n

n2n � ��

n�1

��1�n�1

n,

x � 4,

��

n�1

��1�n�1��2�n

n2n � ��

n�1

��1��2n�n2n � �

n�1

��1�n

,

x � 0,

�x � 22 � < 1 ⇒ �2 < x � 2 < 2 ⇒ 0 < x < 4

� limn→��x � 2

2

nn � 1� � �x � 2

2 � limn→��an�1

an � � limn→����1�n�2�x � 2�n�1

�n � 1�2n�1 ���1�n�1�x � 2�n

n2n ���

n�1

��1�n�1�x � 2�n

n2n25. is geometric. It converges if

Interval convergence: 0 < x < 6

�x � 33 � < 1 ⇒ �x � 3� < 3 ⇒ 0 < x < 6.

��

n�1�x � 3

3 �n�1

26.

Interval:

When converges.

When converges.

Therefore, the interval of convergence is �1 ≤ x ≤ 1.

x � �1, ��

n�0

��1�n�1

2n � 1

x � 1, ��

n�0

��1�n

2n � 1

�1 < x < 1

R � 1

� limn→�

��2n � 1��2n � 3�x

2� � �x2�limn→�

�un�1

un � � limn→�

���1�n�1x2n�3

�2n � 3� ��2n � 1�

��1�nx2n�1���

n�0

��1�nx2n�1

2n � 1

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312 Chapter 9 Infinite Series

28.

Therefore, the interval of convergence is �� < x < �.

� limn→�

� x2

n � 1� � 0

limn→�

�un�1

un � � limn→�

���1�n�1x2n�2

�n � 1�! �n!

��1�n x2n� ��

n�0 ��1�n x2n

n!

29.

Therefore, the interval of convergence is �� < x < �.

� limn→�

� x2

�2n � 2��2n � 3�� � 0

limn→�

�un�1

un � � limn→�

� x2n�3

�2n � 3�! ��2n � 1�!

x2n�1 ���

n�0

x2n�1

�2n � 1�!30.

Therefore, the interval of convergence is �� < x < �.

� limn→�

� �n � 1�x�2n � 2��2n � 1�� � 0

limn→�

�un�1

un � � limn→�

��n � 1�!xn�1

�2n � 2�! ��2n�!n!xn �

��

n�1 n!xn

�2n�!

32.

When the series diverges by comparing it to

which diverges. Therefore, the interval of convergence is �1 < x < 1.

��

n�1

12n � 1

x � ±1,

R � 1

limn→�

�un�1

un � � limn→�

� 2 � 4 . . . �2n��2n � 2�x2n�3

3 � 5 � 7 . . . �2n � 1��2n � 3� �3 � 5 . . . �2n � 1�2 � 4 . . . �2n�x2n�1� � lim

n→� ��2n � 2�x2

�2n � 3� � � �x 2�

��

n�1

2 � 4 � 6 . . . �2n�3 � 5 � 7 . . . �2n � 1� �x

2n�1�

31.

Converges if

At diverges.

Interval of convergence: �1 < x < 1

x � ±1,

�x� < 1 ⇒ �1 < x < 1.

limn→��an�1

an � � limn→���n � 2�xn�1

�n � 1�xn � � limn→��n � 2

n � 1x� � �x�

��

n�1 2 � 3 � 4 . . . �n � 1�xn

n!� �

n�1�n � 1�xn

27.

Interval:

When the series diverges

by the nth Term Test.

When the alternating series diverges.

Therefore, the interval of convergence is �12

< x < 12

.

��

n�1 ��1�n�1n

n � 1x �

12

,

��

n�1

nn � 1

x � �12

,

�12

< x < 12

R �12

� limn→�

���2x��n � 1�2

n�n � 2� � � 2�x�

limn→�

�un�1

un � � limn→�

��n � 1���2x�n

n � 2�

n � 1n��2x�n�1�

��

n�1

nn � 1

��2x�n�1

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Section 9.8 Power Series 313

33.

Center:

Therefore, the series converges only for x � 3.

x � 3

R � 0

� limn→�

��4n � 3��x � 3�4 � � �

limn→�

�un�1

un � � limn→�

���1�n�2 � 3 � 7 � 11 . . . �4n � 1��4n � 3��x � 3�n�1

4n�1 �4n

��1�n�1 � 3 � 7 � 11 . . . �4n � 1��x � 3�n���

n�1 ��1�n�1 3 � 7 � 11 . . . �4n � 1��x � 3�n

4n

34.

Converges if

At diverges.

At diverges.

Interval of convergence: �3 < x < 1

an �n!��2�n

1 � 3 . . . �2n � 1� � ��1�n 2 � 4 . . . 2n1 � 3 . . . �2n � 1�, x � �3,

an �n!2n

1 � 3 � 5 . . . �2n � 1� �2 � 4 � 6 . . . 2n

1 � 3 � 5 . . . �2n � 1� > 1, x � 1,

12�x � 1� < 1 ⇒ �2 < x � 1 < 2 ⇒ �3 < x < 1.

� limn→���n � 1��x � 1�

2n � 1 � �12�x � 1�

limn→��an�1

an � � limn→�� �n � 1�!�x � 1�n�1

1 � 3 � 5 . . . �2n � 1��2n � 1�� �n�!�x � 1�n

1 � 3 � 5 . . . �2n � 1����

n�1

n!�x � 1�n

1 � 3 � 5 . . . �2n � 1�

35.

Center:

Interval: or

When the series diverges.

When the series diverges.

Therefore, the interval of convergence is 0 < x < 2c.

��

n�1 1x � 2c,

��

n�1��1�n�1x � 0,

0 < x < 2c�c < x � c < c

x � c

R � c

�1c�x � c�lim

n→� �un�1

un � � limn→�

��x � c�n

cn �cn�1

�x � c�n�1���

n�1 �x � c�n�1

cn�1 36. is a positive integer.

Converges if �x�kk < 1 ⇒ R � kk.

� �x�kk

� limn→�� �n � 1�kx

�k � nk��k � 1 � nk� . . . �1 � nk�� limn→��an�1

an � � limn→����n � 1�!kxn�1

�k�n � 1�! ��n!�kxn

�kn�! ���

n�0 �n!�kxn

�kn�! , k

37.

Since the series is geometric, it converges only if or .�k < x < k.�x�k� < 1

��

n�0�x

k�n

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314 Chapter 9 Infinite Series

38.

Center:

Interval: or

When the p-series diverges.

When the alternating series converges. Therefore, the interval of convergence is 0 < x ≤ 2c.��

n�1 ��1�n�1

nx � 2c,

��

n�1 �1n

x � 0,

0 < x < 2c�c < x � c < c

x � c

R � c

� limn→�

�n�x � c�c�n � 1�� �

1c�x � c�lim

n→� �un�1

un � � limn→�

���1�n�2�x � c�n�1

�n � 1�cn�1 �ncn

��1�n�1�x � c�n���

n�1 ��1�n�1�x � c�n

ncn

39.

When the series diverges and the interval of convergence is

k�k � 1� . . . �k � n � 1�1 � 2 . . . n

≥ 1��1 < x < 1.x � ±1,

R � 1

limn→�

�un�1

un � � limn→�

�k�k � 1� . . . �k � n � 1��k � n�xn�1

�n � 1�! �n!

k�k � 1� . . . �k � n � 1�xn� � limn→�

��k � n�xn � 1 � � �x�

��

n�1 k�k � 1� . . . �k � n � 1�xn

n!

40.

Interval: or

The series diverges at the endpoints. Therefore, the interval of convergence is

n!�c � 2 � c�n

1 � 3 � 5 . . . �2n � 1� �n!22

1 � 3 � 5 . . . �2n � 1�� 2 � 4 � 6 . . . �2n�

1 � 3 � 5 . . . �2n � 1� > 1�

c � 2 < x < c � 2.

c � 2 < x < c � 2�2 < x � c < 2

R � 2

limn→�

�un�1

un � � limn→�

� �n � 1�!�x � c�n�1

1 � 3 � 5 . . . �2n � 1��2n � 1� �1 � 3 � 5 . . . �2n � 1�

n!�x � c� � � limn→�

��n � 1��x � c�2n � 1 � �

12

�x � c�

��

n�1

n!�x � c�n

1 � 3 � 5 . . . �2n � 1�

41. � ��

n�1

xn�1

�n � 1�! ��

n�0 xn

n!� 1 �

x1

�x2

2� . . . 42. �

n�0��1�n�1�n � 1�xn � �

n�1��1�n�n�xn�1

43.

Replace n with n � 1.

��

n�0

x2n�1

�2n � 1�! � ��

n�1

x2n�1

�2n � 1�!44. �

n�0

��1�nx2n�1

2n � 1� �

n�1

��1�n�1x2n�1

2n � 1

45. (a) (Geometric)

(b)

(c)

(d) �2 ≤ x < 2� f �x� dx � ��

n�0

2n � 1

�x2�

n�1

,

�2 < x < 2f ��x� � ��

n�2�n

2��n � 1

2 ��x2�

n�2

,

f��x� � ��

n�1�n

2��x2�

n�1

, �2 < x < 2

f �x� � ��

n�0�x

2�n

, �2 < x < 2 46. (a)

(b)

(c)

(d) 0 ≤ x ≤ 10� f �x� dx � ��

n�1 ��1�n�1�x � 5�n�1

n�n � 1�5n ,

0 < x < 10f ��x� � ��

n�2 ��1�n�1�n � 1��x � 5�n�2

5n ,

f��x� � ��

n�1 ��1�n�1�x � 5�n�1

5n , 0 < x < 10

f �x� � ��

n�1 ��1�n�1�x � 5�n

n5n , 0 < x ≤ 10

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Section 9.8 Power Series 315

48. (a)

(b)

(c)

(d) �f �x� dx � ��

n�1

��1�n�1�x � 2�n�1

n�n � 1� , 1 ≤ x ≤ 3

f��x� � ��

n�2��1�n�1�n � 1��x � 2�n�2, 1 < x < 3

f��x� � ��

n�1��1�n�1�x � 2�n�1, 1 < x < 3

f �x� � ��

n�1

��1�n�1�x � 2�n

n, 1 < x ≤ 347. (a)

(b)

(c)

(d) 0 ≤ x ≤ 2� f �x� dx � ��

n�1 ��1�n�1�x � 1�n�2

�n � 1��n � 2� ,

f ��x� � ��

n�1��1�n�1n�x � 1�n�1, 0 < x < 2

f��x� � ��

n�0 ��1�n�1�x � 1�n, 0 < x < 2

f �x� � ��

n�0 ��1�n�1�x � 1�n�1

n � 1, 0 < x ≤ 2

49.

Matches (c).S1 � 1, S2 � 1.33.

g�1� � ��

n�0�1

3�n

� 1 �13

�19

� . . . 50.

Matches (a).S1 � 1, S2 � 1.67.

g�2� � ��

n�0�2

3�n

� 1 �23

�49

� . . .

51. diverges. Matches (b).g�3.1� � ��

n�0�3.1

3 �n

52. alternating. Matches (d).g��2� � ��

n�0��

23�

n

53. converges

Matches (b).S1 � 1, S2 � 1.25, S3 � 1.3125

g�18� � �

n�02�1

8��n

� ��

n�0�1

4�n

� 1 �14

�1

16� . . .,

54. converges

Matches (c).S1 � 1, S2 � 0.75, S3 � 0.8125

g��18� � �

n�02��

18��

n

� ��

n�0��

14�

n

� 1 �14

�1

16� . . .,

55. diverges

Matches (d).S1 � 1, S2 �178

g� 916� � �

n�02� 9

16��n

� ��

n�0�9

8�n

, 56. converges

Matches (a).S1 � 1, S2 � 1.75

g�38� � �

n�02�3

8��n

� ��

n�0�3

4�n

,

57. A series of the form

is called a power series centered at c.

��

n�0an�x � c�n

58. The set of all values of x for which the power series converges is the interval of convergence. If the powerseries converges for all x, then the radius of convergenceis If the power series converges at only c, then

Otherwise, according to Theorem 8.20, thereexists a real number (radius of convergence) suchthat the series converges absolutely for anddiverges for �x � c� > R.

�x � c� < RR > 0

R � 0.R � �.

59. A single point, an interval, or the entire real line. 60. You differentiate and integrate the power series term byterm. The radius of convergence remains the same.However, the interval of convergence might change.

61. Answers will vary.

converges for . At the convergence is conditional because diverges.

converges for . At the convergence is absolute.x � ±1,�1 ≤ x ≤ 1��

n�1

xn

n2

� 1n

x � �1,�1 ≤ x < 1��

n�1

xn

n

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316 Chapter 9 Infinite Series

62. Many answers possible.

(a) Geometric:

(b) converges for �1 < x ≤ 1��

n�1

��1�nxn

n

�x2� < 1 ⇒ �x� < 2��

n�1�x

2�n

(c) Geometric:

(d) converges for �2 ≤ x < 6��

n�1

�x � 2�n

n4n

�2x � 1� < 1 ⇒ �1 < x < 0��

n�1�2x � 1�n

63. (a)

(See Exercise 29.)

(b) f��x� � ��

n�0 ��1�n x2n

�2n�! � g�x�

g�x� � ��

n�0 ��1�n x2n

�2n�! , �� < x < �

f �x� � ��

n�0 ��1�n x2n�1

�2n � 1�! , �� < x < �

64. (a) (See Exercise 11)

(b) � ��

n�1

xn�1

�n � 1�! � ��

n�0 xn

n!� f �x� f��x� � �

n�1 nxn�1

n!

f �x� � ��

n�0 xn

n!, �� < x < � (c)

(d) f �x� � ex

f �0� � 1

f �x� � ��

n�1 xn

n!� 1 � x �

x2

2!�

x3

3!�

x 4

4!� . . .

65.

y� � y � ��

n�1 ��1�n x2n�1

�2n � 1�! � ��

n�1 ��1�n�1x2n�1

�2n � 1�! � 0

y� � ��

n�1 ��1�n �2n�x2n�1

�2n�! � ��

n�1 ��1�x2n�1

�2n � 1�!

y� � ��

n�0 ��1�n �2n � 1�x2n

�2n � 1�! � ��

n�0 ��1�nx2n

�2n�!

y � ��

n�0 ��1�n x2n�1

�2n � 1�! � ��

n�1 ��1�n�1x2n�1

�2n � 1�! 66.

y� � y � ��

n�1 ��1�n x2n�2

�2n � 2�! � ��

n�1 ��1�n�1x2n�2

�2n � 2�! � 0

y� � ��

n�1 ��1�n �2n � 1�x2n�2

�2n � 1�! � ��

n�1 ��1�nx2n�2

�2n � 2�!

y� � ��

n�1 ��1�n �2n�x2n�1

�2n�! � ��

n�1 ��1�nx2n�1

�2n � 1�!

y � ��

n�0 ��1�n x2n

�2n�! � ��

n�1 ��1�n�1x2n�2

�2n � 2�!

