CHAPTER 9 Infinite Series · Infinite Series Section 9.1 Sequences.....258 Section 9.2 Series and Convergence .....272 Section 9.3 ... Section 9.5 Alternating Series.....306 Section

© 2010 Brooks/Cole, Cengage Learning

C H A P T E R 9 Infinite Series

Section 9.1 Sequences............................................................................................258

Section 9.2 Series and Convergence .....................................................................272

Section 9.3 The Integral Test and p-Series ...........................................................287

Section 9.4 Comparisons of Series........................................................................299

Section 9.5 Alternating Series ...............................................................................306

Section 9.6 The Ratio and Root Tests...................................................................315

Section 9.7 Taylor Polynomials and Approximations .........................................328

Section 9.8 Power Series .......................................................................................339

Section 9.9 Representation of Functions by Power Series...................................352

Section 9.10 Taylor and Maclaurin Series ..............................................................360

Review Exercises .......................................................................................................377

Problem Solving .........................................................................................................390

258 © 2010 Brooks/Cole, Cengage Learning

C H A P T E R 9 Infinite Series

Section 9.1 Sequences

1.

1

22

33

44

55

3

3 3

3 9

3 27

3 81

3 243

nna

a

a

a

a

a

1

=

= =

= =

= =

= =

= =

2.

1

2

2

3

3

4

4

5

5

3!

3 31!3 92! 23 27 93! 6 23 81 274! 24 83 243 815! 120 40

n

nan

a

a

a

a

a

=

= =

= =

= = =

= = =

= = =

3. ( )

( )( )( )( )

1

22

33

44

55

14

14

1 14 16

1 14 64

1 14 256

1 14 1024

nna

a

a

a

a

a

= −

= −

= − =

= − = −

= − =

= − = −

4. ( )1

2

3

4

5

23

23

49

827

1681

32243

nna

a

a

a

a

a

= −

= −

=

= −

=

= −

5.

1

2

3

4

5

sin2

sin 12

sin 03sin 12

sin 2 05sin 12

nna

a

a

a

a

a

π

π

ππ

ππ

=

= =

= =

= = −

= =

= =

6.

1

2

3

4

5

23

2 14 2456 168710 58 4

nna

n

a

a

a

a

a

=+

= =

=

= =

=

= =

7.

( )

( )

( )

( )

( )

( )

( 1) 2

2

1

1 2

3

2 2

6

3 2

10

4 2

15

5 2

1

11

1

1 12 4

1 13 9

1 14 16

1 15 25

n n

nan

a

a

a

a

a

+−=

−= = −

−= = −

−= =

−= =

−= = −

Section 9.1 Sequences 259


8.

( ) 1

1

2

3

4

5

21

2 21

2 12

23

2 14 2

25

nna

n

a

a

a

a

a

+ ⎛ ⎞= − ⎜ ⎟⎝ ⎠

= =

= − = −

=

= − = −

=

9. 2

1

2

3

4

5

1 15

5 1 1 51 1 1952 4 41 1 4353 9 91 1 7754 16 161 1 12155 25 25

nan n

a

a

a

a

a

= − +

= − + =

= − + =

= − + =

= − + =

= − + =

10. 2

1

2

3

4

5

2 610

10 2 6 183 2510 12 2

2 2 34103 3 31 3 87102 8 82 6 266105 25 25

nan n

a

a

a

a

a

= + +

= + + =

= + + =

= + + =

= + + =

= + + =

11.

( )( )( )( )( )( )( )( )( )

1 1

2 1

3 2

4 3

5 4

3, 2 1

2 1

2 3 1 4

2 1

2 4 1 6

2 1

2 6 1 10

2 1

2 10 1 18

k ka a a

a a

a a

a a

a a

+= = −

= −

= − =

= −

= − =

= −

= − =

= −

= − =

12. 1 1

2 1

3 2

4 3

5 4

14,2

1 1 42

2 1 62

3 1 122

4 1 302

k kka a a

a a

a a

a a

a a

++⎛ ⎞= = ⎜ ⎟

⎝ ⎠+⎛ ⎞= =⎜ ⎟

⎝ ⎠+⎛ ⎞= =⎜ ⎟

⎝ ⎠+⎛ ⎞= =⎜ ⎟

⎝ ⎠+⎛ ⎞= =⎜ ⎟

⎝ ⎠

13.

( )( )( )( )

1 1

2 1

3 2

4 3

5 4

12

1 12 21 12 21 12 21 12 2

32,

32 16

16 8

8 4

4 2

k ka a a

a a

a a

a a

a a

+= =

= = =

= = =

= = =

= = =

14.

( )( )( )( )

21 1

2 22 1

2 23 2

2 24 3

225 4

13

1 13 3

1 13 3

1 13 3

1 13 3

6,

6 12

12 48

48 768

768 196,608

k ka a a

a a

a a

a a

a a

+= =

= = =

= = =

= = =

= = =

15. 1 210 10 10, 5,

1 1 1 3na a an

= = = =+ +

Matches (c)

16. 1 210 10 20, 5,

1 2 3nna a a

n= = = =

+

Matches (a)

17. ( ) 1 2 31 , 1, 1, 1,nna a a a= − = − = = − …

Matches (d)

18. ( )1 2

1 1 1, 1, .1 2

n

na a an− −

= = = − =

Matches (b)

19. 22, 0, , 1,3

− … matches (b)

1 2 3

42

4 4 4 22 2, 2 0, 2 ,1 2 3 3

nan

a a a

= −

= − = − = − = = − = …

260 Chapter 9 Infinite Series


20. 16, 8, 4, 2,− − … matches (c)

( )

( ) ( )

1

1 1 2 11 2

16 0.5

16 0.5 16, 16 0.5 8,

nna

a a

−

− −

= −

= − = = − = − …

21. 82 43 3 3, , 2, ,…matches (a)

1 2 3

23

62 43 3 3, , 2,

na n

a a a

=

= = = = …

22. 4 3 81, , , ,3 2 5

…matches (d)

1 2 3

21

2 4 6 31, , ,2 3 4 2

nna

n

a a a

=+

= = = = = …

23.

( )( )

5

6

3 1

3 5 1 14

3 6 1 17

na n

a

a

= −

= − =

= − =

Add 3 to preceding term.

24.

5

6

62

5 6 112 2

6 6 62

nna

a

a

+=

+= =

+= =

Add 12

to preceding term.

25.

( )( )

1 1

5

6

2 , 5

2 40 80

2 80 160

n na a a

a

a

+ = =

= =

= =

Multiply the preceding term by 2.

26. 1 1

5 6

12

1 116 32

, 1

,

n na a a

a a

−= − =

= = −

Multiply the preceding term by 12.−

27. ( )

( )

( )

1

5 4

6 5

32

3 3162

3 3322

n na

a

a

−=−

= =−

= = −−

Multiply the preceding term by 1.2

−

28. 1 1

5 6

32

81 24316 32

, 1

,

n na a a

a a

−= − =

= = −

Multiply the preceding term by 32.−

29. ( )( ) ( )( )11 10 9 8!11! 11 10 9 9908! 8!

= = =

30. ( )( )( )( )( )

( )( )( )( )

25 24 23 22 21 20 !25!20! 20!

25 24 23 22 21 6,375,600

=

= =

31. ( ) ( )1 ! ! 11

! !n n n

nn n+ +

= = +

32. ( ) ( )( ) ( )( )2 ! ! 1 21 2

! !n n n n

n nn n+ + +

= = + +

33. ( )( )

( )( ) ( )( ) ( )

2 1 ! 2 1 ! 12 1 ! 2 1 ! 2 2 1 2 2 1

n nn n n n n n− −

= =+ − + +

34. ( )( )

( ) ( )( )( )

( )( )

2 2 ! 2 ! 2 1 2 22 ! 2 !

2 1 2 2

n n n nn n

n n

+ + +=

= + +

35. 2

25lim 5

2n

nn→∞

=+

36. 21lim 5 5 0 5

n n→∞

⎛ ⎞− = − =⎜ ⎟⎝ ⎠

37. ( )2 2

2 2 2lim lim 211 1 1n n

nn n→∞ →∞

= = =+ +

38. ( )2 2

5 5 5lim lim 514 1 4n n

nn n→∞ →∞

= = =+ +

39. 1lim sin 0n n→∞

⎛ ⎞ =⎜ ⎟⎝ ⎠

40. 2lim cos 1n n→∞

⎛ ⎞ =⎜ ⎟⎝ ⎠

41.

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

1 1lim lim lim lim 1 1.nn n x x

n xan x→∞ →∞ →∞ →∞

+ += = = =

−1

12−1

3



42.


3 2 3 21 1lim lim lim 0.n

n n xa

n x→∞ →∞ →∞= = =

43.

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

{ } { }0, 1, 0, 1, 0, 1, .na = − − …

So, lim nn

a→∞

does not exist.

44.


1lim lim 3 3 0 3.2n nn n

a→∞ →∞

⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

45. ( )lim 0.3 1 0 1 1,n

n→∞⎡ ⎤− = − = −⎣ ⎦ converges

46. 3lim 4 4 0 4,n n→∞

⎡ ⎤− = − =⎢ ⎥⎣ ⎦converges

47. 5lim 0,2n n→∞=

+converges

48. 2lim 0,!n n→∞= converges

49. ( )lim 11

n

n

nn→∞

⎛ ⎞− ⎜ ⎟+⎝ ⎠

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

50. ( )lim 1 1 n

n→∞⎡ ⎤+ −⎣ ⎦

does not exist, (alternates between 0 and 2), diverges.

51. 2

23 4 3lim ,

2 1 2n

n nn→∞

− +=

+converges

52. 3

3lim 1,

1n

nn→∞

=+

converges

53. ( )( )

1 3 5 2 1

2

1 3 5 2 1 12 2 2 2 2

n n

na

n

nn n n n n

⋅ ⋅ ⋅ ⋅ ⋅ −=

−= ⋅ ⋅ ⋅ ⋅ ⋅ <

So, lim 0,nn

a→∞

= converges.

54. The sequence diverges. To prove this analytically, you use mathematical induction to show that

13 .2

n

na−

⎛ ⎞≥ ⎜ ⎟⎝ ⎠

Clearly, 0

131 1.2

a ⎛ ⎞= ≥ =⎜ ⎟⎝ ⎠

Assume that

13 .2

k

ka−

⎛ ⎞≥ ⎜ ⎟⎝ ⎠

Then,

( )( )( )1

1

1 3 5 2 1 2 11 !

2 1 3 2 1 .1 2 1

k

k

k

k ka

k

k kak k

+

−

⋅ ⋅ − +=

+

+ +⎛ ⎞ ⎛ ⎞= ≥ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Because

2 1 3,1 2

kk

+≥

+

you have 13 ,2

k

ka +⎛ ⎞≥ ⎜ ⎟⎝ ⎠

which shows that

132

n

na−

⎛ ⎞≥ ⎜ ⎟⎝ ⎠

for all n.

55. ( )1 1lim 0,

n

n n→∞

+ −= converges

56. ( )2

1 1lim 0,

n

n n→∞

+ −= converges

57. ( ) ( )3ln ln3lim lim2 2

3 1lim 0, converges2

n n

n

n nn n

n

→∞ →∞

→∞

=

⎛ ⎞= =⎜ ⎟⎝ ⎠

(L'Hôpital's Rule)

58. ( )1 2 lnlnlim lim

1lim 0, converges2

n n

n

nnn n

n

→∞ →∞

→∞

=

= =

(L'Hôpital's Rule)

−1 12

−1

2

12−1

−2

2

−1 12

−1

4



59. ( )34lim 0,

n

n→∞= converges

60. ( )lim 0.5 0,n

n→∞= converges

61. ( ) ( )1 !lim lim 1 , diverges

!n n

nn

n→∞ →∞

+= + = ∞

62. ( )( )

2 ! 1lim lim 0,! 1n n

nn n n→∞ →∞

−= =

−converges

63. ( )( )

2 2

2

11lim lim1 1

1 2lim 0, converges

n n

n

n nn nn n n n

nn n

→∞ →∞

→∞

− −−⎛ ⎞− =⎜ ⎟− −⎝ ⎠

−= =

−

64. 2 2 2

22 1lim lim ,

2 1 2 1 4 1 2n n

n n nn n n→∞ →∞

⎛ ⎞ −− = = −⎜ ⎟+ − −⎝ ⎠

converges

65. lim 0,p

nn

ne→∞

= converges

( )0, 2p n> ≥

66. 1sinna nn

=

Let ( ) 1sin .f x xx

=

( )

( ) ( )2

2

sin 11lim sin lim1

1 cos 1lim

11lim cos cos 0

1 (L'Hôpital's Rule)

x x

x

x

xx

x x

x xx

x

→∞ →∞

→∞

→∞

=

−=

−

= =

=

or,

( ) ( )0

sin 1 sinlim lim 1.

1x y

x yx y+→∞ →

= = Therefore

1lim sin 1,n

nn→∞= converges.

67. 1 0lim 2 2 1,nn→∞

= = converges

68. 1lim 3 lim 0,3

nnn n

−

→∞ →∞

−− = = converges

69.

( )10

1

lim 1 lim 1

n

n

n ku kn u

kan

k u en→∞ →

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎡ ⎤+ = + =⎜ ⎟ ⎣ ⎦⎝ ⎠

where ,u k n= converges

70.

12

2 2

1

1 11 1

lim 1, converges

nn n

n

nn

n

an n

a e→∞

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= + = +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= =

71. ( )sin 1lim lim sin 0,n n

n nn n→∞ →∞

= =

converges (because ( )sin n is bounded)

72. 2coslim 0,

n

nnπ

→∞= converges

73. 3 2na n= −

74. 4 1na n= −

75. 2 2na n= −

76. ( ) 1

2

1 n

nan

−−=

77. 12n

nan+

=+

78. ( ) 1

2

12

n

n na−

−

−=

79. 1 11nna

n n+

= + =

80.

1

2 112

2 12

n

n n

n

n

a

+

−= +

−=

81. ( )( )1 2n

nan n

=+ +

82. 1!na

n=

83. ( )( )

( )( )

1

1

11 3 5 2 1

1 2 !2 !

n

n

n n

an

nn

−

−

−=

⋅ ⋅ −

−=

84. ( )

1

1 !

n

nxa

n

−

=−

85. ( )2 !, 1, 2, 3,na n n= = …

86. ( )2 1 !, 1, 2, 3,na n n= − = …



87. 11 14 4 ,

1n na an n += − < − =

+

Monotonic; 4,na < bounded

88. Let ( ) 3 .2

xf xx

=+

Then ( )( )2

6 .2

f xx

′ =+

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

3,na < bounded

89. ( )

( )

?

2 1 2

?3 2

?

12 2

2 2 1

2 11

n n

n n

n n

n n

n nn

+ + +

+ +

+≥

≥ +

≥ +

≥

So,

( )

( )

3 2

2 1 2

1

12 1

2 2 1

12 2

.

n n

n n

n n

nn n

n n

n n

a a

+ +

+ + +

+

≥

≥ +

≥ +

+≥

≥

Monotonic; 1,8na ≤ bounded

90. 2

1

2

3

0.60650.73580.6694

nna ne

aaa

−=

=

=

=

Not monotonic; 0.7358,na ≤ bounded

91. ( )

1

2

3

11

112

13

nna

na

a

a

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= −

=

= −

Not monotonic; 1,na ≤ bounded

92. ( )1

2

3

23

23

49

827

nna

a

a

a

= −

= −

=

= −

Not monotonic; 2,3na ≤ bounded

93. ( ) ( ) 11

2 23 3

n nn na a

++= > =

Monotonic; 23,na ≤ bounded

94. ( ) ( ) 11

3 32 2

n nn na a

++= < =

Monotonic; lim ,nn

a→∞

= ∞ not bounded

95.

1

2

3

4

sin6

0.5000.86601.0000.8660

nna

aaaa

π⎛ ⎞= ⎜ ⎟⎝ ⎠

=

=

=

=


96.

1

2

3

cos2

cos 02

cos 1

3cos 02

nna

a

a

a

π

π

π

π

⎛ ⎞= ⎜ ⎟⎝ ⎠

= =

= = −

⎛ ⎞= =⎜ ⎟⎝ ⎠


97.

1

2

3

4

cos

0.54030.20810.32300.1634

nna

naaaa

=

=

= −

= −

= −


98.

( ) ( )

( ) ( )

( ) ( )

1 4

2 5

3 6

sin

sin 1 sin 40.8415 0.1892

1 4sin 2 sin 5

0.4546 0.19182 5

sin 3 sin 60.0470 0.0466

3 6

nna

n

a a

a a

a a

=

= ≈ = ≈ −

= ≈ = ≈ −

= ≈ = ≈ −




99. (a)

{ }

15

15 6 , bounded

n

n

an

an

= +

+ ≤ ⇒

{ }1

1 15 51

, monotonic

n

n n

an n

a a+

= + > ++

= ⇒

Therefore, { }na converges.

(b)

1lim 5 5n n→∞

⎛ ⎞+ =⎜ ⎟⎝ ⎠

100. (a)

{ }

34

34 4 , bounded

n

n

an

an

= −

− < ⇒

{ }13 44 3 , monotonic

1n n na a an n += − < − = ⇒

+


(b)

3lim 4 4n n→∞

⎛ ⎞− =⎜ ⎟⎝ ⎠

101. (a)

{ }

1 113 3

1 1 11 , bounded3 3 3

n n

nn

a

a

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞− < ⇒⎜ ⎟⎝ ⎠

{ }

1

1

1 1 1 11 13 3 3 3

, monotonic

n n n

n n

a

a a

+

+

⎛ ⎞ ⎛ ⎞= − < −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ⇒


(b)

1 1 1lim 13 3 3nn→∞

⎡ ⎤⎛ ⎞− =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

102. (a)

{ }

142

14 4.5 , bounded2

n n

nn

a

a

= +

+ ≤ ⇒

{ }

1

1

1 14 42 2

, monotonic

n n n

n n

a

a a

+

+

= + > +

= ⇒


(b)

1lim 4 42nn→∞

⎛ ⎞+ =⎜ ⎟⎝ ⎠

103. { }na has a limit because it is a bounded, monotonic

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

2 lim 4nn

a→∞

≤ ≤

104. The sequence { }na could converge or diverge. If { }na is

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

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

105. 112

n

nrA P⎛ ⎞= +⎜ ⎟

⎝ ⎠

(a) Because 0P > and 1 1,12r⎛ ⎞+ >⎜ ⎟

⎝ ⎠the sequence

diverges. lim nn

A→∞

= ∞

(b) 0.05510,000, 0.055, 10,000 112

n

nP r A ⎛ ⎞= = = +⎜ ⎟⎝ ⎠

0

1

2

3

4

5

6

7

8

9

10

10,00010,045.8310,091.8810,138.1310,184.6010,231.2810,278.1710,325.2810,372.6010,420.1410,467.90

AAAAAAAAAAA

=

=

=

=

=

=

=

=

=

=

=

−1

−1

12

7

−10

12

8

−1

−0.1

12

0.4

−1 12

−1

6



106. (a) ( )( )0

1

2

3

4

5

6

100 401 1.0025 1

0100.25200.75301.50402.51503.76605.27

nnA

AAAAAAA

= −

=

=

=

=

=

=

=

(b) 60 6480.83A =

(c) 240 32,912.28A =

107. No, it is not possible. See the "Definition of the Limit of a sequence". The number L is unique.

108. (a) A sequence is a function whose domain is the set of positive integers. (b) A sequence converges if it has a limit. See the

definition. (c) A sequence is monotonic if its terms are

nondecreasing, or nonincreasing. (d) A sequence is bounded if it is bounded below

( )for somena N N≥ and bounded above

( )for some .na M M≤

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

110. (a) 110nan

= −

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

(c) 34 1n

nan

=+

(d) Impossible. An unbounded sequence diverges.

111. (a) ( )0.8 4,500,000,000nnA =

(b) 1

2

3

4

$3,600,000,000$2,880,000,000$2,304,000,000$1,843,200,000

AAAA

=

=

=

=

(c) ( ) ( )lim lim 0.8 4.5 0,nn

n nA

→∞ →∞= = converges

112. ( )1

2

3

4

5

25,000 1.045

$26,125.00$27,300.63$28,529.15$29,812.97$31,154.55

nnP

PPPPP

=

=

=

=

=

=

113. (a) 25.364 608.04 4998.3na n n= − + +

(b) For 122012, 12: $11,522.4 billionn a= ≈

114. (a) 1127.4 17684na n= +

(b) For 222012, 22: $42,487n a= ≈

115. 10!

n

nan

=

(a) 9

9 1010 1,000,000,000 1,562,5009! 362,880 567

a a= = = =

(b) Decreasing (c) Factorials increase more rapidly than exponentials.

116.

1

2

3

4

5

6

11

2.00002.25002.37042.44142.48832.5216

1lim 1

n

n

n

n

an

aaaaaa

en→∞

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

=

≈

≈

≈

≈

⎛ ⎞+ =⎜ ⎟⎝ ⎠

117. 1n nna n n= =

1 11

2

33

44

55

66

1 1

2 1.4142

3 1.4422

4 1.4142

5 1.3797

6 1.3480

a

a

a

a

a

a

= =

= ≈

= ≈

= ≈

= ≈

= ≈

Let 1lim .nn

y n→∞

=

1 ln 1ln lim ln lim lim 0n n n

n ny nn n n→∞ →∞ →∞

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

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

lim 1.n

nn

→∞=

04000

8

10,000

50

17

50,000



118. Because lim 0,n

ns L

→∞= >

there exists for each 0,ε >

an integer N such that ns L ε− < for every .n N>

Let 0Lε = > and you have,

, ,n ns L L L s L L− < − < − < or 0 2ns L< < for

each .n N>

119. True

120. True

121. True

122. True

123. True

124. False. Let ( )1 nna = − and ( ) 11 n

nb += − then { }na and

{ }nb diverge. But { } ( ) ( ){ }11 1n nn na b ++ = − + −

converges to 0.

125. 2 1n n na a a+ += +

(a)

1 7

2 8

3 9

4 10

5 11

6 12

1 8 5 131 13 8 211 1 2 21 13 342 1 3 34 21 553 2 5 55 34 895 3 8 89 55 144

a aa aa aa aa aa a

= = + =

= = + =

= + = = + =

= + = = + =

= + = = + =

= + = = + =

(b)

1

61

72

3 8

4 9

5 10

, 1

131 1.625181212 1.61542131

3 341.5 1.61902 215 551.6667 1.61763 348 891.6 1.61825 55

nn

n

ab na

bb

bb

b b

b b

b b

+= ≥

= == =

= ≈= =

= = = ≈

= ≈ = ≈

= = = ≈

(c) 1 1

1 1 1

1 11 1

1

n n n

n n n nn

n n n

b a aa a a a ba a a

− −

− − +

+ = +

+= + = = =

(d) If lim ,nn

b ρ→∞

= then 1

1lim 1 .n nb

ρ→∞ −

⎛ ⎞+ =⎜ ⎟

⎝ ⎠

Because 1lim lim ,n nn n

b b −→∞ →∞

= you have

( )1 1 .ρ ρ+ =

2

2

1

0 1

1 1 4 1 52 2

ρ ρ

ρ ρ

ρ

+ =

= − −

± + ±= =

Because ,na and therefore ,nb is positive,

1 5 1.6180.2

ρ += ≈

126. 0 11

1 11, , 1, 2,2n n

nx x x n

x−−

= = + = …

1 6

2 7

3 8

4 9

5 10

1.5 1.4142141.41667 1.4142141.414216 1.4141141.414214 1.4142141.414214 1.414214

x xx xx xx xx x

= =

= =

= =

= =

= =

The limit of the sequence appears to be 2. In fact, this sequence is Newton's Method applied to ( ) 2 2.f x x= −



127. (a) 1

2

3

4

5

2 1.4142

2 2 1.8478

2 2 2 1.9616

2 2 2 2 1.9904

2 2 2 2 2 1.9976

a

a

a

a

a

= ≈

= + ≈

= + + ≈

= + + + ≈

= + + + + ≈

(b) 1 12 , 2, 2n na a n a−= + ≥ =

(c) First use mathematical induction to show that 2;na ≤ clearly 1 2.a ≤ So assume 2.ka ≤ Then

1

2 4

2 2

2.

k

k

k

a

a

a +

+ ≤

+ ≤

≤

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

( )( )2

2

1

2 1 0

2 02

2.

n n

n n

n n

n n

n n

a a

a aa a

a aa a +

− + ≤

− − ≤

≤ +

≤ +

≤

Because { }na is a bounding increasing sequence, it converges to some number L, by Theorem 9.5.

( )( ) ( )

2 2lim 2 2 2 0

2 1 0 2 1

nn

a L L L L L L L

L L L L→∞

= ⇒ + = ⇒ + = ⇒ − − =

⇒ − + = ⇒ = ≠ −

128. (a) 1

2

3

4

5

6 2.4495

6 6 2.9068

6 6 6 2.9844

6 6 6 6 2.9974

6 6 6 6 6 2.9996

a

a

a

a

a

= ≈

= + ≈

= + + ≈

= + + + ≈

= + + + + ≈

(b) 1 16 , 2, 6n na a n a−= + ≥ =

(c) First use mathematical induction to show that 3;na ≤ clearly 1 3.a ≤ So assume 3.ka ≤ Then

1

6 9

6 3

3.

k

k

k

a

a

a +

+ ≤

+ ≤

≤

Now show that { }na is an increasing sequence. Because 0na ≥ and 3,na ≤

( )( )2

2

1

3 2 0

6 0

6

6

.

n n

n n

n n

n n

n n

a a

a a

a a

a a

a a +

− + ≤

− − ≤

≤ +

≤ +

≤

Because { }na is a bounded increasing sequence, it converges to some number : lim .nn

L a L→∞

= So,

( )( ) ( )

2 26 6 6 0

3 2 0 3 2

L L L L L L

L L L L

+ = ⇒ + = ⇒ − − =

⇒ − + = ⇒ = ≠ −



129. (a) Use mathematical induction to show that

1 1 4 .2n

ka + +≤

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

11 4 1 1 4 .

2 2k ka k + + +

= ≤ ≤

Before proceeding to the induction step, note that

2

2 2 1 4 4 2 2 1 4 4

1 1 4 1 2 1 4 1 42 4

1 1 4 1 1 42 2

1 1 4 1 1 4 .2 2

k k k k

k k kk

k kk

k kk

+ + + = + + +

+ + + + + ++ =

⎡ ⎤+ + + ++ = ⎢ ⎥

⎢ ⎥⎣ ⎦

+ + + ++ =

So assume 1 1 4 .2n

ka + +≤ Then

1

1 1 42

1 1 42

1 1 4 .2

n

n

n

ka k k

ka k k

ka +

+ ++ ≤ +

+ ++ ≤ +

+ +≤

{ }na is increasing because

2

2

1

1 1 4 1 1 4 02 2

0

.

n n

n n

n n

n n

n n

k ka a

a a k

a a k

a a k

a a +

⎛ ⎞⎛ ⎞+ + − +− − ≤⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

− − ≤

≤ +

≤ +

≤

(b) Because { }na is bounded and increasing, it has a limit L.

(c) lim nn

a L→∞

= implies that

2

2 0

1 1 4 .2

L k L L k L

L L k

kL

= + ⇒ = +

⇒ − − =

± +⇒ =

Because 1 1 40, .2

kL L + +> =



130. (a)

( )

0 0

0 01 1 0 0

1 12 2 1 1

2 23 3 2 2

3 34 4 3 3

4 45 5 4 4

10, 310 3 6.5 10 3 5.4772

2 2

5.9886 5.96672

5.9777 5.97772

5.9777 5.97772

5.9777 5.97772

a ba ba b a b

a ba b a b

a ba b a b

a ba b a b

a ba b a b

= =

+ += = = = = ≈

+= ≈ = ≈

+= ≈ = ≈

+= ≈ = ≈

+= ≈ = ≈

The terms of { }na are decreasing and those of { }nb are increasing. They both seem to approach the same limit.

(b) For 0,n = you need to show 0 1 1 0.a a b b> > > Because 0 00 0 0 0 0 0 1, 2 .

2a ba b a a b a a+

> > + ⇒ > =

Because ( )20 0 0,a b− >

( )

2 2 2 20 0 0 0 0 0 0 0 0 0

20 0 0 0 0 0 0 0 1 1

2 0 2 4

4 2 .

a a b b a a b b a b

a b a b a b a b a b

− + > ⇒ + + >

⇒ + > ⇒ + > ⇒ >

Because 20 0 0 0 0 0 0 0 1 0, .a b a b b a b b b b> > ⇒ > ⇒ > So, you have shown that 0 1 1 0.a a b b> > >

Now assume 1 1 .k k k ka a b b+ +> > > Because 1 1,k ka b+ +>

1 11 1 1 1 22 .

2k k

k k k k ka ba a b a a+ +

+ + + + ++

> + ⇒ > =

Because ( )21 1 0,k ka b+ +− >

( )

2 2 2 21 1 1 1 1 1 1 1 1 1

21 1 1 1

1 1 1 1

2 2

2 0 2 4

4

2

.

k k k k k k k k k k

k k k k

k k k k

k k

a a b b a a b b a b

a b a b

a b a b

a b

+ + + + + + + + + +

+ + + +

+ + + +

+ +

− + > ⇒ + + >

⇒ + >

⇒ + >

⇒ >

Because 21 1 1 1 1 1 1 1 2 1, .k k k k k k k k k ka b a b b a b b b b+ + + + + + + + + +> > ⇒ > ⇒ >

So, you have shown that 1 2 2 1.k k k ka a b b+ + + +> > >

(c) { }na converges because it is decreasing and bounded below by 0. { }nb converges because it is increasing and bounded

above by 0.a

(d) Let lim nn

a A→∞

= and lim .nn

b B→∞

= Then .2

A BA A B+= ⇒ =

131. (a) ( )

( ) ( )( )

( ) ( )

1sin , sin

cos , 0 1

sin 11lim lim sin lim 1 01

n

nn n n

f x x a nn

f x x f

na n f

n n→∞ →∞ →∞

= =

′ ′= =

′= = = =

(b) ( ) ( ) ( ) ( ) ( )( )0 0

0 0 1 10 lim lim lim lim lim1 n

n n nh h

f h f f h f nf nf a

h h n n+ + →∞ →∞ →∞→ →

+ − ⎛ ⎞′ = = = = =⎜ ⎟⎝ ⎠



132. nna nr=

(a) 1 1: 02 2 2

n

n nnr a n⎛ ⎞= = = →⎜ ⎟

⎝ ⎠

(b) 1: ,nr a n= = diverges

(c) 23 3: ,

2 2nr a n⎛ ⎞= = ⎜ ⎟⎝ ⎠

diverges

(d) If 1,r ≥ the sequences diverges. If 1,r <

( ) ( )

" "

1 0.ln ln

nn

n

n

nnrr

rr r r

−

−

∞=

∞−

→ = →−

The sequence converges for 1.r <

133. (a)

( ) ( )

1ln ln 2 ln 3 ln

ln 1 2 3 ln !

nx dx n

n n

< + + +

= ⋅ ⋅ =

∫

(b)

( )1

1ln ln 2 ln 3 ln ln !

nx dx n n

+> + + + =∫

(c)

1

ln ln

ln ln 1 ln 1n n

x dx x x x C

x dx n n n n n

= − +

= − + = − +

∫

∫

From part (a): ( )ln 1

1

ln 1 ln !

!

!

n

nn n

n

n

n n n

e n

n ne

− +

−

− + <

<

<

( ) ( )

( )

1

1

1

ln ( 1) ln 1 1 1

ln 1

n

n

x dx n n n

n n

+

+

= + + − + +

= + −

∫

From part (b): ( )( )

( )

1

1ln 1

1

ln( 1) ln !

!

1!

n

nn n

n

n

n n n

e n

nn

e

+

++ −

+

+ − >

>

+>

(d) ( )

( )( )( )

( )( ) ( )

1

1

1

1 1

1 1

1 1

1!

1!

11 !

nn

n n

n nn

n

nn

n

nn ne e

nn nee

nnn nee

+

−

+

−

+

−

+< <

+< <

+< <

( )

( ) ( ) ( ) ( )

( )

1 1

1 1 1

1 1lim

1 1 1lim lim

11

1

nn

n n

n n

ee

n n nne n e

e

e

−→∞

+

→∞ →∞

=

+ + +=

=

=

By the Squeeze Theorem, ! 1lim .n

n

nn e→∞

=

(e) 20

50

100

20!20: 0.41522050!50: 0.3897

50100!100: 0.3799

1001 0.3679

n

n

n

e

= ≈

= ≈

= ≈

≈

x

y

2 43

0.5

1.0

1.5

2.0 y = lnx

x

y

y = lnx

... n

x

y

2 3 4

0.5

1.0

1.5

2.0

n + 1...

y = lnx



134. ( )1

1 11

n

nk

an k n=

=+∑

(a)

( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

1

2

3

4

5

1 1 0.51 1 2

1 1 1 1 2 1 7 0.58332 1 1 2 1 1 2 3 2 12

1 1 1 1 37 0.61673 1 1 3 1 2 3 1 1 60

1 1 1 1 1 533 0.63454 1 1 4 1 2 4 1 3 4 1 1 840

1 1 1 1 1 1 16275 1 1 5 1 2 5 1 3 5 1 4 5 1 1 25

a

a

a

a

a

= = =+⎡ ⎤ ⎡ ⎤= + = + = ≈⎢ ⎥ ⎢ ⎥+ + ⎣ ⎦⎢ ⎥⎣ ⎦

⎡ ⎤= + + = ≈⎢ ⎥

+ + +⎢ ⎥⎣ ⎦

⎡ ⎤= + + + = ≈⎢ ⎥

+ + + +⎢ ⎥⎣ ⎦

⎡ ⎤= + + + + =⎢ ⎥

+ + + + +⎢ ⎥⎣ ⎦0.6456

20≈

(b) ( )

1 1

001

1 1 1lim lim ln 1 ln 21 1

n

nn n k

a dx xn k n x→∞ →∞ =

⎡ ⎤= = = + =⎣ ⎦+ +∑ ∫

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

31

na Ln

ε− = <

whenever .n M> That is,

1 3

3 1 1or .n nε ε

⎛ ⎞> > ⎜ ⎟⎝ ⎠

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

( )1 31 .M ε> For ,n M> you have

1 3

3

3 3

1

1

1 1 0 .

n

n

n n

ε

ε

ε ε

⎛ ⎞> ⎜ ⎟⎝ ⎠

>

> ⇒ − <

So, 31lim 0.

n n→∞=

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

na L r ε− = whenever .n M> That is,

( )ln lnn r ε< or

( )lnln

nrε

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

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

( )ln.

lnM

rε

>

For ,n M> you have

( )

( )

( )

lnln

ln ln

ln ln

0 .

n

n

n

nr

n r

r

r

r

ε

ε

ε

ε

ε

>

<

<

<

− <

So, lim 0nn

r→∞

= for 1 1.r− < <

137. Answers will vary. Sample answer:

{ } ( ){ } { }

{ } ( ){ } { }22

1 1, 1, 1, 1, diverges

1 1, 1, 1, 1, converges

nn

nn

a

a

= − = − −

= − =

…

…

138. Let ( ) ( )sinf x xπ=

( )lim sinx

xπ→∞

does not exist.

( ) ( )sin 0na f n nπ= = = for all n

lim 0,nn

a→∞

= converges



139. If { }na is bounded, monotonic and nonincreasing, then 1 2 3 .na a a a≥ ≥ ≥ ≥ ≥ Then

1 2 3 na a a a− ≤ − ≤ − ≤ ≤ − ≤ is a bounded, monotonic, nondecreasing sequence which converges by the first half of the theorem. Because { }na− converges, then so does { }.na

140. Define 1 1 , 1.n nn

n

x xa nx

+ −+= ≥

( ) ( )

2 21 2 1 1

1 1 1 2

1 1 2

1

1

1n n n n n n

n n n n n n

n n n n

n n

n n

x x x x x x

x x x x x x

x x x xx x

a a

+ + − +

+ + − +

+ − +

+

+

− = = − ⇒

+ = +

+ +=

=

Therefore, 1 2 .a a a= = =… So,

1 1 1.n n n n n nx a x x ax x+ − −= − = −

141. ! 2nnT n= +

Use mathematical induction to verify the formula.