67.

y� � y � 0

y� � ��

n�1 �2n�x2n�1

�2n�! � ��

n�1

x2n�1

�2n � 1�! � y

y� � ��

n�0 �2n � 1�x2n

�2n � 1�! � ��

n�0

x2n

�2n�!

y � ��

n�0

x2n�1

�2n � 1�! � ��

n�1

x2n�1

�2n � 1�! 68.

y� � y � 0

y� � ��

n�1 �2n � 1�x2n�2

�2n � 1�! � ��

n�1

x2n�2

�2n � 2�! � y

y� � ��

n�1 �2n � 2�x2n�1

�2n � 2�! � ��

n�1

x2n�1

�2n � 1�!

y � ��

n�0

x2n

�2n�! � ��

n�1

x2n�2

�2n � 2�!

69.

� 0 � ��

n�0 2�n � 1�x2n ��2n � 1� � �2n � 1�

2n�1�n � 1�!

� ��

n�0�2n � 2��2n � 1�x2n

2n�1�n � 1�! ��2n � 1�x2n

2n n!�

2�n � 1�2�n � 1��

� ��

n�1 2n�2n � 1�x2n�2

2n n!� �

n�0 �2n � 1�x2n

2n n!

y� � xy� � y � ��

n�1 2n�2n � 1�x2n�2

2n n!� �

n�1 2nx2n

2n n!� �

n�0

x2n

2n n!

y� � ��

n�1 2n�2n � 1�x2n�2

2n n!y� � �

n�1 2nx2n�1

2n n!y � �

n�0

x2n

2n n!

(c)

(d) and g�x� � cos xf �x� � sin x

� � ��

n�0 ��1�n x2n�1

�2n � 1�! � �f �x�

g��x� � ��

n�1 ��1�n x2n�1

�2n � 1�! � ��

n�0 ��1�n�1x2n�1

�2n � 1�!

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Section 9.8 Power Series 317

70.

� 0 � ��

n�1

��1�n�1 4�n � 1�x4n�2

22n�2�n � 1�! � 3 � 7 � 11 . . . �4n � 1� � ��

n�1

��1�n�1 x 4n�2

22n n! � 3 � 7 � 11 . . . �4n � 1� 22�n � 1�22�n � 1�

y� � x2y � �x2 � ��

n�2

��1�n 4nx 4n�2

22n n! � 3 � 7 � 11 . . . �4n � 5� � ��

n�1

��1�n x4n�2

22n n! � 3 � 7 � 11 . . . �4n � 1� � x2

� �x2 � ��

n�2

��1�n 4nx4n�2

22n n! � 3 � 7 � 11 . . . �4n � 5� y� � ��

n�1

��1�n 4n�4n � 1�x4n�2

22n n! � 3 � 7 � 11 . . . �4n � 1�

y� � ��

n�1

��1�n 4nx 4n�1

22n n! � 3 � 7 � 11 . . . �4n � 1�

y � 1 � ��

n�1

��1�n x4n

22n n! � 3 � 7 � 11 . . . �4n � 1�

71.

(a)

Therefore, the interval of convergence is

(b)

� ��

k�0 ��1�k x2k�2

4k �k!�2 �4k � 24k � 4

�2

4k � 4�

4k � 44k � 4� � 0

� ��

k�0 ��1�k x2k�2

4k �k!�2 ��1�2�2k � 1�4�k � 1� � ��1� 2

4�k � 1� � 1�

x2J0� � xJ0� � x2J0 � ��

k�0��1�k�1

2�2k � 1� x2k�2

4k�1�k � 1�!k!� �

k�0��1�k�1

2x2k�2

4k�1�k � 1�!k!� �

k�0��1�k

x2k�2

4k �k!�2

� ��

k�0��1�k�1

�2k � 2��2k � 1�x2k

4k�1 ��k � 1�!2 J0� � ��

k�1��1�k

2k�2k � 1�x2k�2

4k �k!�2

� ��

k�0��1�k�1

�2k � 2� x2k�1

4k�1��k � 1�!2 J0� � ��

k�1��1�k

2kx2k�1

4k �k!�2

J0 � ��

k�0��1�k

x2k

4k �k!�2

�� < x < �.

limk→�

�uk�1

uk � � limk→�

� ��1�k�1 x2k�2

22k�2 ��k � 1�!2 �22k �k!�2

��1�k x2k� � limk→�

� ��1�x2

22�k � 1�2� � 0

J0�x� � ��

k�0 ��1�k x2k

22k �k!�2

(d)

(integral is approximately 0.9197304101)

� 1 �1

12�

1320

0.92

� ��

k�0

��1�k

4k�k!�2�2k � 1�

� ��

k�0

��1�k x2k�1

4k�k!�2�2k � 1��1

0

�1

0 J0dx � �1

0 �

k�0 ��1�k x2k

4k �k!�2 dx

72.

(a)

Therefore, the interval of convergence is

—CONTINUED—

�� < x < �.

limk→�

�uk�1

uk � � limk→�

� ��1�k�1 x2k�3

22k�3�k � 1�!�k � 2�! �22k�1k!�k � 1�!

��1�k x2k�1 � � limk→�

� ��1�x2

22 �k � 2��k � 1�� � 0

J1�x� � x ��

k�0

��1�k x2k

22k�1k!�k � 1�! � ��

k�0

��1�k x2k�1

22k�1k!�k � 1�!

(c)

−6 6

−5

3

P6�x� � 1 �x2

4�

x4

64�

x6

2304

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318 Chapter 9 Infinite Series

72. —CONTINUED

(b)

(c)

(d)

Note: J0��x� � �J1�x� �J1�x� � � ��

k�0

��1�kx2k�1

22k�1k!�k � 1�! � ��

k�0

��1�k�1x2k�1

22k�1k!�k � 1�!

J0��x� � ��

k�0 ��1�k�12�k � 1�x2k�1

22k�2�k � 1�!�k � 1�! � ��

k�0

��1�k�1x2k�1

22k�1k!�k � 1�! −6 6

−4

4P7�x� �x2

�1

16 x3 �

1384

x5 �1

18,432 x7

� ��

k�0 ��1�kx2k�3���1� � 1

22k�1k!�k � 1�! � 0

� ��

k�0

��1�k�1x2k�3

22k�1k!�k � 1�! � ��

k�0

��1�kx2k�3

22k�1k!�k � 1�!

� ��

k�1

��1�kx2k�1

22k�1�k � 1�!k!� �

k�0

��1�k x2k�3

22k�1k!�k � 1�!

� ��

k�1 ��1�kx2k�14k�k � 1�

22k�1k!�k � 1�! � ��

k�0

��1�kx2k�3

22k�1k!�k � 1�!

� ��

k�1 ��1�kx2k�1��2k � 1��2k� � �2k � 1� � 1

22k�1k!�k � 1�! � ��

k�0

��1�kx2k�3

22k�1k!�k � 1�!

�x2

� ��

k�1

��1�kx2k�1

22k�1k!�k � 1�!� � ��

k�0

��1�kx2k�3

22k�1k!�k � 1�!

� ��

k�1 ��1�k�2k � 1��2k�x2k�1

22k�1k!�k � 1�! �x2

� ��

k�1 ��1�k�2k � 1�x2k�1

22k�1k!�k � 1�!

� ��

k�0

��1�kx2k�3

22k�1k!�k � 1�! � ��

k�0

��1�k x2k�1

22k�1k!�k � 1�!

x2J1� � xJ1� � �x2 � 1�J1 � ��

k�1 ��1�k �2k � 1��2k�x2k�1

22k�1k!�k � 1�! � ��

k�0 ��1�k�2k � 1�x2k�1

22k�1k!�k � 1�!

J1��x� � ��

k�1 ��1�k �2k � 1��2k�x2k�1

22k�1k!�k � 1�!

J1��x� � ��

k�0 ��1�k �2k � 1�x2k

22k�1k!�k � 1�!

J1�x� � ��

k�0

��1�k x2k�1

22k�1k!�k � 1!�

73.

−2

2

−2� 2�

f �x� � ��

n�0��1�n

x2n

�2n�! � cos x 74.

� sin x−

−2

� �

2 f �x� � ��

n�0��1�n

x2n�1

�2n � 1�!

75. Geometric

−1 10

3

�1

1 � ��x� �1

1 � x for �1 < x < 1

f �x� � ��

n�0��1�nxn � �

n�0��x�n 76.

−2.5 2.5

− �2

�2

f �x� � ��

n�0��1�n

x2n�1

2n � 1� arctan x, �1 ≤ x ≤ 1

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Section 9.8 Power Series 319

77.

(a)

(c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum.

−1 60

1.80

�1

1 � �3�8� �85

� 1.6 ��

n�0�3�4

2 �n

� ��

n�0�3

8�n

��

n�0�x

2�n

(b)

(d) �N

n�0�3

2�n

> M

0.60−1 6

1.10

�1

1 � ��3�8� �8

11 0.7272

��

n�0��3�4

2 �n

� ��

n�0��

38�

n

M 10 100 1000 10,000

N 4 9 15 21

78. converges on

(a)

(c) The alternating series converges more rapidly. The partial sums in (a) approach the sum 2 from below. The partial sums in (b) alternate sides of the horizontal line y �

23.

3

60

−1

��

n�0�3�1

6��n

� ��

n�0�1

2�n

�1

1 � �1�2� � 2x �16

:

��13

, 13�.�

n�0�3x�n

(b)

(d) �N

n�0�3 �

23�

n

� �N

n�02n > M

1.5

6

−0.5

−1

�1

1 � �1�2� �23

��

n�0�3��

16��

n

� ��

n�0��

12�

n

x � �16

:

M 10 100 1000 10,000

N 3 6 9 13

79. False;

converges for but diverges for x � �2.x � 2

��

n�0 ��1�nxn

n2n

81. True; the radius of convergence is for both series.R � 1

80. True; if

converges for then we know that it must convergeon ��2, 2.

x � 2,

��

n�0 an x

n

82. True

� ��

n�0

an

n � 1 �1

0 f �x� dx � �1

0 ��

n�0 an x

n� dx � ��

n�0 an x

n�1

n � 1 �1

0

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320 Chapter 9 Infinite Series

83.

Thus, the series converges for all x : R � �.

limn→�

�an�1

an � � limn→�

� �n � 1 � p�!�n � 1�!�n � 1 � q�! x

n�1� �n � p�!n!�n � q�! x

n� � limn→�

� �n � 1 � p�x�n � 1��n � 1 � q�� � 0

84. (a)

Since each series is geometric,

(b) For g�x� �1

1 � x2 � 2x1

1 � x2 �1 � 2x1 � x2.�x� < 1,

R � 1.

limn→�

S2n � ��

n�0x2n � 2x �

n�0x2n

� �1 � x2 � x 4 � . . . � x2n� � 2x�1 � x2 � x 4 � . . . � x2n�

S2n � 1 � 2x � x2 � 2x3 � x 4 � . . . � x2n � 2x2n�1

g�x� � 1 � 2x � x2 � 2x3 � x 4 � . . .

85. (a)

Each series is geometric, and the interval of convergence is

(b) For f �x� � c01

1 � x3 � c1x1

1 � x3 � c2x2 11 � x3 �

c0 � c1x � c2x2

1 � x3 .�x� < 1,

��1, 1�.R � 1,

limn→�

S3n � c0 ��

n�0x3n � c1x �

n�0x3n � c2x2 �

n�0x3n

S3n � c0�1 � x3 � . . . � x3n� � c1x�1 � x3 � . . . � x3n� � c2x2�1 � x3 � . . . � x3n�

� c0 � c1x � c2x2 � c0x

3 � c1x4 � c2x

5 � c0x6 � . . .

f �x� � ��

n�0cnxn, cn�3 � cn

86. For the series

For the series

limn→�

�bn�1

bn � � limn→�

�cn�1x2n�2

cn x2n � � limn→�

�cn�1

cn x

2� < 1 ⇒ �x2� < � cn

cn�1� � R ⇒ �x� < �R.

�cn x2n,

limn→�

�an�1

an � � limn→�

�cn�1xn�1

cnxn � � limn→�

�cn�1

cn

x� < 1 ⇒ �x� < � cn

cn�1� � R

�cn xn,

87. At the series is which converges.

At the series is which diverges.

Hence, at we have converges, diverges.

The series converges conditionally. ⇒

��

n�0�cn��R�n� � �

n�0cnRn�

n�0cn��R�nx0 � R,

��

n�0cnRn,x0 � R,

��

n�0cn�x0 � x0 � R�n � �

n�0cn��R�nx0 � R,

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Section 9.9 Representation of Functions by Power Series

Section 9.9 Representation of Functions by Power Series 321

1. (a)

This series converges on

(b)

x3

8�

x 4

16

x3

8

x2

4�

x3

8

x2

4

x2

�x2

4

x2

1 �x2

2 � x ) 1

12

�x4

�x2

8�

x3

16� . . .

��2, 2�.

� ��

n�0 12

�x2�

n

� ��

n�0

xn

2n�1

1

2 � x�

1�21 � �x�2� �

a1 � r

2. (a)

This series converges on

(b)

4x3

125�

4x 4

625

4x3

125

425

x2 �4x3

125

425

x2

45

x �4

25x2

45

x

4 �45

x

5 � x ) 4

45

�4

25x �

4125

x2 �4x3

625� . . .

��5, 5�.

� ��

n�0 45

�x5�

n

� ��

n�0

4xn

5n�1

f �x� �4

5 � x�

4�51 � x�5

�a

1 � r

3. (a)

This series converges on

(b)

�x3

8�

x 4

16

�x3

8

x2

4�

x3

8

x2

4

�x2

�x2

4

�x2

1 �x2

2 � x ) 1

12

�x4

�x2

8�

x3

16� . . .

��2, 2�.

� ��

n�0 12

��x2�

n

� ��

n�0 ��1�n xn

2n�1

1

2 � x�

1�21 � ��x�2� �

a1 � r

4. (a)

This series converges on

(b)

��x3 � x 4�x3

x2 � x3x2

�x � x2�x

1 � x1 � x ) 1

1 � x � x2 � x3 � . . .��1, 1�.

� ��

n�0��x�n � �

n�0��1�n xn

1

1 � x�

11 � ��x� �

a1 � r

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322 Chapter 9 Infinite Series

5. Writing in the form we have

which implies that and Therefore, the power series for is given by

� ��

n�0 �x � 5�n

��3�n�1, �x � 5� < 3 or 2 < x < 8.

1

2 � x� �

n�0 arn � �

n�0 �

13��

13

�x � 5�n

f �x�r � ��1�3��x � 5�.a � �1�3

12 � x

�1

�3 � �x � 5� ��1�3

1 � �1�3��x � 5�

a��1 � r�,f �x�

7. Writing in the form we have

which implies that and Therefore, thepower series for is given by

� �3 ��

n�0�2x�n, �2x� < 1 or �

12

< x < 12

.

3

2x � 1� �

n�0 arn � �

n�0 ��3��2x�n

f �x�r � 2x.a � �3

32x � 1

��3

1 � 2x�

a1 � r

a��1 � r�,f �x�

6. Writing in the form we have

Therefore, the power series for is given by

or �9 < x < 5�x � 2� < 7

� ��

n�0

4�x � 2�n

7n�1 .