0

1

2

1 1 21 2 32 4 6

TTT

= + =

= + =

= + =

Assume ! 2 .kkT k= + Then

( ) ( ) ( )( )( ) ( ) ( )( ) ( ) ( )( )( )( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( )( )

1 1 2

1 2

2

2 2

1

1 4 4 1 4 1 8

5 ! 2 4 1 1 ! 2 4 4 2 ! 2

5 1 4 1 1 4 1 2 ! 5 4 8 1 4 1 2

5 4 4 4 1 ! 8 2

1 ! 2 .

k k k k

k k k

k

k

k

T k T k T k T

k k k k k k

k k k k k k k k k k

k k k k

k

+ − −

− −

−

−

+

= + + − + + + −

⎡ ⎤= + + − + − + + − − +⎣ ⎦

= ⎡ + − − + − + − ⎤ − + ⎡ + − + + − ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤= + − − + − + ⋅⎣ ⎦

= + +

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

Section 9.2 Series and Convergence

1. 1

2

3

4

5

141 14 91 1 14 9 161 1 1 14 9 16 25

1

1 1.2500

1 1.3611

1 1.4236

1 1.4636

S

S

S

S

S

=

= + =

= + + ≈

= + + + ≈

= + + + + ≈

2. 1

2

3

4

5

161 16 6

31 16 6 20

31 1 26 6 20 15

3 51 1 26 6 20 15 42

0.1667

0.3333

0.4833

0.6167

0.7357

S

S

S

S

S

= ≈

= + ≈

= + + ≈

= + + + ≈

= + + + + ≈

3. 1

2

3

4

5

929 272 49 27 812 4 89 27 81 2432 4 8 16

3

3 1.5

3 5.25

3 4.875

3 10.3125

S

S

S

S

S

=

= − = −

= − + =

= − + − = −

= − + − + =

4. 1

2

3

4

5

131 13 51 1 13 5 71 1 1 13 5 7 9

1

1 1.3333

1 1.5333

1 1.6762

1 1.7873

S

S

S

S

S

=

= + ≈

= + + ≈

= + + + ≈

= + + + + ≈

Section 9.2 Series and Convergence 273


5. 1

2

3

4

5

323 32 43 3 32 4 83 3 3 32 4 8 16

3

3 4.5

3 5.250

3 5.625

3 5.8125

S

S

S

S

S

=

= + =

= + + =

= + + + =

= + + + + =

6. 1

2

3

4

5

121 12 61 1 12 6 241 1 1 12 6 24 120

1

1 0.5

1 0.6667

1 0.6250

1 0.6333

S

S

S

S

S

=

= − =

= − + ≈

= − + − ≈

= − + − + ≈

7.

{ }

1

1

2 3 4, , , converges to 11 2 3

2 3 4 diverges1 2 3

n

n

nn

nan

a

a∞

=

+=

⎧ ⎫= ⎨ ⎬⎩ ⎭

= + + +∑

…

8. ( ){ } ( ) ( ) ( ){ }

( ) ( ) ( )

2 3

2 3

1

45

4 4 45 5 5

4 4 45 5 5

3

3 , 3 , 3 , converges to 0

3 3 3 converges

nn

n

nn

a

a

a∞

=

=

=

= + + +∑

…

( )45Geometric series, 1r = <

9. ( )0

76

n

n

∞

=∑

Geometric series

76 1r = >

Diverges by Theorem 9.6

10. ( )0

11105

n

n

∞

=∑

Geometric series

1110 1r = >


11. ( )0

1000 1.055 n

n

∞

=∑

Geometric series 1.055 1r = >


12. ( )0

2 1.03 n

n

∞

=

−∑

Geometric series

1.03 1r = >


13. 1 1

lim 1 01

n

n

nn

nn

∞

=

→∞

+

= ≠+

∑


14. 1 2 3

1lim 02 3 2

n

n

nn

nn

∞

=

→∞

+

= ≠+

∑


15. 2

21

2

2

1

lim 1 01

n

n

nn

nn

∞

=

→∞

+

= ≠+

∑


16.

( )

21

2 2

11lim lim 1 0

1 1 1

n

n n

nn

nn n

∞

=

→∞ →∞

+

= = ≠+ +

∑


17. 11

1

2 12

2 1 1 2 1lim lim 02 2 2

n

nn

n n

nn n

∞

+=

−

+→∞ →∞

+

+ += = ≠

∑


18. 1

!2

!lim2

nn

nn

n

n

∞

=

→∞= ∞

∑


19. ( )0

0 1 2

9 91 1 14 4 4 4 16

9 9 5 45 9 214 4 4 16 4 16

1

, , 2.95,

n

n

S S S

∞

=

⎡ ⎤= + + +⎣ ⎦

= = ⋅ = = ⋅ ≈

∑

…

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

( )( )0

9 4 9 49 14 4 1 1 4 3 4 3

n

n

∞

=−= = =∑



20. ( )0

0 1 2

23

2 413 9

51, , 2.11, .3

n

n

S S S

∞

=

= + + +

= = ≈

∑

…

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

( )0

23

1 1 31 2 3 1 3

n

n

∞

=

= = =−∑

21. 0

0 1 2

1 14 16

15 1 15 14 4 4

15 45, , 3.05,4 16

n

n

S S S

∞

=

⎛ ⎞ ⎡ ⎤− = − + −⎜ ⎟ ⎣ ⎦⎝ ⎠

= = ≈

∑

…

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

( )0

15 1 15 4 15 4 34 4 1 1 4 5 4

n

n

∞

=

⎛ ⎞− = = =⎜ ⎟ − −⎝ ⎠∑

22. 0

0 1 3

17 8 17 8 6413 9 3 9 81

17, 0.63, 5.1,3

n

n

S S S

∞

=

⎛ ⎞ ⎡ ⎤− = − + −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

= ≈ ≈

∑

…

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

( )0

17 8 17 3 17 3 33 9 1 8 9 17 9

n

n

∞

=

⎛ ⎞− = = =⎜ ⎟ − −⎝ ⎠∑

23. 0

0 1

17 1 17 1 113 2 3 2 4

17 17, ,3 6

n

n

S S

∞

=

⎛ ⎞ ⎡ ⎤− = − + −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

= =

∑

…

Matches (f ). The series is geometric:

( )0

17 1 17 1 17 2 343 2 3 1 1 2 3 3 9

n

n

∞

=

⎛ ⎞− = = =⎜ ⎟ − −⎝ ⎠∑

24. ( )0

0 1

25

2 415 25

71, ,5

n

n

S S

∞

=

= + + +

= =

∑

…

Matches (e). The series is geometric:

( )0

2 1 55 1 2 5 3

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

25. ( )0

56

n

n

∞

=∑

Geometric series with 56 1r = <

Converges by Theorem 9.6

26. ( )0

122

n

n

∞

=

−∑

Geometric series with 12 1r = − <


27. ( )0

0.9 n

n

∞

=∑

Geometric series with 0.9 1r = <


28. ( )0

0.6 n

n

∞

=

−∑

Geometric series with 0.6 1r = − <


29. ( )

( )

1 1

1

1 1 1 1 1 1 1 1 1 1 11 , 11 1 2 2 3 3 4 4 5 1

1 1lim lim 1 11 1

nn n

nn nn

Sn n n n n

Sn n n

∞ ∞

= =

∞

→∞ →∞=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − + − + − + − + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= = − =⎜ ⎟+ +⎝ ⎠

∑ ∑

∑

30. ( ) ( )

( ) ( ) ( )

1 1

1

1 1 1 1 1 1 1 1 1 1 1 1 12 2 2 2 2 6 4 8 6 10 8 12 10 14

1 1 1 1 1 1 1 3lim lim2 2 4 2 1 2 2 2 4 4

n n

nn nn

n n n n

Sn n n n

∞ ∞

= =

∞

→∞ →∞=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − + − + − + − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟+ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

⎡ ⎤= = + − − = + =⎢ ⎥

+ + +⎢ ⎥⎣ ⎦

∑ ∑

∑

31. (a) ( )1 1

6 1 1 1 1 1 1 1 1 12 2 13 3 4 2 5 3 6 4 7n nn n n n

∞ ∞

= =

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − + − + − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∑ ∑

1 1 1 1 1 1 1 112 1 2 1 3.6672 3 1 2 3 2 3 3nS

n n n⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞= + + − + + = + + = ≈⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥+ + +⎝ ⎠ ⎝ ⎠⎣ ⎦⎝ ⎠

(b)

n 5 10 20 50 100

nS 2.7976 3.1643 3.3936 3.5513 3.6078



(c)

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

32. (a) ( )1 1

4 1 14 4

115

n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

= −

∑ ∑

1 12 6

⎛ ⎞+ −⎜ ⎟

⎝ ⎠

1 13 7

⎛ ⎞+ −⎜ ⎟

⎝ ⎠

1 14 8

⎛ ⎞+ −⎜ ⎟

⎝ ⎠

15

⎛ ⎞+⎜ ⎟

⎝ ⎠

19

−16

⎛ ⎞+⎜ ⎟

⎝ ⎠

110

−

1 1 1 251 2.08332 3 4 12

⎛ ⎞+⎜ ⎟

⎝ ⎠

= + + + = ≈

(b) (c)


33. (a) ( ) ( )1

1 0

22 0.9 2 0.9 201 0.9

n n

n n

∞ ∞−

= =

= = =−∑ ∑

(b)

(c)


34. (a) ( ) 1

1

33 0.85 201 0.85

n

n

∞−

=

= =−∑ (Geometric series)

(b) (c)


n 5 10 20 50 100

nS 1.5377 1.7607 1.9051 2.0071 2.0443

n 5 10 20 50 100

nS 8.1902 13.0264 17.5685 19.8969 19.9995

n 5 10 20 50 100

nS 11.1259 16.0625 19.2248 19.9941 19.999998

00

5

11

00 11

4

00

11

22

00

11

24



35. (a) ( ) 1

1

10 4010 0.25 13.33331 0.25 3

n

n

∞−

=

= = ≈−∑

(b) (c)

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

36. (a) ( )

1

1

1 5 155 3.753 1 1 3 4

n

n

−∞

=

⎛ ⎞− = = =⎜ ⎟ − −⎝ ⎠∑

(b)

(c)

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

37. ( )0

1 1 22 1 1 2

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

38. ( )0

4 66 305 1 4 5

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

(Geometric)

39. ( )0

1 1 33 1 1 3 4

n

n

∞

=

⎛ ⎞− = =⎜ ⎟ −⎝ ⎠∑

40. ( )0

6 3 2137 1 6 7 13

n

n

∞

=

⎛ ⎞− = =⎜ ⎟ − −⎝ ⎠∑

41. 22 2 2

1 1 2 1 2 1 1 11 1 1 2 1 1n n nn n n n n

∞ ∞ ∞

= = =

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟− − + − +⎝ ⎠ ⎝ ⎠∑ ∑ ∑

1 1 1 1 1 1 1 1 1 1 1 1 1 11 12 3 2 4 3 5 2 1 1 2 2 1nS

n n n n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − + − = + − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− − + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

22

1 1 1 1 1 3lim lim 11 2 2 1 4n

n nnS

n n n

∞

→∞ →∞=

⎛ ⎞= = + − − =⎜ ⎟− +⎝ ⎠∑

n 5 10 20 50 100

nS 13.3203 13.3333 13.3333 13.3333 13.3333

n 5 10 20 50 100

nS 3.7654 3.7499 3.7500 3.7500 3.7500

1100

15

00 11

7



42. ( )1 1

4 1 122 2n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠∑ ∑

1 1 1 1 1 1 1 1 1 1 1 12 1 2 13 2 4 3 5 1 1 2 2 1 2nS

n n n n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − + − = + − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− + + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

( )1

4 1 1 1lim lim 2 1 32 2 1 2n

n nnS

n n n n

∞

→∞ →∞=

⎛ ⎞= = + − − =⎜ ⎟+ + +⎝ ⎠∑

43. ( )( )1 1

8 1 181 2 1 2n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ + + +⎝ ⎠∑ ∑

1 1 1 1 1 1 1 1 1 18 82 3 3 4 4 5 1 2 2 2nS

n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

( )( )1

8 1 1lim lim 8 41 2 2 2n

n nnS

n n n

∞

→∞ →∞=

⎛ ⎞= = − =⎜ ⎟+ + +⎝ ⎠∑

44. ( )( )1 1

1 1 1 12 1 2 3 2 2 1 2 3n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ + + +⎝ ⎠∑ ∑

1 1 1 1 1 1 1 1 1 1 1 12 3 5 5 7 7 9 2 1 2 3 2 3 2 3nS

n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

( )( )1

1 1 1 1 1lim lim2 1 2 3 2 3 2 3 6n

n nnS

n n n

∞

→∞ →∞=

⎛ ⎞= = − =⎜ ⎟+ + +⎝ ⎠∑

45. ( )0

1 1 1010 1 1 10 9

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

46. ( )0

3 88 324 1 3 4

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

47. ( )0

1 3 933 1 1 3 4

n

n

∞

=

⎛ ⎞− = =⎜ ⎟ − −⎝ ⎠∑

48. ( )0

1 4 842 1 1 2 3

n

n

∞

=

⎛ ⎞− = =⎜ ⎟ − −⎝ ⎠∑

49.

( ) ( )

0 0 0

1 1 1 12 3 2 3

1 1 3 121 1 2 1 1 3 2 2

n n

n nn n n

∞ ∞ ∞

= = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − = − =− −

∑ ∑ ∑

50. ( ) ( )

( ) ( )

1 0 0

7 90.7 0.9 210 101 1 2

1 7 10 1 9 10

10 3410 23 3

n nn n

n n n

∞ ∞ ∞

= = =

⎛ ⎞ ⎛ ⎞⎡ ⎤+ = + −⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠

= + −− −

= + − =

∑ ∑ ∑

51. Note that ( )sin 1 0.8415 1.≈ < The series ( )1

sin 1n

n

∞

=

⎡ ⎤⎣ ⎦∑

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

( ) ( ) ( ) ( )( )1 0

sin 1sin 1 sin 1 sin 1 5.3080.

1 sin 1n n

n n

∞ ∞

= =

⎡ ⎤ = ⎡ ⎤ = ≈⎣ ⎦ ⎣ ⎦ −∑ ∑

52. ( )( )2

1 1

1 1

1 19 3 2 3 1 3 2

1 1 1 1 19 3 9 6 3 3 1 3 2

1 1 1 1 1 1 1 1 1 1 1 13 2 5 5 8 8 11 3 1 3 2 3 2 3 2

n n

nk k

n n

k k

Sk k k k

k k k k

n n n

= =

= =

= =+ − − +

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥− + − +⎣ ⎦ ⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

∑ ∑

∑ ∑

1 1 1 1lim lim3 2 3 2 6n

n nS

n→∞ →∞

⎛ ⎞= − =⎜ ⎟+⎝ ⎠



53. (a) 0

4 10.410 10

n

n

∞

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

(b) Geometric series with 410

a = and 110

r =

( )

4 10 41 1 1 10 9

aSr

= = =− −

54. (a)

1 0

9 9 90.910 100 1000

1 9 1910 10 10

n n

n n

∞ ∞

= =

= + + +

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑

(b) ( )

9 10.9 110 1 1 10

= =−

55. (a) 0

81 10.81100 100

n

n

∞

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑


a = and 1100

r =

( )

81 100 81 91 1 1 100 99 11

aSr

= = = =− −

56. (a) 1 0

1 1 10.01100 100 100

n n

n n

∞ ∞

= =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑

(b) ( )

1 1 1 100 10.01100 1 1 100 100 99 99

= ⋅ = ⋅ =−

57. (a) 0

3 10.07540 100

n

n

∞

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑


a = and 1100

r =

3 40 51 99 100 66

aSr

= = =−

58. (a) 0

1 3 10.2155 200 100

n

n

∞

=

⎛ ⎞= + ⎜ ⎟⎝ ⎠

∑


a = and 1100

r =

1 1 3 200 715 1 5 99 100 330

aSr

= + = + =−

59. ( )0

1.075 n

n

∞

=∑

Geometric series with 1.075r =


60. 1

31000

n

n

∞

=∑

Geometric series with 3 1.r = >


61. 1

1010 1n

nn

∞

=

++∑

10 1lim 010 1 10n

nn→∞

+= ≠

+


62. 1

4 13 1n

nn

∞

=

+−∑

4 1 4lim 03 1 3n

nn→∞

+= ≠

−


63. 1

1 12n n n

∞

=

⎛ ⎞−⎜ ⎟+⎝ ⎠∑

1 1 1 1 1 1 1 1 1 1 1 11 13 2 4 3 5 1 1 2 2 1 2nS

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

1

1 1 1 1 1 3lim lim 1 , converges2 2 1 2 2n

n nnS

n n n n

∞

→∞ →∞=

⎛ ⎞ ⎛ ⎞− = = + − − =⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠∑

64. 1

1 11 2n n n

∞

=

⎛ ⎞−⎜ ⎟+ +⎝ ⎠∑

1 1 1 1 1 1 1 12 3 3 4 1 2 2 2nS

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

1

1 1 1 1 1lim lim , converges1 2 2 2 2n

n nnS

n n n

∞

→∞ →∞=

⎛ ⎞ ⎛ ⎞− = = − =⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠∑



65. ( )1 1

1 1 1 13 3 3n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠∑ ∑

1 1 1 1 1 1 1 1 1 1 1 1 1 1 13 1 4 2 5 3 6 4 7 2 1 1 2 3

1 1 1 1 1 113 2 3 1 2 3

nSn n n n n n

n n n

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + − + + − − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− + − + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞= + + − − −⎜ ⎟+ + +⎝ ⎠

( )1

1 1 1 1 1 1 1 1 11 11lim lim 1 , converges3 3 2 3 1 2 3 3 6 8n

n nnS

n n n n n

∞

→∞ →∞=

⎛ ⎞ ⎛ ⎞= = + + − − − = =⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠∑

66. ( )1 1

1 1 1 12 1 2 1n nn n n n

∞ ∞

= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠∑ ∑

1 1 1 1 1 1 1 1 1 11 12 2 2 3 3 4 1 2 1nS

n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

( )1

1 1 1 1lim lim 1 , converges2 1 2 1 2n

n nnS

n n n

∞

→∞ →∞=

⎛ ⎞= = − =⎜ ⎟+ +⎝ ⎠∑

67. 1

3 12 1n

nn

∞

=

−+∑

3 1 3lim 02 1 2n

nn→∞

−= ≠

+


68. 31

3n

n n

∞

=∑

( )

( ) ( )

3 2

2 3

ln 2 33lim lim3

ln 2 3 ln 3lim lim

6 6

nn

n n

n n

n n

n n

nn

→∞ →∞

→∞ →∞

=

= = = ∞

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

69. 0 0

4 142 2

n

nn n

∞ ∞

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ ∑

Geometric series with 12

r =


70. 0 0

3 13 ,5

n

nn nS

∞ ∞

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ ∑ convergent


r =

71. Because ( )ln ,n n> the terms ( )lnnnan

= do not

approach 0 as .n → ∞ So, the series ( )2 lnn

nn

∞

=∑ diverges.