4

5 � x� �

n�0arn � �

n�0

47�

17

�x � 2��n

f �x�

45 � x

�4

7 � �x � 2� �4�7

1 � 1�7�x � 2� �a

1 � r.

a��1 � r�,f �x�

8. Writing in the form we have

which implies that and Therefore, the power series for is given by

or 12

< x < 72

.�x � 2� < 32

� ��

n�0 ��2�n�x � 2�n

3n ,

3

2x � 1� �

n�0 arn � �

n�0��

23

�x � 2�n

,

f �x�r � ��2�3��x � 2�.a � 1

32x � 1

�3

3 � 2�x � 2� �1

1 � �2�3��x � 2� �a

1 � r

a��1 � r�,f �x�

9. Writing in the form we have

which implies that and Therefore, the power series for is given by

or �172

< x < 52

.�x � 3� < 112

� � ��

n�0 2n�x � 3�n

11n�1 ,

1

2x � 5� �

n�0 arn � �

n�0��

111��

211

�x � 3�n

f �x�r � �2�11��x � 3�.a � �1�11

��1�11

1 � �2�11��x � 3� �a

1 � r

1

2x � 5�

�111 � 2�x � 3�

a��1 � r�,f �x� 10. Writing in the form we have

which implies that and Therefore,the power series for is given by

or �52

< x < 52

.�x� < 52

12x � 5

� ��

n�0 ar n � �

n�0��

15��

25

x�n

� � ��

n�0 2n xn

5n�1,

f �x�r � �2�5�x.a � �1�5

12x � 5

�1

�5 � 2x�

�1�51 � �2�5�x �

a1 � r

a��1 � r�,f �x�

11. Writing in the form we have

which implies that and Therefore,the power series for is given by

or �2 < x < 2.�x� < 2

� 3 ��

n�0 ��1�n xn

2n�1 �32

��

n�0 ��

x2�

n

,

3

x � 2� �

n�0 arn � �

n�0 32

��12

x�n

f �x�r � ��1�2�x.a � 3�2

3x � 2

�3

2 � x�

3�21 � �1�2�x �

a1 � r

a��1 � r�,f �x� 12. Writing in the form we have

which implies that and Therefore, the power series for is given by

or �23

< x < 143

.�x � 2� < 83

�12

��

n�0 ��3�n �x � 2�n

8n ,

4

3x � 2� �

n�0 arn � �

n�0 12��

38

�x � 2�n

f �x�r � ��3�8��x � 2�.a � 1�2

43x � 2

�4

8 � 3�x � 2� �1�2

1 � �3�8��x � 2� �a

1 � r

a��1 � r�,f �x�

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Section 9.9 Representation of Functions by Power Series 323

13.

Writing as a sum of two geometric series, we have

The interval of convergence is since

� limn→�

��1 � ��2�n�1�x�2 � ��2�n�1 � � �x�.lim

n→� �un�1

un � � limn→�

��1 � ��2�n�1�xn�1

��2�n�1 ���2�n

�1 � ��2�n� xn��1 < x < 1

� ��

n�0 � 1

��2�n � 1xn. 3x

x2 � x � 2� �

n�0��

12

x�n

� ��

n�0��1��x�n

f �x�

�1

1 � �1�2�x ��1

1 � x

3xx2 � x � 2

�2

x � 2�

1x � 1

�2

2 � x�

1�1 � x

14.

Writing as a sum of two geometric series, we have

or �12

< x < 12

.�x� < 12

4x � 72x2 � 3x � 2

� ��

n�0�3

2���12

x�n

� ��

n�0 2�2x�n � �

n�0�3��1�n

2n�1 � 2n�1xn,

f �x�

4x � 72x2 � 3x � 2

�3

x � 2�

22x � 1

�3

2 � x�

2�1 � 2x

�3�2

1 � �1�2�x �2

1 � 2x

16. First finding the power series for we have

Now replace x with

The interval of convergence is or since

limn→�

�un�1

un � � limn→�

���1�n�1x2n�2

4n�1 �4n

��1�n x2n� � ��x2

4 � � �x2�4

.

�2 < x < 2�x2� < 4

4

4 � x2 � ��

n�0 ��1�n x2n

4n .

x2.

1

1 � �1�4�x � ��

n�0��

14

x�n

� ��

n�0 ��1�n xn

4n

4��4 � x�,

15.

Writing as a sum of two geometric series, we have

The interval of convergence is or since limn→�

�un�1

un � � limn→�

�2x2n�2

2x2n � � �x2�.�1 < x < 1�x2� < 1

21 � x2 � �

n�0 xn � �

n�0��x�n � �

n�0�1 � ��1�n�xn � �

n�0 2x2n.

f �x�

21 � x2 �

11 � x

�1

1 � x

17.

(See Exercise 15.) � 2 � 0x � 2x2 � 0x3 � 2x4 � 0x5 � 2x6 � . . . � ��

n�0 2x2n, �1 < x < 1

h�x� ��2

x2 � 1�

11 � x

�1

1 � x� �

n�0��1�n xn � �

n�0xn � �

n�0��1�n � 1�xn

11 � x

� ��

n�0��1�n ��x�n � �

n�0��1�2n xn � �

n�0 xn

11 � x

� ��

n�0��1�n xn

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324 Chapter 9 Infinite Series

18.

� � ��

n�0x2n�1, 1 < x < 1 �

12 �

n�0��2�x2n�1

�12

0 � 2x � 0x2 � 2x3 � 0x4 � 2x5 � . . .� �12 �

n�0��1�n � 1�xn

�12 �

n�0��1�nxn �

12 �

n�0xn h�x� �

xx2 � 1

�1

2�1 � x� �1

2�1 � x�

19. By taking the first derivative, we have Therefore,

� ��

n�0��1�n�1�n � 1�xn, �1 < x < 1.

�1

�x � 1�2 �ddx� �

n�0��1�n xn � �

n�1��1�n nxn�1

ddx�

1x � 1 �

�1�x � 1�2.

20. By taking the second derivative, we have Therefore,

�ddx

� ��

n�1��1�nnxn�1 � �

n�2��1�n n�n � 1�xn�2 � �

n�0��1�n �n � 2��n � 1�xn, �1 < x < 1.

2

�x � 1�3 �d 2

dx2 � ��

n�0��1�n xn

d 2

dx2 � 1x � 1 �

2�x � 1�3.

21. By integrating, we have Therefore,

To solve for C, let and conclude that Therefore,

ln�x � 1� � ��

n�0 ��1�n xn�1

n � 1, �1 < x ≤ 1.

C � 0.x � 0

�1 < x ≤ 1.ln�x � 1� � �� ��

n�0��1�n xn dx � C � �

n�0 ��1�n xn�1

n � 1,

� 1

x � 1 dx � ln�x � 1�.

22. By integrating, we have

and

Therefore,

To solve for C, let and conclude that Therefore,

ln�1 � x2� � � ��

n�0 x2n�2

n � 1, �1 < x < 1.

C � 0.x � 0

� C � ��

n�0 ��1�n � 1�xn�1

n � 1� C � �

n�0 �2x2n�2

2n � 2� C � �

n�0 ��1�x2n�2

n � 1

� � � ��

n�0��1�n xn dx � � � �

n�0 xn dx � �C1 � �

n�0 ��1�n xn�1

n � 1 � �C2 � ��

n�0

xn�1

n � 1

ln�1 � x2� � � 1

1 � x dx � �

11 � x

dx

f �x� � ln�1 � x2� � ln�1 � x� � �ln�1 � x��.

� 1

1 � x dx � �ln�1 � x� � C2.�

11 � x

dx � ln�1 � x� � C1

23.1

x2 � 1� �

n�0��1�n�x2�n � �

n�0��1�n x2n, �1 < x < 1

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Section 9.9 Representation of Functions by Power Series 325

24. (See Exercise 23.)

Since we have

To solve for C, let and conclude that Therefore,

�1 ≤ x ≤ 1.ln�x2 � 1� � ��

n�0 ��1�n x2n�2

n � 1,

C � 0.x � 0

�1 ≤ x ≤ 1.ln�x2 � 1� � � � ��

n�0��1�n 2x2n�1 dx � C � �

n�0 ��1�n x2n�2

n � 1,

ddx

�ln�x2 � 1�� �2x

x2 � 1,

� ��

n�0��1�n 2x2n�1

2x

x 2 � 1� 2x �

n�0��1�n x2n

25. Since, we have �12

< x < 12

.1

4x2 � 1� �

n�0��1�n�4x2�n � �

n�0��1�n 4n x2n � �

n�0��1�n �2x�2n,

1x � 1

� ��

n�0��1�n xn,

26. Since we can use the result of Exercise 25 to obtain

To solve for C, let and conclude that Therefore,

�12

< x ≤ 12

.arctan�2x� � 2 ��

n�0 ��1�n 4nx 2n�1

2n � 1,

C � 0.x � 0

�12

< x ≤ 12

.arctan�2x� � 2� 1

4x2 � 1 dx � 2�� �

n�0��1�n 4n x2n dx � C � 2 �

n�0 ��1�n 4n x2n�1

2n � 1,

� 1

4x2 � 1 dx �

12

arctan�2x�,

27.

−4 8

−3

S3f

S2

5

x �x2

2 ≤ ln�x � 1� ≤ x �

x2

2�

x3

3x 0.0 0.2 0.4 0.6 0.8 1.0

0.000 0.180 0.320 0.420 0.480 0.500

0.000 0.182 0.336 0.470 0.588 0.693

0.000 0.183 0.341 0.492 0.651 0.833x �x2

2�

x3

3

ln�x � 1�

x �x2

2

28.

≤ x �x2

2�

x3

3�

x 4

4�

x5

5−4 8

−3

f

S5

S4

5

x �x2

2�

x3

3�

x 4

4 ≤ ln�x � 1�

x 0.0 0.2 0.4 0.6 0.8 1.0

0.0 0.18227 0.33493 0.45960 0.54827 0.58333

0.0 0.18232 0.33647 0.47000 0.58779 0.69315

0.0 0.18233 0.33698 0.47515 0.61380 0.78333x �x 2

2�

x3

3�

x 4

4�

x5

5

ln�x � 1�

x �x 2

2�

x3

3�

x 4

4

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326 Chapter 9 Infinite Series

29.

(a)

(b) From Example 4,

� ln x, 0 < x ≤ 2, R � 1.

��

n�1

��1�n�1�x � 1�n

n� �

n�0

��1�n�x � 1�n�1

n � 1

3

−3

0 4

n = 1n = 3

n = 6n = 2

��

n�1

��1�n�1�x � 1�n

n�

�x � 1�1

��x � 1�2

2�

�x � 1�3

3� . . .

30.

(a)

(b) ��

n�0

��1�nx2n�1

�2n � 1�! � sin x, R � �

4

6

−4

−6

��

n�0

��1�nx2n�1

�2n � 1�! � x �x3

3!�

x5

5!� . . .

(c)

(d) This is an approximation of

The error is approximately 0.

sin�12�.

��

n�0

��1�n�1�2�2n�1

�2n � 1�! 0.4794255386

31. line

Matches (c)

g�x� � x 32. cubic with 3 zeros.

Matches (d)

g�x� � x �x3

3,

33.

Matches (a)

g�x� � x �x3

3�

x5

5 34.

Matches (b)

g�x� � x �x3

3�

x5

5�

x7

7,

In Exercises 35-38, arctan x � ��

n�0��1�n

x2n�1

2n � 1.

35.

Since we can approximate the series by its first two terms: arctan 14 14 �

1192 0.245.1

5120 < 0.001,

�14

�1

192�

15120

� . . . arctan 14

� ��

n�0��1�n

�1�4�2n�1

2n � 1� �

n�0

��1�n

�2n � 1�42n�1

(c)

(d) This is an approximation of The error is approxi-mately 0. The error is less than the first omitted term,1��51 � 251� 8.7 � 10�18.�

ln�1

2�.��

n�1

��1�n�1��1�2�n

n� �

n�1

��1�2�n

n �0.693147

x � 0.5:

36.

Since we can approximate the series by its first two terms: 0.134.177,147�230,686,720 < 0.001,

�27192

�2187

344,064�

177,147230,686,720

�3�4

0 arctan x2 dx � �

n�0��1�n

�3�4�4n�3

�4n � 3��2n � 1� � ��

n�0��1�n 34n�3

�4n � 3��2n � 1�44n�3

� arctan x2 dx � ��

n�0��1�n

x 4n�3

�4n � 3��2n � 1� � C, C � 0

arctan x2 � ��

n�0��1�n

x4n�2

2n � 1

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Section 9.9 Representation of Functions by Power Series 327

37.

Since we can approximate the series by its first term: �1�2

0 arctan x2

x dx 0.125.

11152

< 0.001,

�1�2

0 arctan x2

x dx � �

n�0��1�n

1�4n � 2��2n � 1�24n�2 �

18

�1

1152� . . .

� arctan x2

x dx � �

n�0��1�n

x 4n�2

�4n � 2��2n � 1� � C �Note: C � 0�

arctan x2

x�

1x �

n�0��1�n

�x2�2n�1

2n � 1� �

n�0��1�2

x 4n�1

2n � 1

38.

Since we can approximate the series by its first term: �1�2

0x2 arctan x dx 0.016.

11152

< 0.001,

�1�2

0 x2 arctan x dx � �

n�0��1�n

1�2n � 4��2n � 1�22n�4 �

164

�1

1152� . . .

� x2 arctan x dx � ��

n�0��1�n

x2n�4

�2n � 4��2n � 1�

x2 arctan x � ��

n�0��1�n

x2n�3

2n � 1

In Exercises 39–43, use 1

1 � x� �

n�0 xn, �x� < 1.

39.1

�1 � x�2 �ddx�

11 � x �

ddx� �

n�0xn � �

n�1nxn�1, �x� < 1 40.

x�1 � x�2 � x �

n�1nxn�1 � �

n�1nxn, �x� < 1

41.

� ��

n�0�2n � 1�xn, �x� < 1

� ��

n�1n�xn�1 � xn�, �x� < 1

1 � x

�1 � x�2 �1

�1 � x�2 �x

�1 � x�2 42.

(See Exercise 41.)

x�1 � x��1 � x�2 � x �

n�0�2n � 1�xn � �

n�0�2n � 1�xn�1, �x� < 1

43.

Since the probability of obtaining a head on a single toss is it is expected that, on average, a head will be obtained in two tosses.

12 ,

�12

1

1 � �1�2��2 � 2

E�n� � ��

n�1 nP�n� � �

n�1 n�1

2�n

�12

��

n�1 n�1

2�n�1

P�n� � �12�

n

44. (a)

(b) �9

100�

11 � �9�10��2 � 9

110

��

n�1 n� 9

10�n

�9

100 �

n�1 n� 9

10�n�1

13

��

n�1 n�2

3�n

�29

��

n�1 n�2

3�n�1

�29

1

1 � �2�3��2 � 2

In Exercise 44,1

1 � x� �

n�0 xn, �x�1.

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328 Chapter 9 Infinite Series

45. Replace x with ��x�. 46. Replace x with x2. 47. Replace x with andmultiply the series by 5.

��x� 48. Integrate the series andmultiply by ��1�.