72. ( )

( )1 1

1ln ln

0 ln 2 ln 3 ln

n n

nk k

S kk

n= =

⎛ ⎞= = −⎜ ⎟⎝ ⎠

= − − − −

∑ ∑

Because lim nn

S→∞

diverges, 1

1lnn n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ diverges.

73. For 0,k ≠

lim 1 lim 1

0.

kn n k

n n

k

k kn n

e

→∞ →∞

⎡ ⎤⎛ ⎞ ⎛ ⎞+ = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= ≠

For ( )0, lim 1 0 1 0.n

nk

→∞= + = ≠

So, 1

1n

n

kn

∞

=

⎡ ⎤+⎢ ⎥⎣ ⎦∑ diverges.

74. 1 1

1 nn

n ne

e

∞ ∞−

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ ∑ converges because it is geometric

with

1 1.re

= <

75. lim arctan 02n

n π→∞

= ≠

So, 1arctan

nn

∞

=∑ diverges.



76.

( ) ( ) ( )( )( ) ( ) ( )

1

1ln

2 3 1ln ln ln1 2

ln 2 ln1 ln 3 ln 2 ln 1 ln

ln 1 ln 1 ln 1

n

nk

kSk

nn

n n

n n

=

+⎛ ⎞= ⎜ ⎟⎝ ⎠

+⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − + − + + + −

= + − = +

∑

Diverges

77. See definitions on page 608.

78. lim 5nn

a→∞

= means that the limit of the sequence { }na is

5.

1 21

5nn

a a a∞

=

= + + =∑ means that the limit of the

partial sums is 5.

79. The series given by

2

0, 0n n

nar a ar ar ar a

∞

=

= + + + + + ≠∑

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

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

1.r ≥

80. If lim 0,nn

a→∞

≠ then 1

nn

a∞

=∑ diverges.

81. (a) 1 2 31

nn

a a a a∞

=

= + + +∑

(b) 1 2 31

kk

a a a a∞

=

= + + +∑

These are the same. The third series is different, unless 1 2a a a= = = is constant.

(c) 1

k k kn

a a a∞

=

= + +∑

82. (a) Yes, the new series will still diverge. (b) Yes, the new series will converge.

83. 1 1 02 2 2 2

n nn

nn n n

x x x x∞ ∞ ∞

= = =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

Geometric series: converges for 12x

< or 2x <

( )

( )

02 21 2 , 2

2 1 2 2 2 2

n

n

x xf x

x x x xx x x

∞

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

= = = <− − −

∑

84. ( ) ( ) ( )1 0

3 3 3n n

n nx x x

∞ ∞

= =

=∑ ∑

Geometric series: converges for 3 13

x x 1< ⇒ <

( ) ( ) ( ) ( )0

1 33 3 3 ,1 3 1 3 3

n

n

xf x x x x xx x

∞

=

1= = = <

− −∑

85. ( ) ( ) ( )1 0

1 1 1n n

n nx x x

∞ ∞

= =

− = − −∑ ∑

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

( ) ( ) ( )

( ) ( )

01 1

1 11 , 0 21 1 2

n

nf x x x

xx xx x

∞

=

= − −

−= − = < <

− − −

∑

86. 0

344

n

n

x∞

=

−⎛ ⎞⎜ ⎟⎝ ⎠

∑

Geometric series: converges for

3 1 3 4 1 74

x x x−< ⇒ − < ⇒ − < <

( ) ( )

( )

0

3 444 1 3 4

4 16 , 1 74 3 4 7

n

n

xf xx

xx x

∞

=

−⎛ ⎞= =⎜ ⎟ − ⎡ − ⎤⎝ ⎠ ⎣ ⎦

= = − < <− + −

∑

87. ( ) ( )0 0

1 n nn

n nx x

∞ ∞

= =

− = −∑ ∑


( ) ( )

0

1 1 1 1

1 , 1 11

n

n

x x x

f x x xx

∞

=

− < ⇒ < ⇒ − < <

= − = − < <+∑

88. ( ) ( )2 2

0 01

nn n

n nx x

∞ ∞

= =

− = −∑ ∑


( ) ( ) ( )

2

222

0

1 1 1

1 1 , 1 111

n

n

x x

f x x xxx

∞

=

− < ⇒ − < <

= − = = − < <+− −

∑

89. 0

1 n

n x

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Geometric series: converges if 1 1x

<

1 1 or 1x x x⇒ > ⇒ < − >

( ) ( )0

1 1 , 1 or 11 1 1

n

n

xf x x xx x x

∞

=

⎛ ⎞= = = > < −⎜ ⎟ − −⎝ ⎠∑



90. 2 2 2

2 2 21 04 4 4

n n

n n

x x xx x x

∞ ∞

= =

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠

∑ ∑

Geometric series: converges for 2

2 14

xx

<+

2 2 4x x⇒ < + ⇒ converges for all x

( )2 2 2

22 2

2

1 44 4 4 41

4

x x x xf xxx x

x

2+= ⋅ = =

+ +−+

91. ( ) ( )( )

22

2 0 0

1 11 111

nn n

n n nc c

cc

∞ ∞ ∞− − −

= = =

⎛ ⎞+ = + = ⎜ ⎟+⎝ ⎠+∑ ∑ ∑

For convergence, 1 1 1 11

0 or 2

cc

c c

< ⇒ < ++

⇒ > < −

( )

( )

( )

20

2

2

1 1211

1 12 111

112 1

2 2 1 0

n

n cc

ccc

c

cc

c c

∞

=

⎛ ⎞= ⎜ ⎟+⎝ ⎠+

++ = =

⎛ ⎞− ⎜ ⎟+⎝ ⎠

+ =

+ − =

∑

2 4 8 1 34 2

c − + + − ±= =

Because 1 32 0,2

− −− < < the answer is

1 32

c − +=

92. ( )0 0

5

1 51

11545

4ln5

ncn c

n n

c

c

c

e e

e

e

e

c

∞ ∞

= =

= =

=−

− =

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ ∑

93. Neither statement is true. The formula

21 11

x xx= + + +

−

holds for 1 1.x− < <

94. (a) False. The fact that 41 0n

→ is irrelevant to the

convergence of 41

1 .n n

∞

=∑ Furthermore, 4

1

1 0.n n

∞

=

≠∑

(b) False. The fact that 4

1 0n

→ is irrelevant to the

convergence of 4

1 .n∑ In fact,

4

1n∑ diverges.

95. (a) x is the common ratio.

(b) 2

0

11 , 11

n

nx x x x

x

∞

=

+ + + = = <−∑

(c) 1

22 3

2 3 43 5

11

1

1

yx

y S x x

y S x x x x

=−

= = + +

= = + + + +

Answers will vary.

96. (a) 2x⎛ ⎞−⎜ ⎟

⎝ ⎠is the common ratio.

(b)

( )

2 3

01

2 4 8 21

1 2

2 , 22

n

n

x x x x

x

xx

∞

=

⎛ ⎞− + − + = −⎜ ⎟⎝ ⎠

=− −

= <+

∑

(c) 1

2

2 3

2 3 4

3 5

22

12 4

12 4 8 16

yx

x xy S

x x x xy S

=+

= = − +

= = − + − +

Answers will vary.

1.5−1.50

f

S3

S5

3

f

S3

S5

5

5

−5

−5



97. ( ) 1 0.531 0.5

xf x

⎡ ⎤−= ⎢ ⎥−⎣ ⎦

Horizontal asymptote: 6y =

( )

0

132

3 61 1 2

n

n

S

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

= =−

∑

The horizontal asymptote is the sum of the series.

( )f n is the nth partial sum.

98. ( ) 1 0.821 0.8

xf x

⎡ ⎤−= ⎢ ⎥−⎣ ⎦

Horizontal asymptote: 10y =

( )

( )

0

452

1 101 4 5

n

n

S

∞

=

= =−

∑

The horizontal asymptote is the sum of the series. ( )f n is the nth partial sum.

99. ( )

( )( )

2

2

2

1 0.00011

10,000

0 10,000

1 1 4 1 10,0002

n n

n n

n n

n

<+

< +

< + −

− ± − −=

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

1 0.00018

10,000 8

n

n

⎛ ⎞ <⎜ ⎟⎝ ⎠

<

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

100. 1 0.00012

10,000 2

n

n

<

<

This inequality is true when 14.n =

( )0.01 0.0001

10,000 10

n

n

<

<

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

101. ( )1

0

8000 1 0.958000 0.95

1 0.95

160,000 1 0.95 , 0

nni

i

n n

−

=

⎡ ⎤−⎣ ⎦=−

⎡ ⎤= − >⎣ ⎦

∑

102. ( ) ( ) ( )

( ) ( )5475,000 1 0.3 475,000 0.7

5 475,000 0.7 $79,833.25

n nV t

V

= − =

= =

103. ( )0

200 0.75 800 million dollarsi

i

∞

=

=∑

104. ( )0

200 0.60 500 million dollarsi

i

∞

=

=∑

105.

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( )

1

2

up down

2 2 23

2

0

16

0.81 16 0.81 16 32 0.81

16 0.81 16 0.81 32 0.81

16 32 0.81 32 0.81

3216 32 0.81 161 0.81

152.42 feet

n

n

D

D

D

D∞

=

=

= + =

= + =

= + + +

= − + = − +−

≈

∑

106. The ball in Exercise 105 takes the following times for each fall.

( )

( )

( )

( )

( )

21 1

22 2

2 223 3

1 12

16 16 0 if 116 16 0.81 0 if 0.9

16 16 0.81 0 if 0.9

16 16 0.81 0 if 0.9n nn n

s t s ts t s t

s t s t

s t s t− −

= − + = == − + = =

= − + = =

= − + = =

Beginning with 2,s 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 01 2 0.9 1 2 0.9

21 19 seconds.1 0.9

n n

n nt

∞ ∞

= =

= + = − +

= − + =−

∑ ∑

−2 10

−1

y = 67

−2 16

−2

y = 1012



107. ( )

( )2

1 12 2

1 1 122 2 8

n

P n

P

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

( )0

1 1 1 2 12 2 1 1 2

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

108. ( )

( )2

1 23 3

1 2 423 3 27

n

P n

P

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

( )0

1 2 1 3 13 3 1 2 3

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

109. (a) ( )( )1 0

1 1 1 1 1 12 2 2 2 1 1 2

n n

n n

∞ ∞

= =

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠∑ ∑

(b) No, the series is not geometric.

(c) 1

1 22

n

nn

∞

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

∑

110. ( )

( )

( )

4 70

2 5 80

3 6 90

1 1 1 1 1 1 1 4Person 1:2 2 2 2 8 2 1 1 8 7

1 1 1 1 1 1 1 2Person 2:2 2 2 4 8 4 1 1 8 7

1 1 1 1 1 1 1 1Person 3:2 2 2 8 8 81 1 8 7

4 2 1Sum: 17 7 7

n

n

n

n

n

n

∞

=

∞

=

∞

=

⎛ ⎞+ + + = = =⎜ ⎟ −⎝ ⎠

⎛ ⎞+ + + = = =⎜ ⎟ −⎝ ⎠

⎛ ⎞+ + + = = =⎜ ⎟ −⎝ ⎠

+ + =

∑

∑

∑

111. (a) 264 32 16 8 4 2 126 in.+ + + + + =

(b) ( )

2

0

1 6464 128 in.2 1 1 2

n

n

∞

=

⎛ ⎞ = =⎜ ⎟ −⎝ ⎠∑

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

112. (a) 11

1 1 21 1 1

1

1 2 31 2 1 1

1 1

2 3

sin sin

sin sin sin

sin sin sin

sinTotal: sin sin sin1 sin

yy

yy

YY z

zx y

x y Y zY

x yx y x y z

x y

z z z z

θ θ

θ θ θ

θ θ θ

θθ θ θθ

= ⇒ =

= ⇒ = =

= ⇒ = =

+ + + =−

(b) If 1z = and ,6πθ = then

( )1 2total 1.

1 1 2= =

−

113. ( ) ( )20 19 20

11

1 0

11.06

100,000 1 100,000 1 1.06100,000 20, 1.06 $1,146,992.121.06 1.06 1.06 1 1.06

in

n in r

−−

−= =

⎡ ⎤−⎛ ⎞= = = = ≈⎢ ⎥⎜ ⎟ −⎝ ⎠ ⎣ ⎦∑ ∑

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

114. ( ) ( ) ( ) ( )2 22 21 1

3 9Surface area 4 1 9 4 9 4 4π π π π π⎛ ⎞= + + ⋅ + = + + = ∞⎜ ⎟⎝ ⎠

16 in.

16 in.



115. ( ) ( ) ( )1

0

0.01 1 20.01 2 0.01 2 1

1 2

nni n

iw

−

=

−= = = −

−∑

(a) When 29: $5,368,709.11n w= =

(b) When 30: $10,737,418.23n w= =

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

116.

( )( )( ) ( )

12

12 1

0

12

12

121212 1

1212 12

0

1 112

112 1 1

12

12 1 112

12 1 112

1 11 1

t

nt

n

t

t

tr rtt nrr r

n

rPrP

r

rPr

rPr

P e P eP e

e e

−

=

−

=

⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎛ ⎞ ⎣ ⎦+ =⎜ ⎟ ⎛ ⎞⎝ ⎠ − +⎜ ⎟⎝ ⎠

⎡ ⎤⎛⎛ ⎞ ⎛ ⎞⎢ ⎥= − − +⎜⎜ ⎟ ⎜ ⎟⎜⎝ ⎠ ⎝ ⎠⎢ ⎥⎝⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

− −= =

− −

∑

∑

117. 45, 0.03, 20P r t= = =

(a) ( )12 2012 0.0345 1 1 $14,773.59

0.03 12A

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(b) ( )( )0.03 20

0.03 12

45 1$14,779.65

1

eA

e

−= ≈

−

118. 75, 0.055, 25P r t= = =

(a) ( )12 2512 0.05575 1 1 $48,152.81

0.055 12A

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(b) ( )( )0.055 25

0.055 12

75 1$48,245.07

1

eA

e

−= ≈

−

119. 100, 0.04, 35P r t= = =

(a) ( )12 3512 0.04100 1 1 $91,373.09

0.04 12A

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(b) ( )( )0.04 35

0.04 12

100 1$91,503.32

1

eA

e

−= ≈

−

120. 30, 0.06, 50P r t= = =

(a) ( )12 5012 0.0630 1 1 113,615.73

0.06 12A

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(b) ( )( )0.06 50

0.06 12

30 1$114,227.18

1

eA

e

−= ≈

−

121. ( ) ( )

( )

39

39

0

40

50,000 50,000 1.04 50,000 1.04

50,000 1.04

1 1.0450,000 $4,751,275.791 1.04

n

n

T

=

= + + +

=

⎡ ⎤−= ≈⎢ ⎥−⎣ ⎦

∑

122. ( )39

050,000 1.045 $5,351,516.15n

nT

=

= ≈∑

123. False. 1lim 0,n n→∞

= but 1

1

n n

∞

=∑ diverges.

124. True

125. False; 1 1

n

n

aar ar

∞

=

⎛ ⎞= −⎜ ⎟−⎝ ⎠∑

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

126. True

( )

1lim 01000 1 1000n

nn→∞

= ≠+

127. True

( )

3 4

30

3

3

9 90.74999 0.7410 10

9 10.7410 109 10.74

10 1 1 10

9 100.7410 9

10.74 0.75100

n

n

∞

=

= + + +

⎛ ⎞= + ⎜ ⎟⎝ ⎠

= + ⋅−

= + ⋅

= + =

∑

…

128. True



129. By letting 0 0,S = you have

1

11 1

.n n

n k k n nk k

a a a S S−

−= =

= − = −∑ ∑ So,

( )

( )

( ) ( )

11 1

11

11

.

n n nn n

n nn

n nn

a S S

S S c c

c S c S

∞ ∞

−= =

∞

−=

∞

−=

= −

= − + −

= ⎡ − − − ⎤⎣ ⎦

∑ ∑

∑

∑

130. Let { }nS be the sequence of partial sums for the

convergent series

1

.nn

a L∞

=

=∑ Then lim nn

S L→∞

= and because

1

,n k nk n

R a L S∞

= +

= = −∑

you have

( )lim lim lim lim 0.n n nn n n n

R L S L S L L→∞ →∞ →∞ →∞

= − = − = − =

131. Let 01n

na

∞

=

=∑ ∑ and ( )0

1 .nn

b∞

=

= −∑ ∑

Both are divergent series.

( ) ( ) [ ]0 0

1 1 1 1 0n nn n

a b∞ ∞

= =

+ = ⎡ + − ⎤ = − =⎣ ⎦∑ ∑ ∑

132. If ( )n na b∑ + converged, then

( )n n n na b a b∑ + − ∑ = ∑ would converge, which is a

contradiction. So, ( )n na b∑ + diverges.

133. Suppose, on the contrary, that nca∑ converges. Because 0,c ≠

1n nca a

c⎛ ⎞ =⎜ ⎟⎝ ⎠

∑ ∑

converges. This is a contradiction since nca∑ diverged. So, nca∑ diverges.