50. (a) From Exercise 49, we have

(b)

(see part (a).)4 arctan 15

� arctan 1

239� arctan

120119

� arctan 1

239�

4

4 arctan 15

� 2 arctan 15

� 2 arctan 15

� arctan 5

12� arctan

512

� arctan� 2�5�12�1 � �5�12�2 � arctan

120119

2 arctan 15

� arctan 15

� arctan 15

� arctan � 2�1�5�1 � �1�5�2 � arctan

1024

� arctan 5

12

� arctan � �120�119� � ��1�239�1 � �120�119���1�239� � arctan �28,561

28,561� � arctan 1 ��

4

arctan 120119

� arctan 1

239� arctan

120119

� arctan ��1

239�

51. (a)

(b) � � 8 arctan 12

� 4 arctan 17

8�12

��0.5�3

3�

�0.5�5

5�

�0.5�7

7 � 4�17

��1�7�3

3�

�1�7�5

5�

�1�7�7

7 3.14

2 arctan 12

� arctan 17

� arctan 43

� arctan ��17� � arctan� �4�3� � �1�7�

1 � �4�3��1�7� � arctan 2525

� arctan 1 ��

4

2 arctan 12

� arctan 12

� arctan 12

� arctan�12 �

12

1 � �1�2�2 � arctan 43

49. Let Then,

Therefore, for xy � 1.arctan x � arctan y � arctan� x � y1 � xy� arctan� x � y

1 � xy� � .

x � y

1 � xy� tan

tan�arctan x� � tan�arctan y�

1 � tan�arctan x� tan�arctan y� � tan

tan�arctan x � arctan y� � tan

arctan x � arctan y � .

52. (a)

(b)

� 4�12

��1�2�3

3�

�1�2�5

5�

�1�2�7

7 � 4�13

��1�3�3

3�

�1�3�5

5�

�1�3�7

7 4�0.4635� � 4�0.3217� 3.14

� � 4�arctan 12

� arctan 13

arctan 12

� arctan 13

� arctan� �1�2� � �1�3�1 � �1�2��1�3� � arctan�5�6

5�6� ��

4

54. From Exercise 53, we have

� ln�13

� 1� � ln 43

0.2877.

��

n�1��1�n�1

13n n

� ��

n�1 ��1�n�1�1�3�n

n

53. From Exercise 21, we have

Thus,

� ln�12

� 1� � ln 32

0.4055.

��

n�1��1�n�1

12n n

� ��

n�1 ��1�n�1�1�2�n

n

� ��

n�1 ��1�n�1xn

n.

ln�x � 1� � ��

n�0 ��1�n xn�1

n � 1� �

n�1 ��1�n�1xn

n

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Section 9.10 Taylor and Maclaurin Series 329

56. From Example 5, we have

� arctan 1 ��

4� 0.7854

��

n�0��1�n

12n � 1

� ��

n�0��1�n

�1�2n�1

2n � 1

arctan x � ��

n�0��1�n

x2n�1

2n � 1.55. From Exercise 53, we have

� ln�25

� 1� � ln 75

� 0.3365.

��

n�1��1�n�1

2n

5n n� �

n�1 ��1�n�1�2�5�n

n

57. From Exercise 56, we have

� arctan 12

� 0.4636.

��

n�0��1�n

122n�1�2n � 1� � �

n�0��1�n

�1�2�2n�1

2n � 1

58. From Exercise 56, we have

� arctan 13

� 0.3218.

� ��

n�0��1�n �1�3�2n�1

2n � 1

��

n�1��1�n�1

132n�1�2n � 1� � �

n�0��1�n

132n�1�2n � 1�

59. is an odd function (symmetric to the origin).f �x� � arctan x 60. The approximations of degree have relative extrema.�4n � 1, n � 1, 2, . . .�

3, 7, 11, . . . ,

61. The series in Exercise 56 converges to its sum at a slowerrate because its terms approach 0 at a much slower rate.

62. Because the radius of

convergence is the same, 3.

ddx� �

n�0anxn � �

n�1nanxn�1,

64. Using a graphing utility, you obtain the following partialsums for the left hand side. Note that

n � 1: S1 � 0.3183098862

n � 0: S0 � 0.3183098784

1�� � 0.3183098862.

65.

From Example 5 we have

� 3 ��

6� � 0.9068997

� 3 arctan � 13�

� 3 ��

n�0

��1�n�1�3�2n�1

2n � 1

��

n�0

��1�n

3n�2n � 1� � ��

n�0

��1�n

�3�2n�2n � 1� 33

arctan x � ��

n�0��1�n

x2n�1

2n � 1.

��

n�0

��1�n

3n�2n � 1�66.

� sin��

3� �32

� 0.866025

��

n�0

��1�n� 2n�1

32n�1�2n � 1�! � ��

n�0��1�n

���3�2n�1

�2n � 1�!

63. Because the first series is the derivative of the secondseries, the second series converges for andperhaps at the endpoints, x � 3 and x � �5.�

�x � 1� < 4 �

Section 9.10 Taylor and Maclaurin Series

1. For we have:

e2x � 1 � 2x �4x2

2!�

8x3

3!�

16x 4

4!� . . . � �

n�0 �2x�n

n!.

f �n��x� � 2n e2x ⇒ f �n��0� � 2n

f �x� � e2x

c � 0,

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4. For we have:

and so on. Therefore we have:

�22

���

n�0 ��1�n�n�1��2 x � ���4��n�1

�n � 1�! � 1�.

�22

�1 � �x ��

4� � x � ���4��2

2!�

x � ���4��3

3!�

x � ���4��4

4!� . . .

sin x � ��

n�0 f �n����4� x � ���4��n

n!

f �4��x� � sin x f �4���

4� �22

f���x� � �cos x f����

4� � �22

f ��x� � �sin x f ���

4� � �22

f��x� � cos x f���

4� �22

f �x� � sin x f��

4� �22

c � ��4,

330 Chapter 9 Infinite Series

2. For we have:

e3x � 1 � 3x �9x2

2!�

27x3

3!� . . . � �

n�0

�3x�n

n!.

f �n��x� � 3ne3x ⇒ f �n��0� � 3n

f �x� � e3x

c � 0,

3. For we have:

and so on. Therefore, we have:

[Note: 1, 1, 1, . . .]�1,�1,�1,�1,��1�n�n�1��2 � 1,

�22

��

n�0 ��1�n�n�1��2 x � ���4��n

n!.

�22

�1 � �x ��

4� � x � ���4��2

2!�

x � ���4��3

3!�

x � ���4��4

4!� . . .

cos x � ��

n�0 f �n����4� x � ���4��n

n!

f �4��x� � cos�x� f �4���

4� �22

f����x� � sin�x� f�����

4� �22

f � �x� � �cos�x� f ���

4� � �22

f��x� � �sin�x� f���

4� � �22

f �x� � cos�x� f ��

4� �22

c � ��4,

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Section 9.10 Taylor and Maclaurin Series 331

5. For we have,

and so on. Therefore, we have:

� ��

n�0 ��1�n

�x � 1�n�1

n � 1.

� �x � 1� ��x � 1�2

2�

�x � 1�3

3�

�x � 1�4

4�

�x � 1�5

5� . . .

� 0 � �x � 1� ��x � 1�2

2!�

2�x � 1�3

3!�

6�x � 1�4

4!�

24�x � 1�5

5!� . . .

ln x � ��

n�0 f �n��1��x � 1�n

n!

f �5��x� �24x5 f �5��1� � 24

f �4��x� � �6x 4

f �4��1� � �6

f����x� �2x3 f����1� � 2

f ��x� � �1x2 f ��1� � �1

f��x� �1x f��1� � 1

f �x� � ln x f �1� � 0

c � 1,

6. For we have:

ex � ��

n�0 f �n��1��x � 1�n

n!� e �1 � �x � 1� �

�x � 1�2

2!�

�x � 1�3

3!�

�x � 1�4

4!� . . . � e �

n�0 �x � 1�n

n!.

f �n��x� � ex ⇒ f �n��1� � e

f �x� � ex

c � 1,

7. For we have:

and so on. Therefore, we have:

� 2x �8x3

3!�

32x5

5!�

128x7

7!� . . . � �

n�0 ��1�n�2x�2n�1

�2n � 1�! .

� 0 � 2x �0x2

2!�

8x3

3!�

0x 4

4!�

32x5

5!�

0x6

6!�

128x7

7!� . . . sin 2x � �

n�0 f �n��0�xn

n!

f �7��x� � �128 cos 2x f �7��0� � �128

f �6��x� � �64 sin 2x f �6��0� � 0

f �5��x� � 32 cos 2x f �5��0� � 32

f �4��x� � 16 sin 2x f �4��0� � 0

f���x� � �8 cos 2x f���0� � �8

f ��x� � �4 sin 2x f ��0� � 0

f��x� � 2 cos 2x f��0� � 2

f �x� � sin 2x f �0� � 0

c � 0,

Page 101: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

332 Chapter 9 Infinite Series

10. For we have;

tan�x� � ��

n�0 f �n��0�xn

n!� x �

2x3

3!�

16x5

5!� . . . � x �

x3

3�

215

x5 � . . ..

f �5��x� � 8 2 sec6�x� � 11 sec4�x� tan2�x� � 2 sec2�x� tan4�x�� f �5��0� � 16

f �4��x� � 8 sec4�x� tan�x� � sec2�x� tan3�x�� f �4��0� � 0

f���x� � 2 sec4�x� � 2 sec2�x� tan2�x�� f���0� � 2

f ��x� � 2 sec2�x� tan�x� f ��0� � 0

f��x� � sec2�x� f��0� � 1

f �x� � tan�x� f �0� � 0

c � 0,

8. For we have:

and so on. Therefore, we have:

� x2 �x 4

2�

x6

3� . . . � �

n�0 ��1�n x2n�2

n � 1.

� 0 � 0x �2x2

2!�

0x3

3!�

12x 4

4!�

0x5

5!�

240x6

6!� . . . ln�x2 � 1� � �

n�0 f �n��0�xn

n!

f �6��x� ��240�5x6 � 15x 4 � 15x2 � 1�

�x2 � 1�6 f �6��0� � 240

f �5��x� �48x�x 4 � 10x2 � 5�

�x2 � 1�5 f �5��0� � 0

f �4��x� �12��x 4 � 6x2 � 1�

�x2 � 1�4 f �4��0� � �12

f���x� �4x�x2 � 3��x2 � 1�3 f���0� � 0

f ��x� �2 � 2x2

�x2 � 1�2 f ��0� � 2

f��x� �2x

x2 � 1 f��0� � 0

f �x� � ln�x2 � 1� f �0� � 0

c � 0,

9. For we have:

sec�x� � ��

n�0 f �n��0�xn

n!� 1 �

x2

2!�

5x 4

4!� . . ..

f �4��x� � 5 sec5�x� � 18 sec3�x� tan2�x� � sec�x� tan4�x� f �4��0� � 5

f���x� � 5 sec3�x� tan�x� � sec�x� tan3�x� f���0� � 0

f ��x� � sec3�x� � sec�x� tan2�x� f ��0� � 1

f��x� � sec�x� tan�x� f��0� � 0

f �x� � sec�x� f �0� � 1

c � 0,

Page 102: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.10 Taylor and Maclaurin Series 333

11. The Maclaurin series for is

Because or we have for all z. Hence by Taylor’s Theorem,

Since it follows that as Hence, the Maclaurin series for converges to for all x.cos xcos xn →�.Rn�x� → 0limn→�

�x�n�1

�n � 1�! � 0,

0 ≤ �Rn�x�� � �f �n�1��z��n � 1�!x

n�1� ≤ �x�n�1

�n � 1�!.

� f �n�1��z�� ≤ 1±cos x,f �n�1��x� � ±sin x

��

n�0

��1�n x2n

�2n�! .f �x� � cos x

12. The Maclaurin series for is

Hence, by Taylor’s Theorem,

Since it follows that as

Hence, the Maclaurin Series for converges to for all x.e�2xe�2x

n →�.Rn�x� → 0limn→�

���2�n�1xn�1

�n � 1�! � � limn→�

� �2x�n�1

�n � 1�!� � 0,

0 ≤ �Rn�x�� � �f �n�1��z��n � 1�!x

n�1� � ���2�n�1e�2z

�n � 1�! xn�1�.f �n�1��x� � ��2�n�1e�2x.

��

n�0

��2x�n

n!.f �x� � e�2x

13. The Maclaurin series for is

For fixed ,

The argument is the same if . Hence, the Maclaurin series for converges to for all x.sinh xsinh xf �n�1��x� � cosh x��

0 ≤ �Rn�x�� � � f �n�1��z��n � 1�! xn�1� � � sinh�z�

�n � 1�! xn�1� → 0 as n →�.

xf �n�1��x� � sinh x �or cosh x�.

��

n�0

x2n�1

�2n � 1�!.f �x� � sinh x

14. The Maclaurin series for is

For fixed ,

The argument is the same if . Hence, the Maclaurin series for converges to for all x.cosh xcosh xf �n�1��x� � cosh x��

0 ≤ �Rn�x�� � � f �n�1��z��n � 1�! xn�1� � � sinh�z�

�n � 1�! xn�1� → 0 as n →�.

xf �n�1��x� � sinh x �or cosh x�.

��

n�0

x2n

�2n�!.f �x� � cosh x

15. Since we have

� ��

n�0��1�n �n � 1�xn.

�1 � x��2 � 1 � 2x �2�3�x2

2!�

2�3��4�x3

3!�

2�3��4��5�x4

4!� . . . � 1 � 2x � 3x2 � 4x3 � 5x 4 � . . .

�1 � x��k � 1 � kx �k�k � 1�x2

2!�

k�k � 1��k � 2�x3

3!� . . .,

16. Since we have

� 1 � ��

n�1 1 � 3 � 5 . . . �2n � 1�xn

2n n!.

� 1 �x2

��1��3�x2

222!�

�1��3��5�x3

233!� . . .

�1 � ��x��1�2

� 1 � �12�x �

�1�2��3�2�x2

2!�

�1�2��3�2��5�2�x3

3!� . . .

�1 � x��k � 1 � kx �k�k � 1�x2

2!�

k�k � 1��k � 2�x3

3!� . . .,

Page 103: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

334 Chapter 9 Infinite Series

23.

sin 3x � ��

n�0

��1�n�3x�2n�1

�2n � 1�!

sin x � ��

n�0

��1�nx2n�1

�2n � 1�! 24.

� 1 �16x2

2!�

256x4

4!� . . .

cos 4x � ��

n�0

��1�n�4x�2n

�2n�! � ��

n�0

��1�n42nx2n

�2n�!

cos x � ��

n�0

��1�nx2n

�2n�! � 1 �x2

2!�

x4

4!�

x6

6!� . . .

26.

� 2x3 �2x9

3!�

2x15

5!� . . .

� 2�x3 �x9

3!�

x15

5!� . . .

2 sin x3 � 2 ��

n�0

��1�n�x3�2n�1

�2n � 1�!

sin x � ��

n�0

��1�nx2n�1

�2n � 1�!25.

� 1 �x3

2!�

x6

4!� . . .

� ��

n�0 ��1�n x3n

�2n�!

cos x3�2 � ��

n�0 ��1�n �x3�2�2n

�2n�!

cos x � ��

n�0 ��1�n x2n

�2n�! � 1 �x2

2!�

x 4

4!� . . .

17. and since we have

14 � x2

�12

�1 � ��

n�1 ��1�n 1 � 3 � 5 . . . �2n � 1��x�2�2n

2n n! �12

� ��

n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�x2n

23n�1n!.

�1 � x��1�2 � 1 � ��

n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�xn

2nn!,

14 � x2

� �12��1 � �x

2�2

�1�2

19. Since

we have �1 � x2�1�2 � 1 �x2

2� �

n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�x2n

2n n!.

�1 � x�1�2 � 1 �x2

� ��

n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�xn

2n n!

21.

ex2�2 � ��

n�0 �x2�2�n

n!� �

n�0

x2n

2n n!� 1 �

x 2

2�

x 4

222!�

x6

233!�

x8

244!� . . .

ex � ��

n�0 xn

n!� 1 � x �

x2

2!�

x3

3!�

x 4

4!�

x5

5!� . . .

18.

� 1 �14

x � ��

n�2

��1�n�1 3 � 7 � 11 . . . �4n � 5�4nn!

xn

� 1 �14

x �3

422!x2 �

3 � 7433!

x3 �3 � 7 � 11

444!x 4 � . . .

�1 � x�1�4 � 1 �14

x ��1�4���3�4�

2!x2 �

�1�4���3�4���7�4�3!

x3 � . . .

20. Since

we have �1 � x3�1�2 � 1 �x3

2� �

n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�x3n

2n n!.

�1 � x�1�2 � 1 �x2

� ��

n�2 ��1�n�1 1 � 3 � 5 . . . �2n � 3�xn

2n n!

22.

e�3x � ��

n�0 ��3x�n

n!� �

n�0 ��1�n 3n xn

n!� 1 � 3x �

9x2

2!�

27x3

3!�

81x 4

4!�

243x5

5!� . . .

ex � ��

n�0 xn

n!� 1 � x �

x2

2!�

x3

3!�

x 4

4!�

x5

5!� . . .

Page 104: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.10 Taylor and Maclaurin Series 335

27.

� x �x3

3!�

x5

5!�

x7

7!� . . . � �

n�0

x2n�1

�2n � 1�!

sinh�x� �12

�ex � e�x�

ex � e�x � 2x �2x3

3!�

2x5

5!�

2x7

7!� . . .

e�x � 1 � x �x2

2!�

x3

3!�

x 4

4!�

x5

5!� . . .

ex � 1 � x �x2

2!�

x3

3!�

x 4

4!�

x5

5!� . . . 28.

2 cos h�x� � ex � e�x � ��

n�02

x2n

�2n�!

ex � e�x � 2 �2x2

2!�

2x4

4!� . . .

e�x � 1 � x �x2

2!�

x3

3!� . . .

ex � 1 � x �x2

2!�

x3

3!� . . .

29.

�12�1 � �

n�0 ��1�n�2x�2n

�2n�! �12�1 � 1 �

�2x�2

2!�

�2x�4

4!�

�2x�6

6!� . . .

cos2�x� �12

1 � cos�2x��

30. The formula for the binomial series gives which implies that

� x �x3

2 � 3�

1 � 3x5

2 � 4 � 5�

1 � 3 � 5x7

2 � 4 � 6 � 7� . . . .

� x � ��

n�1 ��1�n1 � 3 � 5 . . . �2n � 1�x2n�1

2n�2n � 1�n!

ln�x � x2 � 1� � � 1

x2 � 1 dx

�1 � x2��1�2 � 1 � ��

n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�x2n

2n n!

�1 � x��1�2 � 1 � ��

n�1 ��1�n 1 � 3 � 5 . . . �2n � 1�xn

2n n!,

31.

� ��

n�0

��1�nx2n�2

�2n � 1�!

� x2 �x4

3!�

x6

5!� . . .

x sin x � x�x �x3

3!�

x5

5!� . . .� 32.

� ��

n�0

��1�nx2n�1

�2n�!

� x �x3

2!�

x5

4!� . . .

x cos x � x�1 �x2

2!�

x4

4!� . . .�

33.

� ��

n�0

��1�nx2n

�2n � 1�!, x 0 � 1 �x2

2!�

x4

4!� . . .

sin x

x�

x � �x3�3!� � �x5�5!� � . . .

x34.

� ��

n�0

�2n�!x2n

�2nn!�2�2n � 1�, x 0

arcsin x

x� �

n�0

�2n�!x2n�1

�2nn!�2�2n � 1� �1x

35.

eix � e�ix

2i� x �

x3

3!�

x5

5!�

x7

7!� . . . � �

n�0 ��1�n x2n�1

�2n � 1�! � sin�x�

eix � e�ix � 2ix �2ix3

3!�

2ix5

5!�

2ix7

7!� . . .

e�ix � 1 � ix ���ix�2

2!�

��ix�3

3!�

��ix�4

4!� . . . � 1 � ix �

x2

2!�

ix3

3!�

x 4

4!�

ix5

5!�

x6

6!� . . .

eix � 1 � ix ��ix�2

2!�

�ix�3

3!�

�ix�4

4!� . . . � 1 � ix �

x2

2!�

ix3

3!�

x 4

4!�

ix5

5!�

x6

6!� . . .

Page 105: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

336 Chapter 9 Infinite Series

40.

−3 3

−3

f P5

3

� x �x2

2�

x3

3�

3x5

40� . . .

� x � �x2 �x2

2 � � �x3

3�

x3

2�

x3

2 � � ��x 4

4�

x 4

3�

x 4

4�

x 4

6 � � �x5

5�

x5

4�

x5

6�

x5

12�

x5

24� � . . .

� �1 � x �x2

2�

x3

6�

x 4

24� . . .��x �

x2

2�

x3

3�

x 4

4�

x5

5� . . .�

f �x� � ex ln�1 � x�

37.

� x � x2 �x3

3�

x5

30� . . .

� x � x2 � �x3

2�

x3

6 � � �x 4

6�

x 4

6 � � � x5

120�

x5

12�

x5

24� � . . .

� �1 � x �x2

2�

x3

6�

x4

24� . . .��x �

x3

6�

x5

120� . . .�

−6 6

−2

P5

f

14 f �x� � ex sin x

36. (See Exercise 35.)

eix � e�ix

2� 1 �

x2

2!�

x 4

4!�

x6

6!� . . . � �

n�0 ��1�n x2n

�2n�! � cos�x�

eix � e�ix � 2 �2x2

2!�

2x 4

4!�

2x6

6!� . . .

38.

� 1 � x �x3

3�

x 4

6� . . .

� 1 � x � �x2

2�

x2

2 � � �x3

6�

x3

2 � � �x 4

24�

x 4

4�

x 4

24� � . . .

� �1 � x �x2

2�

x 4

6�

x 4

24� . . .��1 �

x2

2�

x 4

24� . . .�

−6 6

−40

g

P4

8 g�x� � ex cos x

39.

� x �x2

2�

x3

6�

3x5

40� . . .

� x �x2

2� �x3

3�

x3

2 � � �x 4

4�

x 4

4 � � �x5

5�

x5

6�

x5

24� � . . .

� �1 �x2

2�

x 4

24� . . .��x �

x2

2�

x3

3�

x 4

4�

x5

5� . . .� −3 9

−4

h

P5

4 h�x� � cos x ln�1 � x�

Page 106: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Section 9.10 Taylor and Maclaurin Series 337

41. Divide the series for sin x by

�5x4

6�

5x5

6

�5x4

6�

x5

120

5x3

6�

5x4

6

5x3

6� 0x4

�x2 � x3

�x2 � x3

6

x � x2

1 � x� x � 0x2 � x3

6� 0x4 �

x5

120� . . .

x � x2 �5x3

6�

5x4

6�

�1 � x�.g�x� �sin x

1 � x.

−6 6

−4

g

P4

4

g�x� � x � x2 �5x3

6�

5x 4

6� . . .

42. Divide the series for by

3x4

8�

3x5

8

3x4

8�

x5

120

�x3

3�

x4

3

�x3

3�

x4

24

x2

2�

x3

2

0 �x2

2�

x3

6

1 � x

1 � x� 1 � x �x2

2�

x3

6�

x4

24�

x5

120� . . .

1 �x2

2�

x3

3�

3x4

8� . . .

�1 � x�.exf �x� �ex

1 � x.

−6 6

−6

fP4

6

f �x� � 1 �x2

2�

x3

3�

3x4

8� . . .

43.

Matches (c)

f �x� � x sin x

y � x2 �x 4

3!� x�x �

x3

3!�

45.

Matches (a)

f �x� � xex

y � x � x2 �x3

2!� x�1 � x �

x2

2!�

44.

Matches (d)

f �x� � x cos x

y � x �x3

2!�

x5

4!� x�1 �

x2

2!�

x 4

4!�

46.

Matches (b)

f �x� � x 2 1

1 � x

y � x2 � x3 � x 4 � x2�1 � x � x2�

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338 Chapter 9 Infinite Series

53. Since

we have limx→0

1 � cos xx

� limx→0

��

n�0 ��1�x2n�1

�2n � 2�! � 0.

1 � cos

x�

x2!

�x3

4!�

x5

6!�

x7

8!� . . . � �

n�0

��1�nx2n�1

�2n � 2�!

1 � cos x �x2

2!�

x 4

4!�

x6

6!�

x8

8!� . . . � �

n�0

��1�nx2n�2

�2n � 2�!

cos x � ��

n�0

��1�nx2n

�2n�! � 1 �x2

2!�

x 4

4!�

x6

6!�

x8

8!� . . .

47.

� �x

0 � �

n�0 ��1�n�1t2n�2

�n � 1�! dt � � ��

n�0

��1�n�1 t2n�3

�2n � 3��n � 1�!x

0� �

n�0

��1�n�1x2n�3

�2n � 3��n � 1�!

�x

0 �e�t 2 � 1� dt � �x

0 ���

n�0 ��1�n t2n

n! � � 1 dt

49. Since

we have (10,001 terms)ln 2 � 1 �12

�13

�14

� . . . � ��

n�1��1�n�1

1n

� 0.6931.

�0 < x ≤ 2�ln x � ��

n�0 ��1�n�x � 1�n�1

n � 1� �x � 1� �

�x � 1�2

2�

�x � 1�3

3�

�x � 1�4

4� . . .,

50. Since we have

(4 terms)sin�1� � ��

n�0

��1�n

�2n � 1�! � 1 �13!

�15!

�17!

� . . . � 0.8415.

sin�x� � ��

n�0 ��1�nx2n�1

�2n � 1�! � x �x3

3!�

x5

5!�

x7

7!� . . . ,

52. Since we have

and (6 terms)e � 1

e� 1 � e�1 � 1 �

12!

�13!

�14!

�15!

�17!

� . . . � ��

n�1 ��1�n�1

n!� 0.6321.

e�1 � 1 � 1 �12!

�13!

�14!

�15!

� . . .ex � ��

n�0 xn

n!� 1 � x �

x2

2!�

x3

3!�

x 4

4!�

x5

5!� . . . ,

48.

� x �x 4

8� �

n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�x3n�1

�3n � 1�2nn!

� �t �t4

8� �

n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�t3n�1

�3n � 1�2n n! x

0

�x

0 1 � t3 dt � �x

0 �1 �

t3

2� �

n�2 ��1�n�11 � 3 � 5 . . . �2n � 3�t3n

2n n! dt

51. Since

we have (12 terms)e2 � 1 � 2 �22

2!�

23

3!� . . . � �

n�0 2n

n!� 7.3891.

ex � ��

n�0 xn

n!� 1 � x �

x2

2!�

x3

3!� . . . ,

54. Since

we have limx→0

sin x

x� lim

x→0 �

n�0 ��1�nx2n

�2n � 1�! � 1.

sin x

x� 1 �

x2

3!�

x 4

5!�

x6

7!� . . . � �

n�0 ��1�nx2n

�2n � 1�!

sin x � ��

n�0 ��1�n x2n�1

�2n � 1�! � x �x3

3!�

x5

5!�

x7

7!� . . .

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Section 9.10 Taylor and Maclaurin Series 339

55.

Since we need three terms:

three nonzero

Note: We are using limx→0�

sin xx

� 1.

terms��using �1

0 sin x

x dx � 1 �

13 � 3!

�1

5 � 5!� . . . � 0.9461.

1��7 � 7!� < 0.0001,

�1

0 sin x

x dx � �1

0 � �

n�0 ��1�nx2n

�2n � 1�! dx � � ��

n�0

��1�nx2n�1

�2n � 1��2n � 1�!1

0� �

n�0

��1�n

�2n � 1��2n � 1�!

56.

Since we have

Note: We are using limx→0�

arctan x

x� 1.

�1�2

0 arctan x

x dx � �1

2�

13223 �

15225 �

17227 �

19229� � 0.4872.

1��9229� < 0.0001,

�1�2

0 arctan x

x dx � �1�2

0 �1 �

x2

3�

x 4

5�

x6

7� . . .� dx � �x �

x3

32 �x5

52 �x7

72 � . . .1�2

0

57.

Since we need two terms.

�0.3

0.1 1 � x3 dx � ��0.3 � 0.1� �

18

�0.34 � 0.14� � 0.201.

156 �0.37 � 0.17� < 0.0001,

�0.3

0.1 1 � x3 dx � �0.3

0.1 �1 �

x3

2�

x6

8�

x9

16�

5x12

128� . . .� dx � �x �

x 4

8�

x7

56�

x10

160�

5x13

1664� . . .

0.3

0.1

58.

Since �1�4

0x ln�x � 1� dx �

�1�4�3

3�

�1�4�4

8� 0.00472.

�1�4�5

15< 0.0001,

� �x3

3�

x4

4 � 2�

x5

5 � 3�

x6

6 � 4� . . .

1�4

0 �14

0x ln�x � 1� dx � �1�4

0�x2 �

x3

2�

x4

3�

x5

4� . . .� dx

59.

Since we need five terms.

�1

0 x cos x dx � 2����2�3�2

3�

���2�7�2

14�

���2�11�2

264�

���2�15�2

10,800�

���2�19�2

766,080 � 0.7040.

2���2�23�2��23 � 10!� < 0.0001,

� � ��

n�0 ��1�nx�4n�3��2

�4n � 32 ��2n�!

��2

0

� � ��

n�0 ��1�n2x�4n�3��2

�4n � 3��2n�! ��2

0���2

0 x cos x dx � ���2

0 � �

n�0 ��1�nx�4n�1��2

�2n�! dx

60.

Since we have

�1

0.5 cos x dx � ��1 � 0.5� �

14

�1 � 0.52� �1

72 �1 � 0.53� �

12880

�1 � 0.54� �1

201,600 �1 � 0.5�5 � 0.3243.

1201,600 �1 � 0.55� < 0.0001,

�1

0.5 cos x dx � �1

0.5 �1 �

x2!

�x2

4!�

x3

6!�

x4

8!� . . .� dx � �x �

x2

2�2!� �x3

3�4!� �x 4

4�6!� �x5

5�8!� � . . .1

0.5

61. From Exercise 21, we have

�1

2� �1 �

12 � 1 � 3

�1

22 � 2! � 5�

123 � 3! � 7 � 0.3414.