134. If na∑ converges, then lim 0.nn

a→∞

= So, 1lim 0,n na→∞

≠

which implies that 1na∑ diverges.

135. (a) 3 1 2

1 2 2 3 1 2 3 1 2 3 1 3

1 1 1n n n

n n n n n n n n n n n n

a a aa a a a a a a a a a a a

+ + +

+ + + + + + + + + + + +

−− = = =

(b) 1 30

1 2 2 30

1 2 2 3 2 3 3 4 1 2 2 3 1 2 2 3 2 3

1

1 1

1 1 1 1 1 1 1 1 11

n

nk kk

n

k k k kk

n n n n n n n n

Sa a

a a a a

a a a a a a a a a a a a a a a a a a

+ +=

+ + + +=

+ + + + + + + +

=

⎡ ⎤= −⎢ ⎥

⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + − + + − = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∑

∑

1 3 2 30

1 1lim lim 1 1nn nn n n nn

Sa a a a

∞

→∞ →∞+ + + +=

⎡ ⎤= = − =⎢ ⎥

⎣ ⎦∑

136.

( ) ( ) ( ) ( )

2 3 2 2 12

2 4 2 3 2 1 2 2

0 0

1 2 2 2

1 2 2 2 2

n nn

n nk kn n

k n

S x x x x x

x x x x x x x x x

+

+

= =

= + + + + + +

= + + + + + + + + = +∑ ∑

As ,n → ∞ the two geometric series converge for 1:x <

2 2 2 21 1 1 2lim 2 , 1

1 1 1nn

xS x xx x x→∞

+⎛ ⎞= + = <⎜ ⎟− − −⎝ ⎠

137. ( )2 3

0

1 1 1 1 1 1 1 1since 11 1 1

n

n

rr r r r r r r r

∞

=

⎛ ⎞⎛ ⎞+ + + = = = <⎜ ⎟⎜ ⎟ − −⎝ ⎠ ⎝ ⎠∑

This is a geometric series which converges if

1 1 1.rr

< ⇔ >



138. ( )( )

( )( ) ( ) ( )

11 2 1 2

1 1 2 2 1n n

A B Cn n n n n n

A n n B n C n

= + ++ + + +

= + + + + + +

When 0, 1 2 1 2n A A= = ⇒ =

When 1, 1 1n B B= − = − ⇒ = − When 2, 1 2 1 2n C C= − = ⇒ =

( )( )1

1 1 2 1 1 21 2 1 2n n n n n n n

∞

=

−= + +

+ + + +∑

1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 21 2 3 2 3 4 3 4 4 1 2

1 2 1 1 2 1 1 1 1 1 2 1 1 21 2 2 3 3 4 4 1 2

nSn n n

n n n

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + − + + − + + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + − + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

As , 1 4nn S→ ∞ →

( )( )1

1 11 2 4n n n n

∞

=

=+ +∑

139. (a)

(b) ( )

( )

( )

121

00

12 31 20

03 41 2 3

0

112 2

1 1 12 3 2 3 6

1 1 13 4 3 4 12

xx dx x

x xx x dx

x xx x dx

⎡ ⎤− = − =⎢ ⎥

⎣ ⎦

⎡ ⎤− = − = − =⎢ ⎥

⎣ ⎦⎡ ⎤

− = − = − =⎢ ⎥⎣ ⎦

∫

∫

∫

(c) ( )111 1

00

1

1 11 1

1 1 11 12 2 3

n nn n

n

nn

x xa x x dxn n n n

a

+−

∞

=

⎡ ⎤= − = − = −⎢ ⎥+ +⎣ ⎦

⎛ ⎞ ⎛ ⎞= − + − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

∑

Note that the square has area 1.=

140. The entire rectangle has area 2 because the height is 1

and the base is 1 11 2.2 4

+ + + = The squares all lie

inside the rectangle, and the sum of their areas is

2 2 21 1 11 .2 3 4

+ + + +

So, 21

1 2.n n

∞

=

<∑

141. 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

( )12 n ktkt ktnT P Pe Pe Pe −= + + + + where

( )ln 2 .k H= − One time interval after the last dose is

administered is given by 2 3

1 .kt kt kt nktnT Pe Pe Pe Pe+ = + + + + Two time

intervals after the last dose is administered is given by ( )12 3 4

2n ktkt kt kt

nT Pe Pe Pe Pe ++ = + + + + and so

on. Because 0, 0n sk T +< → as ,s → ∞ where s is an integer.

142. The series is telescoping:

( )( )1 11

1

1 11

1

1 1

63 2 3 2

3 33 2 3 2

333 2

kn

n k k k kk

k kn

k k k kk

n

n n

S+ +

=

+

+ +=

+

+ +

=− −

⎡ ⎤= −⎢ ⎥− −⎣ ⎦

= −−

∑

∑

lim 3 1 2nn

S→∞

= − =

y

x12

1

12

y

x12

1

12

1

y

x12

12

1

1

Section 9.3 The Integral Test and p-Series 287


143. ( ) ( ) ( ) ( )1 0, 2 1, 3 2, 4 4,f f f f= = = = …

In general: ( ) ( )2

2

4, even

1 4, odd.

n nf n

n n

⎧⎪= ⎨−⎪⎩

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

( ) ( ) ( ) ( )2 2

.4 4

x y x yf x y f x y xy

+ −+ − − = − =

If both odd,

( ) ( ) ( ) ( )2 21 1.

4 4x y x y

f x y f x y xy+ − − −

+ − − = − =

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

then ( ) 2 4f k k= and

( ) ( ) ( )22 2 1 121 .2 4 2 4 4

kk k k k kf k f k+ −+

+ = + = + = =

The argument is similar if k is odd.

Section 9.3 The Integral Test and p-Series

1. 1

13n n

∞

= +∑

Let ( ) 1 .3

f xx

=+

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

( ) 11

1 ln 33

dx xx

∞ ∞= ⎡ + ⎤ = ∞⎣ ⎦+∫


2. 1

23 5n n

∞

= +∑

Let ( ) 2 .3 5

f xx

=+


( )1

1

2 2 ln 3 53 5 3

dx xx

∞∞ ⎡ ⎤= + = ∞⎢ ⎥+ ⎣ ⎦∫


3. 1

12n

n

∞

=∑

Let ( ) 1 .2xf x =


( )1

1

1 1 12 ln 2 2 2 ln 2x xdx

∞∞ ⎡ ⎤−

= =⎢ ⎥⎢ ⎥⎣ ⎦

∫

Converges by Theorem 9.10.

4. 13 n

n

∞−

=∑

Let ( ) 1 .3xf x =


( )1

1

1 1 13 ln 3 3 3 ln 3x xdx

∞∞ ⎡ ⎤−

= =⎢ ⎥⎢ ⎥⎣ ⎦

∫

Converges by Theorem 9.10.

5. 1

n

ne

∞−

=∑

Let ( ) .xf x e−=


11

1x xe dx ee

∞ ∞− −⎡ ⎤= − =⎣ ⎦∫




6. 2

1

n

nne

∞−

=∑

Let ( ) 2.xf x xe−=

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

( ) 22 02 x

xf xe−′ = < for 3.x ≥

( )2 2 3 233

2 2 10x xxe dx x e e∞ ∞− − −⎡ ⎤= − + =⎣ ⎦∫


7. 21

11n n

∞

= +∑

Let ( ) 21 .

1f x

x=

+


[ ]2 11

1 arctan1 4

dx xx

π∞ ∞= =+∫


8. 1

12 1n n

∞

= +∑

Let ( ) 1 .2 1

f xx

=+


1 1

1 ln 2 12 1

dx xx

∞∞ ⎡ ⎤= + = ∞⎣ ⎦+∫


9. ( )1

ln 11n

nn

∞

=

++∑

Let ( ) ( )ln 1.

1x

f xx

+=

+

f is positive, continuous, and decreasing for

2x ≥ because ( ) ( )( )2

1 ln 10

1

xf x

x

− +′ = <

+for 2.x ≥

( ) ( ) 2

11

ln 1ln 11 2

xxdx

x

∞

∞ ⎡ ⎤⎡ + ⎤+ ⎣ ⎦⎢ ⎥= = ∞⎢ ⎥+⎣ ⎦

∫


10. 2

ln

n

nn

∞

=∑

Let ( ) ( ) 3 2ln 2 ln, .

2x xf x f x

xx−′= =

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

( )2 2

ln 2 ln 2 , divergesx dx x xx

∞∞ ⎡ ⎤= − = ∞⎣ ⎦∫

So, the series diverges by Theorem 9.10.

11. ( )1

11n n n

∞

= +∑

Let ( ) ( )1 ,

1f x

x x=

+

( )( )23 2

1 2 0.2 1

xf xx x

+′ = − <+


( ) ( )1 1

1 2 ln 1 , diverges1

dx xx x

∞∞ ⎡ ⎤= + = ∞⎣ ⎦+∫


12. 21 3n

nn

∞

= +∑

Let ( ) 2 .3

xf xx

=+

( )f x is positive, continuous, and decreasing for

2x ≥ since

( ) ( )

2

2

221 1

3 0 for 2.3

ln 33

xf x xx

x dx xx

∞∞

−′ = < ≥+

⎡ ⎤= + = ∞⎣ ⎦+∫


13. 1

12n n

∞

= +∑

Let ( ) ( )( )3 2

1 1, 02 2 2

f x f xx x

−′= = <+ +


( )1 21 1

1 2 2 , diverges.2

dx xx

∞∞ ⎡ ⎤= + = ∞⎣ ⎦+∫




14. 32

ln

n

nn

∞

=∑

Let ( ) ( )3 4ln 1 3 ln, .x xf x f xx x

−′= =

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

( )2 42

2

2 ln 1ln4

2 ln 2 1, converges16

xx dxx x

∞∞ ⎡ + ⎤

= −⎢ ⎥⎣ ⎦

+=

∫

So, the series converges by Theorem 9.10.

15. 21

ln

n

nn

∞

=∑

Let ( ) ( )2 3ln 1 2 ln, .x xf x f xx x

−′= =

f is positive, continuous, and decreasing for 1 2 1.6.x e> ≈

( )21

1

ln 1ln 1, convergesxx dx

x x

∞∞ ⎡− + ⎤

= =⎢ ⎥⎣ ⎦

∫


16. 2

1lnn n n

∞

=∑

Let ( ) ( )( )3 22

1 2 ln 1, .ln 2 ln

xf x f xx x x x

+′= = −


2 2

1 2 ln , divergesln

dx xx x

∞∞ ⎡ ⎤= = ∞⎣ ⎦∫


17. 21

arctan1n

nn

∞

= +∑

Let ( ) 2arctan ,

1xf x

x=

+

( )( )22

1 2 arctan 0 for 1.1

x xf x xx

−′ = < ≥+


( )2 2

211

arctanarctan 3 , converges1 2 32

xx dxx

π∞

∞ ⎡ ⎤⎢ ⎥= =

+ ⎢ ⎥⎣ ⎦∫


18. ( )3

1ln ln lnn n n n

∞

=∑

Let ( ) ( )1 ,

ln ln lnf x

x x x=

( ) ( ) ( )( ) ( )( )222

ln 1 ln ln 1.

ln ln ln

x xf x

x x x

−⎡ + + ⎤⎣ ⎦′ =


( ) ( )( )

3 3

1 ln ln ln , divergesln ln ln

dx xx x x

∞ ∞⎡ ⎤= = ∞⎣ ⎦∫


19. ( )31

12 3n n

∞

= +∑

Let ( ) ( ) ( )( )

34

62 3 , 02 3

f x x f xx

− −′= + = <+


( )( )

321

1

1 12 3 , converges.1004 2 3

x dxx

∞∞ − ⎡ ⎤−

⎢ ⎥+ = =⎢ ⎥+⎣ ⎦

∫

So, ( )31

12 3n n

∞

= +∑ converges by Theorem 9.10.

20. 1

21n

nn

∞

=

++∑

Let ( ) 2 11 ,1 1

xf xx x+

= = ++ +

( )( )2

1 01

f xx−′ = <+


( ) 11

2 ln 1 diverges.1

x dx x xx

∞ ∞+= ⎡ + + ⎤⎣ ⎦+∫

So, 1

21n

nn

∞

=

++∑ diverges by Theorem 9.10.

[Note: 2lim 1 0,1n

nn→∞

+= ≠

+so the series diverges.]

21. 21

42 1n

nn

∞

= +∑

Let ( ) ( ) ( )( )

2

22 2

4 2 14 , 02 1 2 1

xxf x f xx x

− −′= = <

+ +

for 1x ≥


( )221 1

4 ln 2 1 , diverges.2 1

x dx xx

∞∞ ⎡ ⎤= + = ∞⎣ ⎦+∫

So, 21

42 1n

nn

∞

= +∑ diverges by Theorem 9.10.



22. 41 1n

nn

∞

= +∑

Let ( ) ( )( )

4

24 4

1 3, 01 1

x xf x f xx x

−′= = <+ +

for 1.x >

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

( )241

1

1 arctan , converges1 2 8

x dx xx

π∞∞ ⎡ ⎤= =⎢ ⎥+ ⎣ ⎦∫


23. ( )3 2

1 4 5n

nn

∞

= +∑

Let ( )( )

( ) ( )( )3 2 5 2

2 5, 0

4 5 4 5

xxf x f xx x

− −′= = <

+ +for

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

( )3 21

1

2 54 4 54 5

x xxx

∞∞ +⎡ ⎤= = ∞⎢ ⎥++ ⎣ ⎦∫

(Integration by parts: ( ) 3 2, 4 5u x dv x dx−= = + )

So, ( )3 2

1 4 5n

nn

∞

= +∑ diverges by Theorem 9.10.

24. ( )24 2 21 12 1 1n n

n nn n n

∞ ∞

= =

=+ + +

∑ ∑

Let ( )( )

( ) ( )( )

2

2 32 2

3 1, 0

1 1

xxf x f xx x

− −′= = <

+ +for

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

( ) ( )2 21 2

1

1 1, converges42 11

x dxxx

∞∞ ⎡ ⎤−⎢ ⎥= =

+⎢ ⎥+ ⎣ ⎦∫

So, 4 21 2 1n

nn n

∞

= + +∑ converges by Theorem 9.10.

25. 1

1

k

kn

nn c

−∞

= +∑

Let ( )1

.k

kxf x

x c

−

=+

f is positive, continuous, and decreasing for ( )1kx c k> − because

( )( )

( )

2

2

10

k k

k

x c k xf x

x c

− ⎡ ⎤− −⎣ ⎦′ = <+

for ( )1 .kx c k> −

( )1

11

1 lnk

kkx dx x c

x c k

∞−∞ ⎡ ⎤= + = ∞⎢ ⎥+ ⎣ ⎦∫


26. 1

k n

nn e

∞−

=∑

Let ( ) .k

xxf xe

=

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

( ) ( )1

0k

x

x k xf x

e

− −′ = < for .x k>

Use integration by parts.

( )

111 1

11 !

k x k x k xx e dx x e k x e dx

k kk ke e e e

∞ ∞∞− − − −⎡ ⎤= − +⎣ ⎦

−= + + + +

∫ ∫


27. Let ( ) ( ) ( )1, .

x

nf x f n ax−

= =

The function f is not positive for 1.x ≥

28. Let ( ) ( )cos , .xnf x e x f n a−= =

The function f is not positive for 1.x ≥

29. Let ( ) ( )2 sin , .nxf x f n a

x+

= =

The function f is not decreasing for 1.x ≥

30. Let ( ) ( )2sin , .n

xf x f n ax

⎛ ⎞= =⎜ ⎟⎝ ⎠

The function f is not decreasing for 1.x ≥

31. 31

1

n n

∞

=∑

Let ( ) 31 .f xx

=


3 211

1 1 12 2

dxx x

∞∞ ⎡ ⎤= − =⎢ ⎥⎣ ⎦∫


32. 1 31

1

n n

∞

=∑

Let ( ) 1 31 .f x

x=


2 31 31

1

1 32

dx xx

∞∞ ⎡ ⎤= = ∞⎢ ⎥⎣ ⎦∫




33. Let ( ) ( )1 4 5 41 1, 0

4f x f x

x x−′= = < for 1x ≥

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

3 4

1 411

1 4 , diverges3xdx

x

∞∞ ⎡ ⎤

= = ∞⎢ ⎥⎣ ⎦

∫


34. Let ( ) ( )4 51 4, 0f x f xx x

−′= = < for 1x ≥

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

4 311

1 1 1, converges3 3

dxx x

∞∞ −⎡ ⎤= =⎢ ⎥⎣ ⎦∫

So, 41

1

n n

∞

=∑ converges by Theorem 9.10.

35. 1 551 1

1 1

n n nn

∞ ∞

= =

=∑ ∑

Divergent p-series with 1 15

p = <

36. 5 31

3

n n

∞

=∑

Convergent p-series with 5 13

p = >

37. 1 21

1

n n

∞

=∑


p = <

38. 21

1

n n

∞

=∑

Convergent p-series with 2 1p = >

39. 3 21

1

n n

∞

=∑

Convergent p-series with 3 12

p = >

40. 2 31

1

n n

∞

=∑


p = <

41. 1.041

1

n n

∞

=∑

Convergent p-series with 1.04 1p = >

42. 1

1

n nπ∞

=∑

Convergent p-series with 1p π= >

43. 3 41

2

n n

∞

=∑

1

2

3

4

23.18924.06664.7740

SSSS

=

≈

≈

≈

Matches (c), diverges

44. 1

2

n n

∞

=∑

1

2

3

4

233.66674.1667

SSSS

=

=

≈

≈

Matches (f), diverges

45. 21 1

2 2

n n nn ππ

∞ ∞

= =

=∑ ∑

1

2

3

4

22.67323.02933.2560

SSSS

=

≈

≈

≈

Matches (b), converges Note: The partial sums for 45 and 47 are very similar

because 2 3 2.π ≈

46. 2 51

2

n n

∞

=∑

1

2

3

23.51574.8045

SSS

=

≈

≈

Matches (a), diverges

47. 3 21 1

2 2

n n nn n

∞ ∞

= =

=∑ ∑

1

2

3

4

22.70713.09203.3420

SSSS

=

≈

≈

≈

Matches (d), converges Note: The partial sums for 45 and 47 are very similar

because 2 3 2.π ≈



48. 21

2

n n

∞

=∑

1

2

3

22.52.7222

SSS

=

=

≈

Matches (e), converges

49. (a)

The partial sums approach the sum 3.75 very rapidly.