�1

2� �

n�0

��1�n

2nn!�2n � 1� 1

2� �1

0 e�x2�2 dx �

12�

�1

0 �

n�0 ��1�nx2n

2nn! dx �

12�

� ��

n�0

��1�nx2n�1

2nn!�2n � 1�1

0

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340 Chapter 9 Infinite Series

66.

The polynomial is a reasonable approximation on the interval �0.48, 1.75�.

� 1.3025 ��x � 1�4

4! � 5.5913 ��x � 1�5

5!

P5�x� � 0.7854 � 0.7618�x � 1� � 0.3412 ��x � 1�2

2! � 0.0424 ��x � 1�3

3! −2 4

−1

f

P5

3 f �x� � 3x � arctan x, c � 1

67. See Guidelines, page 680. 68. (odd coefficients are zero)a2n�1 � 0

69. (a) Replace x with (c) Multiply series by x.

(b) Replace x with (d) Replace x with then replace x with and add thetwo together.

�2x,2x,3x.

��x�.

70. The binomial series is The radius of convergence is R � 1.�1 � x�k � 1 � kx �k�k � 1�

2!x2 �

k�k � 1��k � 2�3!

x3 � . . . .

62. From Exercise 21, we have

�8191

26 � 6! � 13�

32,76727 � 7! � 15

�131,071

28 � 8! � 17�

524,28729 � 9! � 19 � 0.1359.

�1

2� �1 �

72 � 1 � 3

�31

22 � 2! � 5�

12723 � 3! � 7

�511

24 � 4! � 9�

204725 � 5! � 11

�1

2� �

n�0 ��1�n�2n�1 � 1�

2nn!�2n � 1�

1

2� �2

1 e�x2�2dx �

12�

�2

1 �

n�0 ��1�nx2n

2nn! dx �

12�

� ��

n�0

��1�nx2n�1

2nn!�2n � 1�2

1

63.

The polynomial is a reasonable approximation on the

interval �34 , 34�.

−3 3

−2

f

P5

2

P5�x� � x � 2x3 �2x5

3

f �x� � x cos 2x � ��

n�0 ��1�n4nx2n�1

�2n�!

65.

The polynomial is a reasonable approximation on the interval 14 , 2�.

P5�x� � �x � 1� ��x � 1�3

24�

�x � 1�4

24�

71�x � 1�5

1920 −2 4

−2

gP5

3 f �x� � x ln x, c � 1

64.

The polynomial is a reasonable approximation on theinterval ��0.60, 0.73�.

−4 8

−4

f

P5

4

P5�x� �x2

2�

x3

4�

7x 4

48�

11x5

96

f �x� � sin x2

ln�1 � x�

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Section 9.10 Taylor and Maclaurin Series 341

71.

� �tan ��x �gx2

2v02 cos2 �

�kgx3

3v03 cos3 �

�k2gx 4

4v04 cos4 �

� . . .

� �tan ��x �gx

kv0 cos ��

gxkv0 cos �

�gx2

2v02 cos2 �

�gkx3

3v03 cos3 �

�gk2x4

4v04 cos4 �

� . . .

� �tan ��x �gx

kv0 cos ��

gk2��

kxv0 cos �

�12�

kxv0 cos ��

2

�13�

kxv0 cos ��

3

�14�

kxv0 cos ��

4

� . . .

y � �tan � �g

kv0 cos ��x �gk2 ln�1 �

kxv0 cos ��

72.

� 3x � 32 ��

n�2

2n xn

n�64�n �16�n�2 � 3x � 32 ��

n�2

xn

n�32�n �16�n�2

� 3x � 32� 22x2

2�64�2 �23x3

3�64�316�

24x 4

4�64�4�16�2 � . . .

y � 3x �32x2

2�64�2�1�2�2 ��1�16��32�x3

3�64�3�1�2�3 ��1�16�2�32�x 4

4�64�4�1�2�4 � . . .

� � 60�, v0 � 64, k �1

16, g � �32

73.

(a) (b)

Let Then

Thus, and we have

(c) This series converges to f at only.x � 0��

n�0 f �n��0�

n! xn � f �0� �

f��0�x1!

�f ��0�x2

2!� . . . � 0 f �x�

f��0� � 0.y � e�� � 0

ln y � limx→0

ln�e�1�x2

x � � limx→0�

��1x2 � ln x � lim

x→0� ��1 � x2 ln x

x2 � ��.

y � limx→0

e�1�x2

x.

f��0� � limx→0

f �x� � f �0�

x � 0� lim

x→0 e�1�x2

� 0x

21 3

2

1

−1−2−3x

y

f �x� � �e�1�x2,

0, x 0 x � 0

74. (a) (b)

From Exercise 8, you obtain:

(c)

(d) The curves are nearly identical for Hence, the integrals nearly agree on that interval.0 < x < 1.

G�x� � �x

0P8�t� dt

F�x� � �x

0 ln�t2 � 1�

t 2 dt

P8 � 1 �x2

2�

x4

3�

x6

4�

x8

5.

P �1x2 �

n�0 ��1�nx2n�2

n � 1� �

n�0 ��1�nx2n

n � 1

0 2

−0.5

1.5f �x� �ln�x2 � 1�

x2 .

x 0.25 0.50 0.75 1.00 1.50 2.00

0.2475 0.4810 0.6920 0.8776 1.1798 1.4096

0.2475 0.4810 0.6924 0.8865 1.6878 9.6063G�x�

F�x�

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342 Chapter 9 Infinite Series

82. Assume is rational. Let and form the following.

Set a positive integer. But,

a contradiction. �1

N � 1 �1 �1

N � 1�

1�N � 1�2 � . . . �

1N � 1� 1

1 � � 1N � 1� �

1N

,

a � N!� 1�N � 1�! �

1�N � 2�! � . . . �

1N � 1

�1

�N � 1��N � 2� � . . . < 1

N � 1�

1�N � 1�2 � . . .

a � N!�e � �1 � 1 � . . . �1N!�,

e � �1 � 1 �12!

� . . . �1

N! �1

�N � 1�! �1

�N � 2�! � . . .

N > qe � p�q

83.

Equating coefficients,

etc.

In general, The coefficients are the Fibonacci numbers.an � an�1 � an�2.

a4 � a3 � a2 � 3,

a3 � a2 � a1 � 0 ⇒ a3 � 2

a2 � a1 � a0 � 0 ⇒ a2 � 1

a1 � a0 � 1 ⇒ a1 � 1

a0 � 0

x � a0 � �a1 � a0�x � �a2 � a1 � a0�x2 � �a3 � a2 � a1�x3 � . . .

x � �1 � x � x2��a0 � a1x � a2x2 � . . .�

g�x� �x

1 � x � x2 � a0 � a1x � a2x2 � . . .

75. By the Ratio Test: which shows that converges for all x.��

n�0 xn

n!lim

n →� � xn�1

�n � 1�! �n!xn� � lim

n →� �x�n � 1

� 0

76.

�ln 3 � 1.098612�

� 1 �1

12�

180

�1

448� 1.098065 ln 3 � ln�1 � 1�2

1 � 1�2� � 2�12��1 �

�1�2�2

3�

�1�2�4

5�

�1�2�6

7

� 2x ��

n�0

x2n

2n � 1, R � 1 � 2x � 2

x3

3� 2

x5

5� . . . �

� �x �x2

2�

x3

3� . . . � � ��x �

x2

2�

x3

3� . . . �

ln�1 � x1 � x� � ln�1 � x� � ln�1 � x�

80.

��91729

� �0.12483

��1�35 � �

��1�3���4�3���7�3���10�3���13�3�5!

81.

Example: �1 � x�2 � ��

n�0�2n�xn � 1 � 2x � x2

�1 � x�k � ��

n�0�kn�xn

77. �53� �

5 � 4 � 33!

�606

� 10 78. ��22 � �

��2���3�2!

� 3 79.

� �0.0390625 � �5

128

�0.54 � �

�0.5���0.5���1.5���2.5�4!

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Review Exercises for Chapter 9 343

Review Exercises for Chapter 9

1. an �1n!

2. an �n

n2 � 13. 6, 5, 4.67, . . .

Matches (a)

an � 4 �2n

:

4. 3.5, 3, . . .

Matches (c)

an � 4 �n2

: 5. 10, 3, . . .

Matches (d)

an � 10�0.3�n�1:6. . . .

Matches (b)

6, �4,an � 6��23�

n�1

:

7.

The sequence seems to converge to 5.

� limn→�

�5 �2n� � 5

limn→�

an � limn→�

5n � 2

n120

0

8an �5n � 2

n 8.

The sequence seems to diverge(oscillates).

1, 0, , 0, 1, 0, . . .�1sin n�

2:

0 12

−2

2an � sin n�

2

9.

Converges

limn→�

n � 1

n2 � 0 10.

Converges

limn→�

1

�n� 0 11. lim

n→�

n3

n2 � 1� � 12.

Diverges

� �

limn→�

n

ln�n� � limn→�

1

1�n

13. Converges� limn→�

1

�n � 1 � �n� 0lim

n→� ��n � 1 � �n� � lim

n→� ��n � 1 � �n��n � 1 � �n

�n � 1 � �n

14.

Converges; k � 2n

limn→�

�1 �1

2n�n

� limk→�

��1 �1k�

k

�1�2

� e1�2 15.

Converges

limn→�

sin �n�n

� 0

84. Assume the interval is Let

Hence,

Since and

Thus,

Note: Let Then and f� �1� � 2. f � �x� � 1 f��x� ≤ 1,f �x� �12�x � 1�2 � 1.

f��x� ≤ 2.

� 3 � x2 ≤ 4.

≤ 1 � 1 �12�1 � x2� �

12�1 � x�2

2 f��x� ≤ f �1� � f ��1� �12�1 � x�2 f ��c� �

12�1 � x�2 f ��d�

f � �x� ≤ 1, f �x� ≤ 1

2 f��x� � f �1� � f ��1� �12�1 � x�2f ��c� �

12�1 � x�2f ��d�.

f �1� � f ��1� � 2 f��x� �12�1 � x�2 f ��c� �

12�1 � x�2 f ��d�

f ��1� � f �x� � ��1 � x� f�(x) �12��1 � x�2 f �(d), d � ��1, x�.

f �1� � f �x� � �1 � x� f�(x) �12�1 � x�2 f �(c), c � �x, 1�

x � �1, 1�,�1, 1�.

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22. (a)

The series converges, by the Limit Comparison Test with � 1n2.

21. (a)

The series converges by the Alternating Series Test.

20. (a)

The series converges by the Alternating Series Test.

19. (a)

The series diverges geometric r �32 > 1�.�

344 Chapter 9 Infinite Series

k 5 10 15 20 25

13.2 113.3 873.8 6448.5 50,500.3Sk

(b)

120

−10

120

k 5 10 15 20 25

0.4597 0.4597 0.4597 0.4597 0.4597Sk

(b)

120

−0.1

1

23. Converges. Geometric series, r � 0.82, r < 1.

25. Diverges. nth-Term Test. limn→�

an � 0.

k 5 10 15 20 25

0.3917 0.3228 0.3627 0.3344 0.3564Sk

(b)

0 120

1

k 5 10 15 20 25

0.8333 0.9091 0.9375 0.9524 0.9615Sk

(b)

0 120

1

24. Diverges. Geometric series, r � 1.82 > 1.

26. Diverges. nth-Term Test, limn→�

an �23

.

27. Geometric series with and

S �a

1 � r�

11 � �2�3� �

11�3

� 3

r �23

.a � 1��

n�0�2

3�n

28.

See Exercise 27.

��

n�0

2n�2

3n � 4 ��

n�0�2

3�n

� 4�3� � 12

16. Let

Assume and note that the terms

converge as Hence converges.ann →�.

bn ln b � cn ln cbn � cn �

bn ln bbn � cn �

cn ln cbn � cn

b ≥ c

limn→�

ln y � limn→�

1

bn � cn�bn ln b � cn ln c�.

ln y �ln�bn � cn�

n

y � �bn � cn�1�n

18. (a)

(b) V5 � 120,000�0.70�5 � $20,168.40

n � 1, 2, 3, 4, 5Vn � 120,000�0.70�n,

17.

(a)

(b) A40 8218.10

A4 5254.73 A8 5522.43

A3 5189.85 A7 5454.25

A2 5125.78 A6 5386.92

A1 � 5062.50 A5 5320.41

n � 1, 2, 3

An � 5000�1 �0.05

4 �n

� 5000�1.0125�n

Page 114: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Review Exercises for Chapter 9 345

32. (a)

(b) 0.923076 �0.923076

1 � 0.000001�

923,076999,999

�12�76,923�13�76,923� �

1213

0.923076 � 0.9230761 � 0.000001 � �0.000001�2 � . . .� � ��

n�0 �0.923076��0.000001�n

33.

meters � �8 � ��

n�016�0.7�n � �8 �

161 � 0.7

� 4513

D � 8 � 16�0.7� � 16�0.7�2 � . . . � 16�0.7�n � . . .�

D2 � 0.7�8� � 0.7�8� � 16�0.7�

D1 � 8

35. See Exercise 110 in Section 9.2.

$5087.14

�200�e�0.06��2� � 1�

e0.06�12 � 1

A �P�ert � 1�er�12 � 1

34.

$4,371,379.65

S � �39

n�032,000�1.055�n �

32,000�1 � 1.05540�1 � 1.055

36. See Exercise 110 in Section 9.2.

$14,343.25

� 100� 120.035���1 �

0.03512 �

120

� 1�

A � P�12r ���1 �

r12�

12t

� 1�

37.

By the Integral Test, the series converges.

��

1x�4 ln�x� dx � lim

b→� ��

ln x3x3 �

19x3�

b

1� 0 �

19

�19

39.

Since the second series is a divergent p-series while the first series is a convergent p-series, the differencediverges.

��

n�1� 1

n2 �1n� � �

n�1 1n2 � �

n�1 1n

38.

Divergent p-series, p �34 < 1

��

n�1

14�n3

� ��

n�1

1n3�4

40.

The first series is a convergent p-series and the secondseries is a convergent geometric series. Therefore, theirdifference converges.

��

n�1� 1

n2 �12n� � �

n�1 1n2 � �

n�1 12n

41.

By a limit comparison test with the convergent p-series

the series converges.��

n�1

1n3�2,

limn→�

1��n3 � 2n

1��n3�2� � limn→�

n3�2

�n3 � 2n� 1

��

n�1

1

�n3 � 2n42.

By a limit comparison test with the series diverges.��

n�1 1n

,

limn→�

�n � 1��n�n � 2�

1�n� lim

n→� n � 1n � 2

� 1

��

n�1

n � 1n�n � 2�

29. �1

1 � �1�2� �1

1 � �1�3� � 2 �32

�12�

n�0� 1

2n �13n� � �

n�0�1

2�n

� ��

n�0�1

3�n

30.