(b)

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

50. 1

1 1 1 1 112 3 4

N

nM

n N=

= + + + + + >∑

(a) (b) No. Because the terms are decreasing (approaching

zero), more and more terms are required to increase the partial sum by 2.

51. Let f be positive, continuous, and decreasing for 1x ≥ and ( ).na f n= Then,

( )1

1andn

na f x dx

∞ ∞

=∑ ∫

either both converge or both diverge (Theorem 9.10). See Example 1, page 620.

52. A series of the form 1

1p

n n

∞

=∑ is a p-series, 0.p >

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

53. Your friend is not correct. The series

10,000

1 1 110,000 10,001n n

∞

=

= + +∑

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

54. No. Theorem 9.9 says that if the series converges, then the terms na tend to zero. Some of the series in Exercises 43–48 converge because the terms tend to 0 very rapidly.

55. ( )6 77

11 2

n nn n

a f x dx a= =

≥ ≥∑ ∑∫

n 5 10 20 50 100

nS 3.7488 3.75 3.75 3.75 3.75

n 5 10 20 50 100

nS 1.4636 1.5498 1.5962 1.6251 1.635

110

0

11

120

0

8

M 2 4 6 8

N 4 31 227 1674

1 2 3 4 5 6 7

1

x

y



56. (a)

1

1

1 1

ndx

n x

∞ ∞

=

>∑ ∫

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

Because 1 1

1 2 , diverges,dx xx

∞∞ ⎡ ⎤= = ∞⎣ ⎦∫1

1

n n

∞

=∑ diverges.

(b)

2 212

1 1

ndx

n x

∞ ∞

=

<∑ ∫

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

Because 211

1 1 1, converges,dxx x

∞∞ ⎡ ⎤= − =⎢ ⎥⎣ ⎦∫

2 22 1

1 1 converges and so does .n nn n

∞ ∞

= =

⎛ ⎞⎜ ⎟⎝ ⎠

∑ ∑

57. ( )2

1ln p

n n n

∞

=∑

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

( )

( ) ( ) 1

2 22

ln1 1ln .1ln

pp

p

xdx x dx

x px x

∞− +∞ ∞ − ⎡ ⎤

⎢ ⎥= =− +⎢ ⎥⎣ ⎦

∫ ∫

Converges for 1 0p− + < or 1p >

58. 2

lnp

n

nn

∞

=∑

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

( )

( )1

22 22

ln ln 1 1 ln .1

pp

px xdx x x dx p x

x p

∞− +∞ ∞ −

⎡ ⎤⎢ ⎥= = ⎡− + − + ⎤⎣ ⎦⎢ ⎥− +⎣ ⎦

∫ ∫ (Use integration by parts.)

Converges for 1 0p− + < or 1p >

1 2 3 4

1

x

y

1 2 3 4

1

x

y



59. ( )21 1

pn

n

n

∞

= +∑

If 21

1,1n

npn

∞

=

=+∑ diverges (see Example 1). Let

( )( )

( ) ( )( )

2

2

12

, 11

1 2 1.

1

p

p

xf x px

p xf x

x+

= ≠+

− −′ =

+

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

is positive, continuous, and eventually decreasing.

( ) ( ) ( )

11 2 21

1

1 1 2 2p p

x dxx x p

∞

∞

−

⎡ ⎤⎢ ⎥=⎢ ⎥+ + −⎣ ⎦

∫

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

60. ( )2

11

p

nn n

∞

=

+∑

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

61. 1 1

1 1 ,n

nn np p

∞ ∞

= =

⎛ ⎞= ⎜ ⎟

⎝ ⎠∑ ∑ Geometric series.

Converges for 1 1 1pp

< ⇒ >

62. ( )3

1

ln ln lnp

n n n n

∞

= ⎡ ⎤⎣ ⎦∑

If 1,p = then

( ) ( )( )3 3

1 ln ln ln ,ln ln ln

dx xx x x

∞ ∞⎡ ⎤= = ∞⎣ ⎦⎡ ⎤⎣ ⎦

∫ so the

series diverges by the Integral Test If 1,p ≠

( )

( ) 1

33

ln ln1 .1ln ln ln

p

p

xdx

px x x

∞− +∞ ⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=

⎢ ⎥− +⎡ ⎤⎣ ⎦ ⎣ ⎦∫

This converges for 1 0 1.p p− + < ⇒ >

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

63. ( )2

1ln p

n n n

∞

=∑ converges for 1.p >

So, 2

1 ,lnn n n

∞

=∑ diverges.

64. ( )2

1ln p

n n n

∞


So, ( ) ( )2 3232 2

1 1 ,lnlnn n n nn n

∞ ∞

= =

=∑ ∑ diverges.

65. ( )2

1ln p

n n n

∞


So, ( )2

2

1 ,lnn n n

∞

=∑ converges.

66. ( )2

1ln p

n n n

∞


So, ( ) ( )2

2 2 2

1 1 1 1 , diverges2 ln 2 lnlnn n nn n n nn n

∞ ∞ ∞

= = =

= =∑ ∑ ∑

67.

1 2

1

10

N

N n Nn

N N nn N

S a a a a

R S S a

=

∞

= +

= = + + +

= − = >

∑

∑

( )

1 21

N N n N Nn N

N

R S S a a a

f x dx

∞

+ += +

∞

= − = = + +

≤

∑

∫

So, ( )0 n NR f x dx

∞≤ ≤ ∫

68. From Exercise 67, you have:

( )

( )

( )1 1

0 N N

N N N

N N

n n Nn n

S S f x dx

S S S f x dx

a S a f x dx

∞

∞

∞

= =

≤ − ≤

≤ ≤ +

≤ ≤ +

∫

∫

∑ ∑ ∫

69. 6 4 4 4 4 4

6 4 366

41

1 1 1 1 11 1.08112 3 4 5 6

1 1 0.00153

11.0811 1.0811 0.0015 1.0826n

S

R dxx x

n

∞∞

∞

=

= + + + + + ≈

⎡ ⎤≤ = − ≈⎢ ⎥⎣ ⎦

≤ ≤ + =

∫

∑

1 2 N N + 1

1

f(1) = a1

f(2) = a2

f(N + 1) = aN + 1

f(N) = aN

x

y

...



70. 4 5 5 5

4 5 444

51

1 1 11 1.03632 3 4

1 1 0.00104

11.0363 1.0363 0.0010 1.0373n

S

R dxx x

n

∞∞

∞

=

= + + + ≈

⎡ ⎤≤ = − ≈⎢ ⎥⎣ ⎦

≤ ≤ + =

∫

∑

71.

[ ]

10

10 2 1010

21

1 1 1 1 1 1 1 1 1 1 0.98182 5 10 17 26 37 50 65 82 101

1 arctan arctan 10 0.09971 210.9818 0.9818 0.0997 1.0815

1n

S

R dx xx

n

π∞ ∞

∞

=

= + + + + + + + + + ≈

≤ = = − ≈+

≤ ≤ + =+

∫

∑

72. ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

10 3 3 3 3

10 3 2 31010

31

1 1 1 1 1.98212 ln 2 3 ln 3 4 ln 4 11 ln 11

1 1 1 0.08702 ln 111 ln 1 2 ln 1

11.9821 1.9821 0.0870 2.06911 ln 1n

S

R dxx x x

n n

∞

∞

∞

=

= + + + + ≈

⎡ ⎤⎢ ⎥≤ = − = ≈⎢ ⎥+ ⎡ + ⎤ ⎡ + ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦

≤ ≤ + =+ ⎡ + ⎤⎣ ⎦

∫

∑

73. 4 4 9 16

162 2 84 4

4

2 8

1

1 2 3 4 0.4049

1 5.6 102 2

0.4049 0.4049 5.6 10

x x

n

n

Se e e e

eR xe dx e

ne

∞ −∞ − − −

∞− −

=

= + + + ≈

⎡ ⎤≤ = − = ≈ ×⎢ ⎥⎣ ⎦

≤ ≤ + ×

∫

∑

74. 4 2 3 4

4 44

0

1 1 1 1 0.5713

0.0183

0.5713 0.5713 0.0183 0.5896

x x

n

n

Se e e e

R e dx e

e

∞ ∞− −

∞−

=

= + + + ≈

⎡ ⎤≤ = − ≈⎣ ⎦

≤ ≤ + =

∫

∑

75. 4 3 31 1 10 0.001

3 3N NN

R dxx x N

∞∞ ⎡ ⎤≤ ≤ = − = <⎢ ⎥⎣ ⎦∫

3

3

1 0.003

333.336.937

NNNN

<

>

>

≥

76. 3 2 1 21 2 20 0.001N N

NR dx

x x N

∞∞ ⎡ ⎤≤ ≤ = − = <⎢ ⎥⎣ ⎦∫

1 2 0.0005

20004,000,000

N

NN

− <

>

≥

77. 5

5 51 0.0015 5

Nx x

N NN

eR e dx e∞ −∞ − −⎡ ⎤≤ = − = <⎢ ⎥⎣ ⎦∫

5

5

1 0.005

2005 ln 200

ln 2005

1.05972

N

N

ee

N

N

NN

<

>

>

>

>

≥

78. 2 22

22 0.001x xN NNN

R e dx ee

∞ ∞− −⎡ ⎤≤ = − = <⎣ ⎦∫

2

2

2 0.001

2000

ln 20002

2 ln 2000 15.216

N

N

ee

N

NN

<

>

>

> ≈

≥



79. [ ]21 arctan

1

arctan 0.0012

N NNR dx x

x

Nπ

∞ ∞≤ =+

= − <

∫

arctan 0.0012

arctan 0.0012

tan 0.0012

1000

N

N

N

N

π

π

π

− < −

> −

⎛ ⎞> −⎜ ⎟⎝ ⎠

≥

80. 22 12 arctan

5 5 5

2 arctan 0.00125 5

n NN

xR dxx

Nπ

∞∞ ⎡ ⎤⎛ ⎞≤ = ⎢ ⎥⎜ ⎟+ ⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞= − <⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∫

arctan 0.0005 52 5

arctan 0.0005 525

arctan 0.0005 525

5 tan 0.0005 52

1000

N

N

N

N

N

π

π

π

π

⎛ ⎞− <⎜ ⎟

⎝ ⎠

⎛ ⎞− < −⎜ ⎟⎝ ⎠

⎛ ⎞ > −⎜ ⎟⎝ ⎠

⎛ ⎞> −⎜ ⎟⎝ ⎠

≥

81. (a) 1.12

1 .n n

∞

=∑ This is a convergent p-series with

2

11.1 1.lnn

pn n

∞

=

= > ∑ is a divergent series. Use the Integral Test.

( ) 1ln

f xx x

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

22

1 ln lnln

dx xx x

∞ ∞⎡ ⎤= = ∞⎣ ⎦∫

(b) 6

1.1 1.1 1.1 1.1 1.1 1.12

6

2

1 1 1 1 1 1 0.4665 0.2987 0.2176 0.1703 0.13932 3 4 5 6

1 1 1 1 1 1 0.7213 0.3034 0.1803 0.1243 0.0930ln 2 ln 2 3 ln 3 4 ln 4 5 ln 5 6 ln 6

n

n

n

n n

=

=

= + + + + ≈ + + + +

= + + + + ≈ + + + +

∑

∑

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

(c) 1.1

1.1

0.1

1 1ln

ln

ln

n n n

n n n

n n

<

<

<

This inequality holds when 153.5 10 .n ≥ × Or, 40.n e> Then ( )0.140 40 4ln 40 55.e e e= < = ≈

82. (a) ( )

1

11010

1 1 , 11 1 10

p

p pxdx p

x p p

∞− +∞

−

⎡ ⎤= = >⎢ ⎥− + −⎣ ⎦

∫

(b) ( ) 1pf x

x=

( )

[ )

1011

1

Area under the graph of over the interval 10,

pn

R pn

f

∞

=

=

≤ ∞

∑

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



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

1

11

1 ln

nn

n

S dxx

S n

− ≤

− ≤

∫

So, 1 ln .nS n≤ + Similarly,

( )1

1

1 ln 1 .n

nS dx nx

+≥ = +∫

So, ( )ln 1 1 ln .nn S n+ ≤ ≤ +

(b) Because ( )ln 1 1 ln ,nn S n+ ≤ ≤ + you have ( )ln 1 ln ln 1.nn n S n+ − ≤ − ≤ Also, because ln x is an increasing

function, ( )ln 1 ln 0n n+ − > for 1.n ≥ So, 0 ln 1nS n≤ − ≤ and the sequence { }na is bounded.

(c) [ ] ( )1

1 11 1ln ln 1 0

1n

n n n n na a S n S n dx

x n+

+ +− = − − ⎡ − + ⎤ = − ≥⎣ ⎦ +∫

So, 1n na a +≥ and the sequence is decreasing.

(d) Because the sequence is bounded and monotonic, it converges to a limit, .γ

(e) ( )100 100 ln 100 0.5822 Actually 0.577216.a S γ= − ≈ ≈

84. ( )( ) ( ) ( )2

2 2 22 2 2 2

1 11 1ln 1 ln ln ln 1 ln 1 2 ln

ln 3

n n n n

n nn n n nn n n

∞ ∞ ∞ ∞

= = = =

+ −⎛ ⎞−⎛ ⎞− = = = ⎡ + + − − ⎤⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠

=

∑ ∑ ∑ ∑

ln 1+( )2 ln 2 ln 4− + ln 2 2 ln 3+ −( ) ln 5+ ln 3+ 2 ln 4−( ) ln 6+ ln 4+ 2 ln 5−( )ln 7+ ln 5+ 2 ln 6−( ) ln 8+ ln 6+ 2 ln 7−( ) ln 9+ ln 7+ 2 ln 8−( ) ln 2+ = −

85. ln

2

n

nx

∞

=∑

(a) ln

2 21: 1 1, divergesn

n nx

∞ ∞

= =

= =∑ ∑

(b) ln

ln

2 2 2

1 1 1: , divergesn

n

n n nx e

e e n

∞ ∞ ∞−

= = =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

∑ ∑ ∑

(c) Let x be given, 0.x > Put ln .px e x p−= ⇔ = −

ln ln

2 2 2 2

1n p n pp

n n n nx e n

n

∞ ∞ ∞ ∞− −

= = = =

= = =∑ ∑ ∑ ∑

This series converges for 11 .p xe

> ⇒ <

86. ( )1 1

1xx

n nx n

nξ

∞ ∞−

= =

= =∑ ∑

Converges for 1x > by Theorem 9.11

1 2 3 ...n−1

n n+1x

1

12

y



87. Let ( ) ( )( )2

1 3, 03 2 3 2

f x f xx x

−′= = <− −

for 1x ≥


1

1

1 1 ln 3 23 2 3

dx xx

∞∞ ⎡ ⎤= − = ∞⎢ ⎥− ⎣ ⎦∫

So, the series 1

13 2n n

∞

= −∑

diverges by Theorem 9.10.

88. 2

2

11n n n

∞

= −∑

Let ( )2

1 .1

f xx x

=−


[ ]22 2

1 arcsec2 31

dx xx x

π π∞ ∞= = −−

∫


89. 5 441 1

1 1

n n nn n

∞ ∞

= =

=∑ ∑

p-series with 54

p =


90. 0.951

13n n

∞

=∑

p-series with 0.95p =


91. ( )0

23

n

n

∞

=∑

Geometric series with 23r =


92. ( )0

1.042 n

n

∞

=∑ is geometric with 1.042 1.r = > Diverges

by Theorem 9.6.

93. 2

1 1n

nn

∞

= +∑

( )2 2

1lim lim 1 01 1 1n n

nn n→∞ →∞

= = ≠+ +


94. 2 3 2 31 1 1

1 1 1 1

n n nn n n n

∞ ∞ ∞

= = =

⎛ ⎞− = −⎜ ⎟⎝ ⎠

∑ ∑ ∑

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

95. 1

11n

n n

∞

=

⎛ ⎞+⎜ ⎟⎝ ⎠

∑

1lim 1 0n

ne

n→∞

⎛ ⎞+ = ≠⎜ ⎟⎝ ⎠

Fails nth-Term Test Diverges by Theorem 9.9

96. ( )2

lnn

n∞

=∑

( )lim lnn

n→∞

= ∞


97. ( )32

1lnn n n

∞

=∑

Let ( )( )3

1 .ln

f xx x

=


( )

( )

( )

( ) ( )

332 2

2

2

2 2

2

1 1lnln

ln2

1 12 ln 2 ln 2

dx x dxxx x

x

x

∞ ∞ −

∞−

∞

=

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦

⎡ ⎤= ⎢− ⎥ =⎢ ⎥⎣ ⎦

∫ ∫

Converges by Theorem 9.10. See Exercise 57.

98. 32

ln

n

nn

∞

=∑

Let ( ) 3ln .xf xx

=

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

( ) 41 3 ln 0xf x

x−′ = < for 2.x ≥

( )

3 2 32 22

22

ln ln 1 12 2

ln 2 18 4

ln 2 1 Use integration by parts.8 16

x xdx dxx x x

x

∞∞ ∞

∞

⎡ ⎤= − +⎢ ⎥⎣ ⎦

⎡ ⎤= + −⎢ ⎥⎣ ⎦

= +

∫ ∫

Converges by Theorem 9.10. See Exercise 36.

Section 9.4 Comparisons of Series 299


Section 9.4 Comparisons of Series

1. (a) 13 2 3 21

13 2 3 21

121

6 6 6 ; 61 2

6 6 6 3;3 4 2 3 2

6 6 6 6; 4.91 1.5 2 4.5 1.50.5

n

n

n

Sn

Sn

Sn n

∞

=

∞

=

∞

=

= + + =

= + + =+ +

= + + = ≈+

∑

∑

∑

(b) The first series is a p-series. It converges 3 1 .2

p⎛ ⎞= >⎜ ⎟⎝ ⎠

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

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

2. (a) 11

11

11

2 22 22

2 2 2 40.50.5 2 0.5

4 4 4 3.30.5 1.5 2.5

n

n

n

Sn

Sn

Sn

∞

=

∞

=

∞

=

= + + =

= + + =− −

= + + ≈+

∑

∑

∑

(b) The first series is a p-series. It diverges 1 1 .2

p⎛ ⎞= <⎜ ⎟⎝ ⎠

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

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

3. 2 21 10

1n n< <

+

Therefore,

21

11n n

∞

= +∑

converges by comparison with the convergent p-series

21

1 .n n

∞

=∑

4. 2 21 1

3 2 3n n<

+

Therefore,

21

13 2n n

∞

= +∑


21

1 1 .3n n

∞

=∑

n

2

1

2

4

3

4

6

5

6 8 10

6n3/2

6n2 + 0.5

an =

6n3/2 + 3

an =

an =n

an

n

2

2

4

4

6

6 8

8

10

10

12

Σk = 1

n

Σk = 1

n

Σ 6k2 + 0.5kk = 1

n

Sn6

k3/2

6k3/2 + 3

n

2

2

4

4 6 8 10

1

3an =

n + 0.54

an =n − 0.5

2

an = 2n

an

n4

4

8

8

10

12

16

20

62

n + 0.54

2n

n − 0.52Σ

Σ

Σ

Sn



5. 1 1 02 1 2n n

> >−

for 1n ≥

Therefore,

1

12 1n n

∞

= −∑

diverges by comparison with the divergent p-series 1

1 1.2 n n

∞

=∑

6. 1 11n n>

−for 2n ≥

Therefore,

2

11n n

∞

= −∑

diverges by comparison with the divergent p-series

2

1 .n n

∞

=∑

7. 1 14 1 4n n<

+

Therefore,

0

14 1n

n

∞

= +∑

converges by comparison with the convergent geometric series

0

1 .4n

n

∞

=∑

8. 4 45 3 5

nn

n⎛ ⎞< ⎜ ⎟+ ⎝ ⎠

Therefore,

0

45 3

n

nn

∞

= +∑


0

4 .5

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

9. For ln 13, 0.1 1

nnn n

≥ > >+ +

Therefore,

1

ln1n

nn

∞

= +∑

diverges by comparison with the divergent series

1

1 .1n n

∞

= +∑

Note: 1

11n n

∞

= +∑ diverges by the Integral Test.

10. 3 23

1 11 nn<

+

Therefore,

3

1

11n n

∞

= +∑


3 21

1 .n n

∞

=∑

11. For 21 13, 0.

!n

n n> > >

Therefore,

0

1!n n

∞

=∑


21

1 .n n

∞

=∑

12. 3 4

1 14 1 4n n

>−

Therefore,

3

1

14 1n n

∞

= −∑

diverges by comparison with the divergent p-series

4

1

1 1 .4 n n

∞

=∑

13. 21 10 nn ee

< ≤

Therefore,

20

1nn e

∞

=∑


0

1 .n

n e

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

14. 3 32 1 2

nn

n⎛ ⎞> ⎜ ⎟− ⎝ ⎠

for 1n ≥

Therefore,

1

32 1

n

nn

∞

= −∑

diverges by comparison with the divergent geometric series

1

3 .2

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑



15. ( )2 2

2

1lim lim 1

1 1n n

n n nn n→∞ →∞

+= =

+

Therefore,

21 1n

nn

∞

= +∑

diverges by a limit comparison with the divergent p-series

1

1.n n

∞

=∑

16. ( )5 4 1 5 4lim lim 51 4 4 1

n n

n nn n→∞ →∞

+ ⋅= =

+

Therefore,

1

54 1n

n

∞

= +∑

converges by a limit comparison with the convergent geometric series

1

1 .4

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

17. 2

2

1 1lim lim 11 1n n

n nn n→∞ →∞

+= =

+

Therefore,

2

0

11n n

∞

= +∑


1

1.n n

∞

=∑

18. ( ) ( )

( )2 1 5 1 2 1 5lim lim 1

5 1 22 5

n n n n

n n nn n→∞ →∞

+ + += ⋅ =

+

Therefore,

1

2 15 1

n

nn

∞

=

++∑


1

2 .5

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

19.

2

5 35

3 5

2 12 23 2 1lim lim

1 3 2 1 3n n

nn nn n

n n n→∞ →∞

−−+ + = =

+ +

Therefore,

2

51

2 13 2 1n

nn n

∞

=

−+ +∑

converges by a limit comparison with the convergent p-series

31

1 .n n

∞

=∑

20. ( ) ( )3 2

2 3

5 2 3 5lim lim1 2 3 1

1

n n

n n n n nn n n→∞ →∞

+ − + ⎛ ⎞+⎛ ⎞= ⎜ ⎟⎜ ⎟− +⎝ ⎠⎝ ⎠=

Therefore,

31

52 3n

nn n

∞

=

+− +∑


21

1 .n n

∞

=∑

21. ( ) ( ) ( )

( )2 2

2 2

3 4 3lim lim 1

1 4n n

n n n n nn n n→∞ →∞

+ + += =

+

Therefore,

( )2

1

34n

nn n

∞

=

++

∑


21

1 .n n

∞

=∑

22. ( )( )

2 3

3 2

1 3lim lim 1

1 3n n


+= =

+

Therefore,

( )2

1

13n n n

∞

= +∑


31

1 .n n

∞

=∑

23. ( )2

2

2 2

1 1lim lim 1

1 1n n


+= =

+

Therefore,

2

1

11n n n

∞

= +∑


21

1 .n n

∞

=∑



24. ( )( )

1

1

1 2lim lim 1

11 2

n

nn n

n n nn

−

−→∞ →∞

⎡ ⎤+⎣ ⎦ = =+

Therefore,

( ) 1

1 1 2nn

nn

∞

−= +∑


1

1

1 .2

n

n

−∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

25. ( ) ( )1 1

lim lim 11 1

k k k

kn n

n n nn n

−

→∞ →∞

+= =

+

Therefore,

1

1 1

k

kn

nn

−∞

= +∑


1

1.n n

∞

=∑

26. ( )2

2

5 4 5 5lim lim1 24n n


+ += =

+ +

Therefore,

2

1

54n n n

∞

= + +∑

diverges by a limit comparison with the divergent harmonic series

1

1.n n

∞

=∑

27. ( ) ( ) ( )2

2

1 cos 1sin 1 1lim lim lim cos 11 1n n n

n nnn n n→∞ →∞ →∞

− ⎛ ⎞= = =⎜ ⎟− ⎝ ⎠

Therefore,

1

1sinn n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑


1

1.n n

∞

=∑

28. ( ) ( ) ( )2 2

2

2

1 sec 1tan 1lim lim

1 1

1lim sec 1

n n

n

n nnn n

n

→∞ →∞

→∞

−=

−

⎛ ⎞= =⎜ ⎟⎝ ⎠

Therefore,

1

1tann n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑


1

1.n n

∞

=∑

29. 1 3

2 31 1

1

n n

nn n

∞ ∞

= =

=∑ ∑

Diverges;

p-series with 23

p =

30. 0

177

n

n

∞

=

⎛ ⎞−⎜ ⎟⎝ ⎠

∑

Converges;


r = −

31. 1

15 1n

n

∞

= +∑

Converges; Direct comparison with convergent geometric series

1

15

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

32. 33

18n n

∞

= −∑

Converges; limit comparison with 33

1

n n

∞

=∑

33. 1

23 2n

nn

∞

= −∑

Diverges; nth-Term Test

2 2lim 03 2 3n

nn→∞

= ≠−

34. 1

1 1 1 1 1 1 1 1 11 2 2 3 3 4 4 5 2n n n

∞

=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − + − + − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑

Converges; telescoping series

35. ( )221 1n

n

n

∞

= +∑

Converges; Integral Test



36. ( )1

33n n n

∞

= +∑

Converges; telescoping series

1

1 13n n n

∞

=

⎛ ⎞−⎜ ⎟+⎝ ⎠∑

37. lim lim .1

nn

n n

a nan→∞ →∞= By given conditions lim n

nna

→∞is

finite and nonzero. Therefore,

1

nn

a∞

=∑

diverges by a limit comparison with the p-series

1

1.n n

∞

=∑

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

( ) ( )lim 0,1 k jn

P n Q nL

n −→∞= > the series ( )

( )1n

P nQ n

∞

=∑

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

( )( )1n

P nQ n

∞

=∑

diverges by the Limit Comparison Test.

39. 21

1 2 3 4 5 ,2 5 10 17 26 1n

nn

∞

=

+ + + + + =+∑

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

40. 22

1 1 1 1 1 1 ,3 8 15 24 35 1n n

∞

=

+ + + + + =−∑

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

41. 31

11n n

∞

= +∑

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

42. 2

31 1n

nn

∞

= +∑

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

43. 3 4

4 41lim lim 0

5 3 5 3 5n n

n nnn n→∞ →∞

⎛ ⎞= = ≠⎜ ⎟+ +⎝ ⎠

Therefore, 3

41 5 3n

nn

∞

= +∑ diverges.

44. 1lim lim lim 0ln 1n n n

n nn n→∞ →∞ →∞= = = ∞ ≠

Therefore, 2

1lnn n

∞

=∑ diverges.

45. 1

1 1 1 1200 400 600 200n n

∞

=

+ + + = ∑

diverges, (harmonic)

46. 0

1 1 1 1200 210 220 200 10n n

∞

=

+ + + =+∑

diverges

47. 21

1 1 1 1 1201 204 209 216 200n n

∞

=

+ + + =+∑

converges

48. 31

1 1 1 1 1201 208 227 264 200n n

∞

=

+ + + + =+∑

converges

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 626. One example is

21

11n n

∞

= +∑ converges because 2 21 1

1n n<

+and

21

1

n n

∞

=∑ converges (p-series).

51. See Theorem 9.13, page 628. One example is

2

11n n

∞

= −∑ diverges because 1 1lim 1

1n

nn→∞

−= and

2

1

n n

∞

=∑ diverges (p-series).

52.

For 20 1, 0 1.n n na a a< < < < < So, the lower terms

are those of 2.naΣ

0.2

0.4

0.6

0.8

1.0

n4 8 12 16 20

Σ ann = 1

Σ ann = 1

2

Terms of

Terms of



53. (a) ( )2 2

1 1

1 14 4 12 1n n n nn

∞ ∞

= =

=− +−

∑ ∑

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

(c) ( )

2

223

1 0.122682 1n

Sn

π∞

=

= − ≈−

∑

(d) ( )

2

9210

1 0.027782 1n

Sn

π∞

=

= − ≈−

∑

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

( )1000

1 1 1 diverges harmonic1000 1001 n n

∞

=

+ + = ∑

and ( )21

1 1 11 converges -series .4 9 n

pn

∞

=

+ + + = ∑

55. False. Let 31

nan

= and 21 . 0n n nb a bn

= < ≤ and both

31

1

n n

∞

=∑ and 2

1

1

n n

∞

=∑ converge.

56. True

57. True

58. False. Let 21 , 1 , 1 .n n na n b n c n= = = Then,

,n n na b c≤ + but 1

nn

c∞

=∑ converges.

59. True

60. False. 1

nn

a∞

=∑ could converge or diverge.

For example, let 1 1

1 ,nn n

bn

∞ ∞

= =

=∑ ∑ which diverges.

1 10n n

< < and 1

1

n n

∞

=∑ diverges, but

21 10n n

< < and 21

1

n n

∞

=∑ converges.

61. Because 1

nn

b∞

=∑ converges, lim 0.n

nb

→∞= There exists N

such that 1nb < for .n N> So, n n na b a< for

n N> and 1

n nn

a b∞

=∑ converges by comparison to the

convergent series 1

.ni

a∞

=∑

62. Because 1

nn

a∞

=∑ converges, then

2

1 1n n n

n na a a

∞ ∞

= =

=∑ ∑ converges by Exercise 61.

63. 21n∑ and 3

1n∑ both converge, and therefore, so does

2 3 51 1 1 .n n n⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑

64. 21n∑ converge, and therefore, so does

2

2 41 1 .n n⎛ ⎞ =⎜ ⎟⎝ ⎠

∑ ∑

65. Suppose lim 0n

n n

ab→∞

= and nbΣ converges.

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

0 1n

n

ab

− <

whenever .n M> So, n na b< for .n M> From the Comparison Test, naΣ converges.

66. Suppose lim n

n n

ab→∞

= ∞ and nbΣ diverges. From the

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

1n

n

ab

>

for .n M> So, n na b> for .n M> By the Comparison Test, naΣ diverges.

n 5 10 20 50 100

nS 1.1839 1.2087 1.2212 1.2287 1.2312



67. (a) Let ( )3

1 ,1

nan

=+

∑ ∑ and 21 ,nbn

=∑ ∑

converges.

( )( ) ( )

32

32

1 1lim lim lim 0

1 1n

n n nn

na nb n n→∞ →∞ →∞

⎡ ⎤+⎣ ⎦= = =+

By Exercise 65, ( )31

11n n

∞

= +∑ converges.

(b) Let 1 ,n na

nπ=∑ ∑ and 1 ,n nb

π=∑ ∑

converges.

( )( )

1 1lim lim lim 01

nn

nn n nn

nab n

π

π→∞ →∞ →∞= = =

By Exercise 65, 1

1n

n nπ

∞

=∑ converges.

68. (a) Let ln ,nna

n=∑ ∑ and 1,nb

n=∑ ∑ diverges.

( )lnlim lim lim ln

1n

n n nn

n na nb n→∞ →∞ →∞

= = = ∞

By Exercise 66, 1

ln

n

nn

∞

=∑ diverges.

(b) Let 1 ,lnna

n=∑ ∑ and 1,nb

n=∑ ∑ diverges.

lim limln

n

n nn

a nb n→∞ →∞

= = ∞

By Exercise 66, 1ln n∑ diverges.

69. Because lim 0,nn

a→∞

= the terms of ( )sin naΣ are positive

for sufficiently large n. Because

( )sinlim 1 andn

nn n

aa

a→∞= ∑

converges, so does ( )sin .naΣ

70. ( )

( )

1 1

1

1 11 2 1 2

21

n n

n

n n n

n n

∞ ∞

= =

∞

=

=+ + + ⎡ + ⎤⎣ ⎦

=+

∑ ∑

∑

Because 21 nΣ converges, and

( )( ) ( )

2

2

2 1 2lim lim 2,11n n

n n nn nn→∞ →∞

⎡ + ⎤⎣ ⎦ = =+

11 2 n+ + +∑ converges.

71. First note that ( ) 1 4ln 0f x x x= − = when

5503.66.x ≈ That is,

1 4ln for 5504n n n< >

which implies that

3 2 5 4ln 1 for 5504.n nn n

< >

Because 5 41

1

n n

∞

=∑ is a convergent p-series,

3 21

ln

n

nn

∞

=∑

converges by direct comparison.

72. The series diverges. For 1,n >

( )

1

1

1

2

21 1

21 1

2

n

n

n

n n

n

n

n

nn +

<

<

>

>

Because 12n∑ diverges, so does ( )1

1 .n nn +∑

73. Consider two cases:

If 11 ,

2n na +≥ then ( )( )1 1

1 11

1 1,2 2

nn

n na+

++

⎛ ⎞≥ =⎜ ⎟⎝ ⎠

and

( )( )

11 1 2 .n n n

n nnn

aa aa

++

= ≤

If 11 ,

2n na +≤ then ( )( )1

11

1 1 ,2 2

n nn n

n n na+

++

⎛ ⎞≤ =⎜ ⎟⎝ ⎠

and

combining, ( )1 12 .2

n nn n na a+ ≤ +

Because 1

122n n

na

∞

=

⎛ ⎞+⎜ ⎟⎝ ⎠

∑ converges, so does ( )1

1

n nn

na

∞+

=∑

by the Comparison Test.



Section 9.5 Alternating Series

1. 21

6

n n

∞

=∑

1

2

3

67.58.1667

SSS

=

=

≈

Matches (d).

2. ( ) 1

21

1 6n

n n

−∞

=

−∑

1

2

3

64.45.1667

SSS

=

=

≈

Matches (f).

3. 1

3!n n

∞

=∑

1

2

3

34.55.0

SSS

=

=

=

Matches (a).

4. ( ) 1

1

1 3!

n

n n

−∞

=

−∑

1

2

3

31.52.0

SSS

=

=

=

Matches (b).

5. 1

102n

n n

∞

=∑

1

2

56.25

SS

=

=

Matches (e).

6. ( )1

1 102

n

nn n

∞

=

−∑

1

2

53.75

SS

=

=

Matches (c).

7. ( ) 1

1

10.7854

2 1 4

n

n nπ

−∞

=

−= ≈

−∑

(a) (b)

(c) The points alternate sides of the horizontal line4

y π= 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.

n 1 2 3 4 5 6 7 8 9 10

nS 1 0.6667 0.8667 0.7238 0.8349 0.7440 0.8209 0.7543 0.8131 0.7605

0.60 11

1.1

Section 9.5 Alternating Series 307


8. ( )( )

1

1

1 1 0.36791 !

n

n n e

−∞

=

−= ≈

−∑

(a)

(b)

(c) The points alternate sides of the horizontal line 1ye

= 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.

9. ( ) 1 2

21

10.8225

12

n

n nπ

−∞

=

−= ≈∑

(a)

(b)

(c) The points alternate sides of the horizontal line2

12y π= 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.