�1

1 � �2�3� � ��1 �12� � �1

2�

13� � �1

3�

14� � . . .� � 3 � 1 � 2

��

n�0��2

3�n

�1

�n � 1��n � 2�� � ��

n�0�2

3�n

� ��

n�0� 1

n � 1�

1n � 2�

31. (a)

(b) 0.09 �0.09

1 � 0.01�

111

0.09 � 0.09 � 0.0009 � 0.000009 � . . . � 0.09�1 � 0.01 � 0.0001 � . . .� � ��

n�0 �0.09��0.01�n

Page 115: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

346 Chapter 9 Infinite Series

45. Converges by the Alternating Series Test. (Conditional convergence)

47. Diverges by the nth-Term Test.

46.

By the Alternating Series Test, the series converges.

limn→�

�n

n � 1� 0

an�1 ��n � 1n � 2

≤ �n

n � 1� an

��

n�1

��1�n�nn � 1

48. Converges by the Alternating Series Test.

an�1 �3 ln�n � 1�

n � 1<

3 ln nn

� an, limn→�

3 ln n

n� 0

49.

By the Ratio Test, the series converges.

� �0��1� � 0 < 1

� limn→�

� 1e2n�1��n � 1

n �

� limn→�

en2�n � 1�en2�2n�1n

limn→�

an�1

an � limn→�

n � 1e�n�1�2

en2

n ��

n�1

nen2

51.

Therefore, by the Ratio Test, the series diverges.

� limn→�

2n3

�n � 1�3 � 2limn→�

an�1

an � limn→�

2n�1

�n � 1�3 n3

2n��

n�1 2n

n3

50.

By the Ratio Test, the series diverges.

� limn→�

n � 1

e� �

limn→�

an�1

an � limn→�

�n � 1�!en�1

en

n!��

n�1 n!en

52.

By the Ratio Test, the series converges.

limn→�

an�1

an � limn→�

1 3 . . . �2n � 1��2n � 1�2 5 . . . �3n � 1��3n � 2�

2 5 . . . �3n � 1�1 3 . . . �2n � 1� � lim

n→� 2n � 13n � 2

�23

��

n�1 1 3 5 . . . �2n � 1�2 5 8 . . . �3n � 1�

43.

Since diverges (harmonic series),

so does the original series.

��

n�1

12n

�12

��

n�1 1n

� �32

54

. . . 2n � 12n � 2�

12n

>1

2n

an �1 3 5 . . . �2n � 1�

2 4 6 . . . �2n�

��

n�1 1 3 5 . . . �2n � 1�

2 4 6 . . . �2n�44. Since converges, converges by the

Limit Comparison Test.

��

n�1

13n � 5�

n�1

13n

Page 116: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

54. (a) The series converges by the Alternating Series Test.

(b) (c)

(d) The sum is approximately 0.0714. 0 120

0.3

Review Exercises for Chapter 9 347

x 5 10 15 20 25

0.0871 0.0669 0.0734 0.0702 0.0721Sn

55. (a) ��

N

1x2 dx � ��

1x�

N�

1N

N 5 10 20 30 40

1.4636 1.5498 1.5962 1.6122 1.6202

0.2000 0.1000 0.0500 0.0333 0.0250��

N

1x2 dx

�N

n�1 1n2

(b)

The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums.

��

N

1x5 dx � ��

14x4�

N�

14N4

N 5 10 20 30 40

1.0367 1.0369 1.0369 1.0369 1.0369

0.0004 0.0000 0.0000 0.0000 0.0000��

N

1x5 dx

�N

n�1

1n5

56. No. Let then The series is a convergent p-series.��

n�1

3937.5n2a75 � 0.7.an �

3937.5n2 ,

57.

f����x� � �18

e�x�2 f����0� � �18

� 1 �12

x �18

x2 �1

48x3 f��x� �

14

e�x�2 f��0� �14

� 1 �12

x �14

x2

2!�

18

x3

3! f��x� � �

12

e�x�2 f��0� � �12

P3�x� � f �0� � f��0�x � f��0�x2

2!� f����0�x3

3! f �x� � e�x�2 f �0� � 1

58.

f����x� � 4 sec2 x tan2 x � 2 sec4 x f������

4� � 16

f��x� � 2 sec2 x tan x f����

4� � �4

f��x� � sec2 x f����

4� � 2

P3�x� � �1 � 2�x ��

4� � 2�x ��

4�2

�83�x �

4�3

f �x� � tan x f ���

4� � �1

53. (a) Converges

(b) (c)

(d) The sum is approximately 3.75.−1

0 12

4

� limn→�

�n � 1n ��3

5� �35

< 1,Ratio Test: limn→�

an�1

an � limn→�

�n � 1��3�5�n�1

n�3�5�n

x 5 10 15 20 25

2.8752 3.6366 3.7377 3.7488 3.7499Sn

Page 117: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

348 Chapter 9 Infinite Series

60. cos�0.75� 1 ��0.75�2

2!�

�0.75�4

4!�

�0.75�6

6! 0.7317

61. ln�1.75� �0.75� ��0.75�2

2�

�0.75�3

3�

�0.75�4

4�

�0.75�5

5�

�0.75�6

6� . . . �

�0.75�14

14 0.559062

62. e�0.25 1 � 0.25 ��0.25�2

2!�

�0.25�3

3!�

�0.25�4

4! 0.779

63.

(a)

This inequality is true for

(b)

This inequality is true for

(c)

This inequality is true for

(d)

This inequality is true for n � 10.

Rn�x� ≤ 2n�1

�n � 1�! < 0.0001

n � 5.

Rn�x� ≤ �0.5�n�1

�n � 1�! < 0.0001

n � 6.

Rn�x� ≤ �1�n�1

�n � 1�! < 0.001

n � 4.

Rn�x� ≤ �0.5�n�1

�n � 1�! < 0.001

f �n�1��z� ≤ 1 ⇒ Rn�x� ≤ xn�1

�n � 1�!

Rn�x� �f �n�1��z��n � 1�!x

n�1

c � 0f �x� � cos x, 64.

−10 10

−10

f

P4

P10P6

10

P10�x� � 1 �x2

2!�

x4

4!�

x6

6!�

x8

8!�

x10

10!

P6�x� � 1 �x2

2!�

x4

4!�

x6

6!

P4�x� � 1 �x2

2!�

x4

4!

f �x� � cos x

65.

Geometric series which converges only if or �10 < x < 10.

x�10 < 1

��

n�0� x

10�n

66.

Geometric series which converges only if or�

12 < x < 12 .

2x < 1

��

n�0�2x�n

67.

Center: 2

Since the series converges when and when ,the interval of convergence is 1 ≤ x ≤ 3.

x � 3x � 1

R � 1

� x � 2

limn→�

un�1

un � limn→�

��1�n�1�x � 2�n�1

�n � 2�2 �n � 1�2

��1�n�x � 2�n ��

n�0

��1�n�x � 2�n

�n � 1�2 68.

Center: 2

Since the series converges at and diverges at , theinterval of convergence is 53 ≤ x < 73 .

73

53

R �13

� 3x � 2

limn→�

un�1

un � limn→�

3n�1�x � 2�n�1

n � 1

n3n�x � 2�n

��

n�1

3n�x � 2�n

n

59. Since

sin 95 � sin�95�

180� 95�

180�

�95��3

18033!�

�95��5

18055!�

�95��7

18077! 0.99594

�95��9

1809 9!< 0.001, use four terms.

Page 118: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Review Exercises for Chapter 9 349

69.

which implies that the series converges only at the centerx � 2.

limn→�

un�1

un � limn→�

�n � 1�!�x � 2�n�1

n!�x � 2�n � �

��

n�0n!�x � 2�n 70.

Geometric series which converges only if

or 0 < x < 4.x � 22 < 1

��

n�0 �x � 2�n

2n � ��

n�0�x � 2

2 �n

71.

� ��

n�0���1�n�11

4n�n!�2 � ��1�n 1

4n�n!�2�x2n�2 � 0

� ��

n�0���1�n�14�n � 1�2

4n�1�n � 1�!�2 � ��1�n 1

4n�n!�2�x2n�2

� ��

n�0���1�n�1�2n � 2��2n � 1 � 1�

4n�1�n � 1�!�2 � ��1�n 1

4n�n!�2�x2n�2

� ��

n�0���1�n�1

�2n � 2��2n � 1�4n�1�n � 1�!�2 �

��1�n�1�2n � 2�4n�1�n � 1�!�2 �

��1�n

4n�n!�2�x2n�2

x2y� � xy� � x2y � ��

n�0 ��1�n�1�2n � 2��2n � 1�x2n�2

4n�1�n � 1�!�2 � ��

n�0

��1�n�1�2n � 2�x2n�2

4n�1�n � 1�!�2 � ��

n�0��1�n

x2n�2

4n�n!�2

y� � ��

n�0 ��1�n�1�2n � 2��2n � 1�x2n

4n�1�n � 1�!�2

y� � ��

n�1 ��1�n�2n�x2n�1

4n�n!�2 � ��

n�0 ��1�n�1�2n � 2�x2n�1

4n�1�n � 1�!�2

y � ��

n�0��1�n

x2n

4n�n!�2

72.

� ��

n�1 ��1�n3n�1x2n

2nn!�2n � 2n� � 0

� ��

n�1 ��1�n�13n�1x2n

2nn!�2n� � �

n�1 ��1�n3n�1x2n

2n�1�n � 1�! 2n2n

� ��

n�0 ��1�n3n�1x2n

2nn!��2n� � �

n�0 ��1�n�13n�2x2n�2

2nn!

� ��

n�0 ��1�n3n�1x2n

2nn!��2n � 1� � 1� � �

n�0 ��1�n�13n�2x2n�2

2nn!

� ��

n�0 ��1�n�13n�1�2n � 2�x2n

2nn!� �

n�0 ��1�n�13n�2x2n�2

2nn!� �

n�0 ��1�n3n�1x2n

2nn!

y� � 3xy� � 3y � ��

n�0 ��3�n�1�2n � 2��2n � 1�x2n

2n�1�n � 1�! � ��

n�0

��1�n�13n�2�2n � 2�x2n�2

2n�1�n � 1�! � ��

n�0 ��1�n3n�1x2n

2nn!

y� � ��

n�0 ��3�n�1�2n � 2��2n � 1�x2n

2n�1�n � 1�!

y� � ��

n�1 ��3�n�2n�x2n�1

2nn!� �

n�0 ��3�n�1�2n � 2�x2n�1

2n�1�n � 1�!

y � ��

n�0 ��3�nx2n

2nn!

73.

��

n�0 23 �

x3�

n

� ��

n�0

2xn

3n�1

23 � x

�2�3

1 � �x�3� �a

1 � r74.

��

n�0

32��

x2�

n

� ��

n�0

��1�n3xn

2n�1

32 � x

�3�2

1 � �x�2� �3�2

1 � ��x�2� �a

1 � r

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350 Chapter 9 Infinite Series

75. Power series

Derivative:

� ��

n�0 29

�n � 1��x3�

n

��

n�1 23

n�x3�

n�1

�13� � �

n�1 29

n�x3�

n�1

��

n�0 23�

x3�

n

g�x� �2

3 � x. 76. Integral: �

n�0

��1�n3xn�1

�n � 1�2n�1

77. �32

< x < 32

1 �23

x �49

x2 �8

27x3 � . . . � �

n�0�2x

3 �n

�1

1 � �2x�3� �3

3 � 2x,

79.

��22 �

n�0 ��1�n�n�1��2x � �3��4��n

n! �

�22

��22 �x �

3�

4 � ��2

2 2!�x �3�

4 �2

� . . .

sin�x� � ��

n�0 f �n��x�x � �3��4��n

n!

f���x� � �cos x, . . .

f��x� � �sin x

f��x� � cos x

f �x� � sin x

78.

�1 < x < 7 �32

4 � �x � 3� �32

1 � x,

8 � 2�x � 3� �12

�x � 3�2 �18

�x � 3�3 � . . . � ��

n�0 8���x � 3�

4 �n

�8

1 � ��x � 3��4�

80.

��22 �1 � �x �

4� � ��

n�1 ��1�n�n�1���2x � ���4��n�1

�n � 1�! �

��22

��22 �x �

4� ��2

2 2!�x ��

4�2

��2

2 3!�x ��

4�3

��2

2 4!�x ��

4�4

� . . .cos x � ��

n�0 f �n�����4�x � ���4��n

n!

f���x� � sin x

f��x� � �cos x

f��x� � �sin x

f �x� � cos x

81. and since we have � 1 � x ln 3 �x2 ln 3�2

2!�

x3 ln 3�3

3!�

x4 ln 3�4

4!� . . . . 3x � �

n�0 �x ln 3�n

n!ex � �

n�0 xn

n!,3x � �eln�3��x � ex ln�3�

82.

� 1 �12!�x �

2�2

�54!�x �

2�4

� . . .csc�x� � ��

n�0 f �n����2�x � ���2��n

n!

f �4��x� � 5 csc5�x� � 15 csc3�x� cot2�x� � csc�x� cot4�x�

f���x� � �5 csc3�x� cot�x� � csc�x� cot3�x�

f��x� � csc3�x� � csc�x� cot2�x�

f��x� � �csc�x� cot�x�

f �x� � csc�x�

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Review Exercises for Chapter 9 351

83.

� ��

n�0 �n!�x � 1�n

n!� � �

n�0�x � 1�n, �2 < x < 0

1x

� ��

n�0

f �n���1��x � 1�n

n!

f����x� � �6x4, . . .

f��x� �2x3

f��x� � �1x2

f �x� �1x

84.

� 2 ��x � 4�

22 � ��

n�2

��1�n�11 3 5 . . . �2n � 3��x � 4�n

23n�1n!

� 2 ��x � 4�

22 ��x � 4�2

252!�

1 3�x � 4�3

283!�

1 3 5�x � 4�4

2114!� . . . �x � �

n�0 f �n��4��x � 4�n

n!

f �4��x� � ��12��

12��

32��

52�x�7�2, . . .

f���x� � �12��

12��

32�x�5�2

f��x� � ��12��

12�x�3�2

f��x� �12

x�1�2

f �x� � x1�2

85.

� 1 �x5

�2

25x2 �

6125

x3 � . . .

� 1 �x5

� ��

n�2

��1�n�14 9 14 . . . �5n � 6�xn

5nn!

� 1 �15

x �1 4x2

522!�

1 4 9x3

533!� . . .

�1 � x�1�5 � 1 �x5

��1�5���4�5�x2

2!�

1�5��4�5���9�5�x3

3!� . . .

�1 � x�k � 1 � kx �k�k � 1�x2

2!�

k�k � 1��k � 2�x3

3!� . . .

86.

� ��

n�0 ��1�n�n � 2�!xn

2n!� �

n�0 ��1�n�n � 2��n � 1�xn

2

1�1 � x�3 � 1 � 3x �

12x2

2!�

60x3

3!�

360x4

4!�

2520x5

5!� . . .

h�5��x� � �2520�1 � x��8

h�4��x� � 360�1 � x��7

h���x� � �60�1 � x��6

h��x� � 12�1 � x��5

h��x� � �3�1 � x��4

h�x� � �1 � x��3

Page 121: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

352 Chapter 9 Infinite Series

87. 0 < x ≤ 2

� ��

n�1��1�n�1

14nn

0.2231

ln�54� � �

n�1��1�n�1��5�4� � 1

n �n

ln x � ��

n�1��1�n�1

�x � 1�n

n,

89.

e1�2 � ��

n�0 �1�2�n

n!� �

n�0

12nn!

1.6487

�� < x < �ex � ��

n�0 xn

n!,

88. 0 < x ≤ 2

� ��

n�1��1�n�1

15nn

0.1823

ln�65� � �

n�1��1�n�1��6�5� � 1

n �n

ln x � ��

n�1��1�n�1

�x � 1�n

n,

90.

� ��

n�0

2n

3nn! 1.9477e2�3 � �

n�0 �2�3�n

n!

�� < x < �ex � ��

n�0 xn

n!,

91.

cos�23� � �

n�0��1�n

22n

32n�2n�! 0.7859

�� < x < � cos x � ��

n�0��1�n

x2n

�2n�!,

93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.

95. (a)

� 1 � 2x � 2x2 �43

x3P�x� � 1 � 2x �4x2

2!�

8x3

3!

f ���x� � 8e2x f ���0� � 8

f��x� � 4e2x f ��0� � 4

f��x� � 2e2x f��0� � 2

f �x� � e2x f �0� � 1 (b)

(c)

P � 1 � 2x � 2x 2 �43

x 3

e x e x � �1 � x �x2

2!� . . .��1 � x �

x2

2!� . . .�

P�x� � 1 � 2x � 2x2 �43

x3

e x � ��

n�0 xn

n!, e2x � �

n�0 �2x�n

n!

92.

sin�13� � �

n�0��1�n

132n�1�2n � 1�! 0.3272

�� < x < � sin x � ��

n�0��1�n

x2n�1

�2n � 1�!,

94.

� 1 �x 3

2� �

n�2

��1�n1 3 . . . �2n � 1�2nn!

x 3n

� 1 �12

�x 3� ���1�2���3�2�

2!x6 �

��1�2���3�2���5�2�3!

x9 � . . .

1

�1 � x 3� �1 � x 3��1�2, k � �

12

96. (a)

—CONTINUED—

� 2x �43

x3 �4

15x5 �

8315

x7 � . . .sin 2x � 0 � 2x �0x2

2!�

8x3

3!�

0x4

4!�

32x5

5!�

0x6

6!�

128x7

7!� . . .

f �7��x� � �128 cos 2x f �7��0� � �128

f �6��x� � �64 sin 2x f �6��0� � 0

f �5��x� � 32 cos 2x f �5��0� � 32

f �4��x� � 16 sin 2x f �4��0� � 0

f���x� � �8 cos 2x f���0� � �8

f��x� � �4 sin 2x f��0� � 0

f��x� � 2 cos 2x f��0� � 2

f �x� � sin 2x f �0� � 0

Page 122: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Review Exercises for Chapter 9 353

97.

� ��

n�0

��1�nx2n�1

�2n � 1��2n � 1�!

�x

0 sin t

t dt � � �

n�0

��1�nt2n�1

�2n � 1��2n � 1�!�x

0

sin t

t� �

n�0 ��1�nt2n

�2n � 1�!

sin t � ��

n�0 ��1�nt2n�1

�2n � 1�!

99.

� ��

n�0 ��1�nxn�1

�n � 1�2�x

0 ln�t � 1�

t dt � � �

n�0 ��1�nt n�1

�n � 1�2 �x

0

ln�t � 1�

t� �

n�0 ��1�nt n

n � 1

ln�1 � t� � � 11 � t

dt � ��

n�0 ��1�ntn�1

n � 1

1

1 � t� �

n�0��1�ntn

96. —CONTINUED—

(b)

(c)

� 2�x �2x3

3�

2x5

15�

4x7

315� . . .� � 2x �

43

x3 �4

15x5 �

8315

x7 � . . .

� 2�x � ��x3

2�

x3

6 � � � x5

24�

x5

12�

x5

120� � ��x7

720�

x7

144�

x7

240�

x7

5040� � . . .�

� 2�x �x3

6�

x5

120�

x7

5040� . . .��1 �

x2

2�

x4

24�

x6

720� . . .�

sin 2x � 2 sin x cos x

� 2x �8x3

6�

32x5

120�

128x7

5040� . . . � 2x �

43

x3 �4

15x5 �

8315

x7 � . . .

sin 2x � ��

n�0 ��1�n�2x�2n�1

�2n � 1�! � 2x ��2x�3

3!�

�2x�5

5!�

�2x�7

7!� . . .

sin x � ��

n�0 ��1�nx2n�1

�2n � 1�!

98.

� ��

n�0

��1�nx n�1

22n�2n�!�n � 1�

�x

0cos

�t2

dt � � ��

n�0

��1�n t n�1

22n�2n�!�n � 1��x

0

cos �t2

� ��

n�0 ��1�n t n

22n�2n�!

cos t � ��

n�0 ��1�nt 2n

�2n�!

100.

�x

0 et � 1

t dt � � �

n�1

tn

n n!�x

0� �

n�1

xn

n n!

et � 1

t� �

n�1 tn�1

n!

et � 1 � ��

n�1 tn

n!

et � ��

n�0 tn

n!

101.

By L’Hôpital’s Rule, limx→0�

arctan x

�x� lim

x→0� � 1

1 � x2�� 1

2�x�� lim

x→0�

2�x1 � x2 � 0.

limx→0�

arctan x

�x� 0

arctan x

�x� �x �

x5�2

3�

x 9�2

5�

x13�2

7�

x17�2

9� . . .

arctan x � x �x3

3�

x5

5�

x7

7�

x9

9� . . .

Page 123: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

354 Chapter 9 Infinite Series

102.

By L’Hôpital’s Rule, limx→0

arcsin x

x� lim

x→0 � 1�1 � x2�

1� 1.

limx→0

arcsin x

x� 1

arcsin x

x� 1 �

x2

2 � 3�

1 � 3x4

2 � 4 � 5�

1 � 3 � 5x6

2 � 4 � 6 � 7� . . .

arcsin x � x �x3

2 � 3�

1 � 3x5

2 � 4 � 5�

1 � 3 � 5x7

2 � 4 � 6 � 7� . . .

Problem Solving for Chapter 9

1. (a)

(b)

(c) limn→�

Cn � 1 � ��

n�0 13 �

23�

n

� 1 � 1 � 0

0, 13

, 23

, 1, etc.

�1�3

1 � �2�3�� 1

1�1

3� � 2�1

9� � 4� 1

27� � . . . � ��

n�0

1

3 �2

3�n

2. Let

Thus, S ��2

6�

14

�2

6�

�2

6 �34� �

�2

8.

� S �122 ��2

6 �. � S �122�1 �

122 �

132 � . . .

� S �122 �

142 � . . .

Then �2

6�

112 �

122 �

132 �

142 � . . .

S � ��

n�1

1

�2n � 1�2�

1

12�

1

32�

1

52� . . ..

3. If there are n rows, then

For one circle,

and

For three circles,

and

For six circles,

and

Continuing this pattern,

limn→�

An ��

2�

14

��

8

An ��

2

n�n � 1�2�3 � 2�n � 1��2

Total Area � ��rn2�an � �� 1

2�3 � 2�n � 1��2

n�n � 1�

2

rn �1

2�3 � 2�n � 1�.

r3 �1

2�3 � 4.

1 � 2�3r3 � 4r3a3 � 6

1 3 r3

r3r32

r2 �1

2 � 2�3.

1 � 2�3r2 � 2r2a2 � 3

1 3 r2

r2r22

r1 �1

3��3

2 � ��3

6�

1

2�3.a1 � 1

1 32

12

r1

an �n�n � 1�

2.

Page 124: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Problem Solving for Chapter 9 355

4. (a) Position the three blocks as indicated in the figure. The bottom block extends overthe edge of the table, the middle block extends over the edge of the bottom block,and the top block extends over the edge of the middle block.

The centers of gravity are located at

bottom block:

middle block:

top block:16

�14

�12

�12

�5

12.

16

�14

�12

� �1

12

16

�12

� �13

1�21�4

1�6

16

512

1112

14

16

12

0

The center of gravity of the top 2 blocks is

which lies over the bottom block. The center of gravity of the 3 blocks is

which lies over the table. Hence, the far edge of the top block lies

beyond the edge of the table.

(b) Yes. If there are n blocks, then the edge of the top block lies from the

edge of the table. Using 4 blocks,

which shows that the top block extends beyond the table.

(c) The blocks can extend any distance beyond the table because the series diverges:

��

i�1 12i

�12�

i�1 1i

� �.

�4

i�1 12i

�12

�14

�16

�18

�2524

�n

i�1 12i

16

�14

�12

�1112

��13

�1

12�

512��3 � 0

��1

12�

512��2 �

16

,

5. (a)

because each series in the second line has

(b)

�Assume all an > 0.� R � 1

� �a0 � a1x � . . . � ap�1x p�1� 11 � x p

� �a0 � a1x � . . . � ap�1xp�1��1 � xp � . . .�

� a0�1 � xp � . . .� � a1x�1 � xp � . . .� � . . . � ap�1xp�1�1 � xp � . . .�

�anxn � �a0 � a1x � . . . � ap�1xp�1� � �a0x p � a1x

p�1 � . . .� � . . .

R � 1.R � 1

� �1 � 2x � 3x2� 11 � x3

� �1 � x3 � x6 � . . .�1 � 2x � 3x2�

� �1 � x3 � x6 � . . .� � 2�x � x4 � x7 � . . .� � 3�x2 � x5 � x8 � . . .�

�anxn � 1 � 2x � 3x2 � x3 � 2x4 � 3x5 � . . .

Page 125: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

356 Chapter 9 Infinite Series

6.

If converges conditionally.

If diverges.

No values of a and b give absolute convergence. implies conditional convergence.a � b

a � b, ��

n�1

��1�n�1�a � b�2n

� ��

n�1

a � b2n

a � b, ��

n�1

��1�n�1�2a�2n

� a ��

n�1

��1�n�1

n

a �b2

�a3

�b4

� . . . � ��

n�1

��1�n�1�a � b� � �a � b�2n

7. (a)

Letting you have Letting

Thus,

(b) Differentiating,

Letting

2e � ��

n�0 n � 1

n!� 5.4366.

x � 1,

xex � ex � ��

n�0

�n � 1�xn n!

.

��

n�1

1�n � 2�n!

�12

.

e � e � 1 � ��

n�0

1�n � 2�n!

�12

� ��

n�1

1�n � 2�n!

.

x � 1,C � 1.x � 0,

xex dx � xex � ex � C � ��

n�0

xn�2

�n � 2�n!

xex � ��

n�0 xn�1

n!

ex � 1 � x �x2

2!� . . . � �

n�0 xn

n!8.

f �12��0�

12!�

16!

⇒ f �12��0� �12!6!

� 665,280

ex2� 1 � x2 �

x4

2!� . . . �

x12

6!� . . .

ex � 1 � x �x2

2!� . . .

9. Let

Then,

Since and this series converges.an�1 < an,limn→�

an � 0

0 sin x

x dx � a1 � a2 � a3 � a4 � . . . .

a1 � �

0 sin x

x dx, a2 � � 2�

sin x

x dx, a3 � 3�

2�

sin x

x dx, etc.

10. (a) If diverges.

If converges.

If diverges.

(b) diverges by part (a).��

n�4

1n ln�n2� �

12 �

n�4

1n ln n

p < 1,

p > 1, �

2

1x�ln x�p dx � lim

b→� ��ln b�1�p

1 � p�

�ln 2�1�p

1 � p

p � 1, �

2

1x ln x

dx � ln ln x�

2

Page 126: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

Problem Solving for Chapter 9 357

11. (a)

[See part (b) for proof.]

(b) Use mathematical induction to show the sequence is increasing. Clearly,

Now assume Then

Use mathematical induction to show that the sequence is bounded above by a. Clearly,

Now assume Then and implies

Hence, the sequence converges to some number L. To find L, assume

Hence, L �1 � �1 � 4a

2.

L �1 ± �1 � 4a

2.

L � �a � L ⇒ L2 � a � L ⇒ L2 � L � a � 0

an�1 � an � L:

a > �an � a � an�1.

a2 > an � a

a2 � a > an

a�a � 1� > an�1�

a � 1 > 1a > anan < a.

a1 � �a < a.

an�1 > an.

�an � a > �an�1 � a

an � a > an�1 � a

an > an�1.

a2 � �a � a1 � �a�a > �a � a1.

limn→�

an �1 � �13

2

a6 � 2.30146

a5 � 2.29672

a4 � 2.27493

a3 � 2.17533

a2 � 1.73205

a1 � 3.0

12. Let

as

By the Root Test, converges converges.⇒ �anr n�bn

Lr <1rr � 1.

n →�. �bn�1�n � �anr n�1�n � an1�n � r → Lr

bn � anrn.

13. (a)

—CONTINUED—

S5 �4532

�1

16�

4732

S4 �118

�1

32�

4532

S3 �98

�14

�118

S1 � 1 �18

�98

S1 �120 � 1

��

n�1

12n���1�n �

121�1 �

122�1 �

123�1 �

124�1 �

125�1 � . . .

Page 127: CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges

358 Chapter 9 Infinite Series

14. (a)

(b)

� 1.0204081632 . . .

� 1 � 0.02 � 0.0004 � . . .

� 1 � 0.02 � �0.02�2 � . . .

1

0.98�

11 � 0.02

� ��

n�0�0.02�n

� 1.010101 . . .

� 1 � 0.01 � �0.01�2 � . . .

1

0.99�

11 � 0.01

� ��

n�0�0.01�n

13. —CONTINUED—

(b)

This sequence is which diverges.

(c)

converges because and and n�2 →1.n�1�2 →1�2��1�n� �12, 2, 12, 2, . . . �

1

2 � n�2��1�n → 1

2< 1

n� 12n���1�n � � 1

2n � 2��1�n�1�n

18, 2, 18, 2, . . .

an�1

an

�2n���1�n

2�n�1����1�n�1 �2��1�n

21���1�n�1

15.

S10 � 1490 � 810 � 440 � 2740

S9 � 810 � 440 � 240 � 1490

S8 � 440 � 240 � 130 � 810

S7 � 240 � 130 � 70 � 440

S6 � 130 � 70 � 40 � 240

16. (a)

(b)

(c)

converges. �43

� ��

n�1

1n3�2

W �43

��1 �1

23�2 �1

33�2 � . . .

S � 4��1 �12

�13

� . . . � 4� ��

n�1

1n

� �

� 2 ��

n�1

1n1�2 � � �p-series, p �

12

< 1�

Height � 2�1 �1�2

�1�3

� . . .

17. (a) diverges (harmonic)

(b) Let By the Mean Value Theorem,

where is between and . Thus,

Because converges, the Comparison Theorem

tells us that converges.��

n�1�sin� 1

2n� � sin� 12n � 1�

��

n�1

12n�2n � 1�

�1

2n�2n � 1�

≤ � 12n

�1

2n � 1� 0 ≤ �sin� 1

2n� � sin� 12n � 1��

yxc

� f �x� � f �y�� � f��c��x � y� � cos�c��x � y� ≤ �x � y�,f �x� � sin x.

��

n�1

12n

�12

��

n�1 1n