10. ( )( ) ( )

1

1

1sin 1 0.8415

2 1 !

n

n n

−∞

=

−= ≈

−∑

(a)

(b)

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

n 1 2 3 4 5 6 7 8 9 10

nS 1 0 0.5 0.3333 0.375 0.3667 0.3681 0.3679 0.3679 0.3679

n 1 2 3 4 5 6 7 8 9 10

nS 1 0.75 0.8611 0.7986 0.8386 0.8108 0.8312 0.8156 0.8280 0.8180

n 1 2 3 4 5 6 7 8 9 10

nS 1 0.8333 0.8417 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415 0.8415

0 110

2

0.60 11

1.1

0 110

2



11. ( ) 1

1

11

n

n n

+∞

=

−+∑

1

1 12 1

1lim lim 01

n n

nn n

a an n

an

+

→∞ →∞

= < =+ +

= =+


12. ( ) 1

1

13 2

n

n

nn

+∞

=

−+∑

1lim3 2 3n

nn→∞

=+

Diverges by nth-Term test

13. ( )1

13

n

nn

∞

=

−∑

1 1

0

1 13 3

1lim 03

n nn n

nn

a a+ +

→

= < =

=


(Note: 1

13

n

n

∞

=

−⎛ ⎞⎜ ⎟⎝ ⎠

∑ is a convergent geometric series)

14. ( )1

1 n

nn e

∞

=

−∑

1 1

1 1

1lim 0

n nn n

nn

a ae e

e

+ +

→∞

= < =

=


(Note: 1

1 n

n e

∞

=

−⎛ ⎞⎜ ⎟⎝ ⎠

∑ is a convergent geometric series)

15. ( ) ( )1

1 5 14 1

n

n

nn

∞

=

− −+∑

5 1 5lim4 1 4n

nn→∞

−=

+


16. ( )( )1

1ln 1

n

n

nn

∞

=

−+∑

( )

limln 1n

nn→∞

= ∞+


17. ( ) 1

1

13 2

n

n n

+∞

=

−+∑

( )11 1

3 1 2 3 2

1lim 03 2

n n

n

a an n

n

+

→∞

= < =+ + +

=+


18. ( )( )1

1ln 1

n

n n

∞

=

−+∑

( ) ( )

( )

11 1

ln 2 ln 1

1lim 0ln 1

n n

n

a an n

n

+

→∞

= < =+ +

=+


19. ( ) 2

21

15

n

n

nn

∞

=

−+∑

2

2lim 15n

nn→∞

=+


20. ( ) 1

21

15

n

n

nn

+∞

=

−+∑

Let ( ) ( ) ( )( )

2

22 2

5, 0

5 5

xxf x f xx x

− −′= = <

+ +for 3x ≥

So, 1n na a+ < for 3n ≥

2lim 05n

nn→∞

=+


21. ( )1

1

1

1 11

1lim 0

n

n

n n

n

n

a an n

n

∞

=

+

→∞

−

= < =+

=

∑


22. ( ) 1 2

21

2

2

14

lim 14

n

n

n

nn

nn

+∞

=

→∞

−+

=+

∑




23. ( ) ( )( )

( ) ( ) ( )

1

1

1 1ln 1

1 1lim lim lim 1ln 1 1 1

n

n

n n n

nn

n nn n

+∞

=

→∞ →∞ →∞

− ++

+= = + = ∞

+ +

∑

Diverges by the nth-Term Test

24. ( ) ( )

( )( )

( )

( ) ( )

1

1

1

1 ln 11

ln 1 1 ln 1for 2

1 1 1

ln 1 1 1lim lim 0

1 1

n

n

n

n n

nn

n na n

n n

n nn

+∞

=

+

→∞ →∞

− ++

⎡ + + ⎤ +⎣ ⎦= < ≥+ + +

+ += =

+

∑


25. ( ) ( ) 1

1 1

2 1sin 1

2n

n n

n π∞ ∞+

= =

⎡ − ⎤= −⎢ ⎥

⎣ ⎦∑ ∑


26. ( ) ( ) 1

1 1

1

2 1 11 sin2

1 11

1lim 0

n

n n

n n

n

nn n

a an n

n

π +∞ ∞

= =

+

→∞

⎡ − ⎤ −=⎢ ⎥

⎣ ⎦

= < =+

=

∑ ∑


27. ( )1 1cos 1 n

n nnπ

∞ ∞

= =

= −∑ ∑


28. ( )1 1

1

11 cos

1 11

1lim 0

n

n n

n n

n

nn n

a an n

n

π∞ ∞

= =

+

→∞

−=

= < =+

=

∑ ∑


29. ( )

( )

0

1

1!

1 11 ! !

1lim 0!

n

n

n n

n

n

a an n

n

∞

=

+

→∞

−

= < =+

=

∑


30. ( )( )

( ) ( )

( )

0

1

12 1 !

1 12 3 ! 2 1 !

1lim 02 1 !

n

n

n n

n

n

a an n

n

∞

=

+

→∞

−+

= < =+ +

=+

∑


31. ( )

( )

1

1

1

12

1 for 21 2 2

lim 02

n

n

n

n

nn

n na nn n

nn

+∞

=

+

→∞

−+

+= < ≥

+ + +

=+

∑


32. ( ) 1

31

1 21 6

1 3

1

lim lim

n

n

n n

nn

n nn

+∞

=

→∞ →∞

−

= = ∞

∑


33. ( )( )

( )( )( ) ( )

1

1

1

1 !1 3 5 2 1

1 ! ! 1 11 3 5 2 1 2 1 1 3 5 2 1 2 1 2 1

n

n

n n n

nn

n n n na a an n n n n

+∞

=

+

−⋅ ⋅ ⋅ ⋅ ⋅ −

+ + +⎛ ⎞= = ⋅ = <⎜ ⎟⋅ ⋅ ⋅ ⋅ ⋅ − + ⋅ ⋅ ⋅ ⋅ ⋅ − + +⎝ ⎠

∑

( ) ( )

! 1 2 3 3 4 5 1lim lim lim lim 2 01 3 5 2 1 1 3 5 2 1 3 5 7 2 3 2 1n

n n n n

n n nan n n n→∞ →∞ →∞ →∞

⋅ ⋅ ⋅ ⋅ ⋅ ⎡ ⎤= = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =⎢ ⎥⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ − − −⎣ ⎦




34. ( ) ( )( )

( )( )( )( )

1

1

1

1 3 5 2 11

1 4 7 3 2

1 3 5 2 1 2 1 2 11 4 7 3 2 3 1 3 1

5 7 9 2 1 1lim lim 3 04 7 10 3 5 3 2

n

n

n n n

nn n

nn

n n na a an n n

nan n

∞+

=

+

→∞ →∞

⋅ ⋅ −−

⋅ ⋅ −

⋅ ⋅ − + +⎛ ⎞= = <⎜ ⎟⋅ ⋅ − + +⎝ ⎠

−⎡ ⎤= ⋅ ⋅ =⎢ ⎥− −⎣ ⎦

∑


35. ( ) ( ) ( ) ( )11

21 1

1 21 21

nn n

n n nn n

e

e e e

++∞ ∞

−= =

−−=

− −∑ ∑

Let ( ) 22 .

1

x

xef x

e=

−Then

( ) ( )( )

2

22

2 10.

1

x x

x

e ef x

e

− +′ = <

−

So, ( )f x is decreasing. Therefore, 1 ,n na a+ < and

2 22 2 1lim lim lim 0.

1 2

n n

n n nn n n

e ee e e→∞ →∞ →∞

= = =−

The series converges by Theorem 9.14.

36. ( ) ( ) ( )11

21 1

1 22 11

nn n

n n nn n

e

e e e

++∞ ∞

−= =

−−=

+ +∑ ∑

Let ( ) 22 .

1

x

xef x

e=

+Then

( ) ( )( )

2 2

22

2 10 for 0.

1

x x

x

e ef x x

e

−′ = < >

+

So, ( )f x is decreasing for 0x > which implies

1 .n na a+ <

2 22 2 1lim lim lim 0

1 2

n n

n n nn n n

e ee e e→∞ →∞ →∞

= = =+

The series converges by Theorem 9.14.

37. ( )5

60

6 6 7

2 10.7333

!0.7333 0.002778 0.7333 0.002778

2 0.0027786!

0.7305 0.7361

n

nS

nS

R S S a

S

=

−= ≈

− ≤ ≤ +

= − ≤ = =

≤ ≤

∑

38. ( )( )

16

61

6 6 7

4 12.7067

ln 1

4 1.9236ln 8

0.7831 4.6303

n

nS

n

R S S a

S

+

=

−= ≈

+

= − ≤ = ≈

≤ ≤

∑

39. ( ) 16

6 21

6 6 7

3 12.4325

2.4325 0.0612 2.4325 0.06123 0.061249

2.3713 2.4937

n

nS

nS

R S S a

S

+

=

−= =

− ≤ ≤ +

= − ≤ = ≈

≤ ≤

∑

40. ( ) 16

61

6 6 7 7

10.1875

27 0.054692

0.1328 0.2422

n

nn

nS

R S S a

S

+

=

−= =

= − ≤ = ≈

≤ ≤

∑

41. ( )0

1!

n

n n

∞

=

−∑

(a) By Theorem 9.15,

( )1

1 0.001.1 !N NR a

N+≤ = <+

This inequality is valid when 6.N =

(b) Approximate the series by

( )6

0

1 1 1 1 1 11 1! 2 6 24 120 720

0.368.

n

n n=

−= − + − + − +

≈

∑

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

42. ( )0

12 !

n

nn n

∞

=

−∑


( )1 11 0.001.

2 1 !n N NR aN+ +≤ = <

+



( )4

0

1 1 1 1 11 0.607.2 ! 2 8 48 348

n

nn n=

−= − + − + ≈∑


43. ( )( )0

12 1 !

n

n n

∞

=

−+∑


( )1

1 0.001.2 1 1 !N NR a

N+≤ = <⎡ + + ⎤⎣ ⎦



( )( )

2

0

1 1 11 0.842.2 1 ! 6 120

n

n n=

−= − + ≈

+∑




44. ( )( )0

12 !

n

n n

∞

=

−∑


( )1

1 0.001.2 2 !N NR a

N+≤ = <+



( )( )

3

0

1 1 1 11 0.540.2 ! 2 24 720

n

n n=

−= − + − ≈∑


45. ( ) 1

1

1!

n

n n

+∞

=

−∑


11 0.001.

1N NR aN+≤ = <

+



( ) 11000

1

1 1 1 1 112 3 4 1000

0.693.

n

n n

+

=

−= − + − + −

≈

∑

(1000 terms)

46. ( ) 1

1

14

n

nn n

+∞

=

−∑


( )1 11 0.001.

4 1N N NR aN+ +≤ = <

+



( ) 13

1

1 1 1 1 0.224.4 4 32 192

n

nn n

+

=

−= − + ≈∑

(3 terms)

47. ( ) 1

31

1 n

n n

+∞

=

−∑

By Theorem 9.15,

( )

( )

1 3

3

1 0.0011

1 1000 1 10.

N NR aN

N N

+≤ = <+

⇒ + > ⇒ + >

Use 10 terms.

48. ( ) 1

21

1 n

n n

+∞

=

−∑

By Theorem 9.15,

( )( )

1 2

2

1 0.0011

1 1000 31.

N NR aN

N N

+≤ = <+

⇒ + > ⇒ =

Use 31 terms.

49. ( ) 1

31

12 1

n

n n

+∞

=

−−∑

By Theorem 9.15,

( )

1 31 0.001.

2 1 1N NR a

N+≤ = <

+ −

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

50. ( ) 1

51

1 n

n n

+∞

=

−∑

By Theorem 9.15,

( )

1 51 0.01

1N NR a

N+≤ = <

+

This inequality is valid when 2.N = Use 2 terms.

51. ( )1

12

n

nn

∞

=

−∑

1

12n

n

∞

=∑ is a convergent geometric series.

Therefore, ( )1

12

n

nn

∞

=

−∑ converges absolutely.

52. ( ) 1

21

1 n

n n

+∞

=

−∑

21

1

n n

∞

=∑ is a convergent p-series.

Therefore, ( ) 1

21

1 n

n n

+∞

=


53. ( )1

1!

n

n n

∞

=

−∑

21 1!n n< for 4n ≥

and 21

1

n n

∞


So, 1

1!n n

∞

=∑ converges, and

( )1

1!

n

n n

∞

=




54. ( ) 1

31

1 n

n

nn

+∞

=

−∑

3 5 21 1

1

n n

nn n

∞ ∞

= =

=∑ ∑ is a convergent p-series. Therefore,

( ) 1

31

1 n

n

nn

+∞

=


55. ( )( )

1

21

1

3

n

n n

+∞

=

−

+∑

( )2

1

13n n

∞

= +∑ converges by a limit comparison to the

p-series 21

1 .n n

∞

=∑ Therefore, ( )

( )

1

21

1

3

n

n n

+∞

=

−

+∑ converges

absolutely.

56. ( ) 1

1

13

n

n n

+∞

=

−+∑

The series converges by the Alternating Series Test. But, the series

1

13n n

∞

= +∑

diverges by comparison to 1

1.n n

∞

=∑

Therefore, ( ) 1

1

13

n

n n

+∞

=

−+∑ converges conditionally.

57. ( ) 1

1

1 n

n n

+∞

=

−∑

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

1

1

n n

∞

=∑

is a divergent p-series. Therefore, the series converges conditionally.

58. ( ) 1

1

1 n

n n n

+∞

=

−∑

3 21 1

1 1

n n nn n

∞ ∞

= =

=∑ ∑ which is a convergent p-series.

Therefore, the given series converges absolutely.

59. ( )( )

1 2

21

1

1

n

n

n

n

+∞

=

−

+∑

( )

2

2lim 11n

nn→∞

=+

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

60. ( ) ( )1

1

1 2 310

2 3lim 210

n

n

n

nn

nn

+∞

=

→∞

− ++

+=

+

∑


61. ( )2

1ln

n

n n n

∞

=

−∑

The series converges by the Alternating Series Test.

Let ( ) 1 .ln

f xx x

=

( ) 22

1 ln lnln

dx xx x

∞ ∞= ⎡ ⎤ = ∞⎣ ⎦∫

By the Integral Test, 2

1lnn n n

∞

=∑ diverges.

So, the series ( )2

1ln

n

n n n

∞

=

−∑ converges conditionally.

62. ( )2

0

1 n

nn e

∞

=

−∑

20

1nn e

∞

=∑ converges by a comparison to the convergent

geometric series 0

1 .n

n e

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ Therefore, the given series

converges absolutely.

63. ( )3

2

15

n

n

nn

∞

=

−−∑

32 5n

nn

∞

= −∑ converges by a limit comparison to the

p-series 22

1 .n n

∞

=∑ Therefore, the given series converges

absolutely.

64. ( ) 1

4 31

1 n

n n

+∞

=

−∑

4 31

1

n n

∞

=∑ is a convergent p-series. Therefore, the given

series converges absolutely.



65. ( )( )

( )

0

0

12 1 !

12 1 !

n

n

n

n

n

∞

=

∞

=

−+

+

∑

∑

is convergent by comparison to the convergent geometric series

0

12

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

because

( )

1 1 for 0.2 1 ! 2n n

n< >

+


66. ( )0

14

n

n n

∞

=

−

+∑

The given series converges by the Alternating Series Test, but

0

14n n

∞

= +∑

diverges by a limit comparison to the divergent p-series

1

1 .n n

∞

=∑

Therefore, the given series converges conditionally.

67. ( )0 0

1cos1 1

n

n n

nn n

π∞ ∞

= =

−=

+ +∑ ∑


0 0

cos 11 1n n

nn n

π∞ ∞

= =

=+ +∑ ∑

diverges by a limit comparison to the divergent harmonic series,

1

1.n n

∞

=∑

( )cos 1lim 1,

1n

n nn

π→∞

+= therefore, the series

converges conditionally.

68. ( ) 1

11 arctann

nn

∞+

=

−∑

lim arctan 02n

n π→∞

= ≠


69. ( )2 2

1 1

1cosn

n n

nn n

π∞ ∞

= =

−=∑ ∑

21

1

n n

∞



70. ( ) ( ) 1

1 1

sin 2 1 2 1 n

n n

nn n

π +∞ ∞

= =

⎡ − ⎤ −⎣ ⎦ =∑ ∑


( )

1 1

sin 2 1 2 1

n n

nn n

π∞ ∞

= =

⎡ − ⎤⎣ ⎦ =∑ ∑

is a divergent p-series. Therefore, the series converges conditionally.

71. An alternating series is a series whose terms alternate in sign.

72. See Theorem 9.14.

73. 1N N NS S R a +− = ≤ (Theorem 9.15)

74. na∑ is absolutely convergent if na∑ converges.

na∑ is conditionally convergent if na∑ diverges, but

na∑ converges.

75. (b). The partial sums alternate above and below the horizontal line representing the sum.

76. (a) False. For example, let ( )1 .n

nan−

=

Then ( )1 n

nan−

=∑ ∑ converges

and ( ) ( ) 11 n

nan

+−− =∑ ∑ converges.

But, 1na

n=∑ ∑ diverges.

(b) True. For if na∑ converged, then so would

na∑ by Theorem 9.16.

77. True. 1001 1 12 3 1001S = − + − + +

Because the next term 1101− is negative, 100S is an

overestimate of the sum.

78. False. Let

( )1 .n

n na bn

−= =∑ ∑ ∑

Then both converge by the Alternating Series Test. But,

1 ,n na bn

=∑ ∑ which diverges.



79. ( )1

11 np

n n

∞

=

−∑

If 0,p = then ( )1

1 n

n

∞

=

−∑ diverges.

If 0,p < then ( )1

1 n p

nn

∞−

=

−∑ diverges.

If 0,p > then 1lim 0pn n→∞= and

( )

11 1 .

1n np pa a

nn+ = < =

+

Therefore, the series converges for 0.p >

80. ( )1

11 n

n n p

∞

=

−+∑

Assume that 0n p+ ≠ so that ( )1na n p= + are defined for all n. For all p,

1lim lim 0nn n

an p→∞ →∞

= =+

11 1 .1n na a

n p n p+ = < <+ + +

Therefore, the series converges for all p.

81. Because

1

nn

a∞

=∑

converges you have lim 0.nn

a→∞

= So, there must exist

an 0N > such that 1Na < for all n N> and it

follows that 2n na a≤ for all .n N> So, by the

Comparison Test,

2

1n

na

∞

=∑

converges. Let 1na n= to see that the converse is false.

82. ( ) 1

1

1 n

n n

−∞

=

−∑ converges, but

1

1

n n

∞

=∑ diverges.

83. 21

1

n n

∞

=∑ converges, and so does 4

1

1 .n n

∞

=∑

84. (a) 1

n

n

xn

∞

=∑

converges absolutely (by comparison) for 1 1,x− < < because

andn

n nx x xn

< ∑

is a convergent geometric series for 1 1.x− < <

(b) When 1,x = − you have the convergent alternating series

( )1

1.

n

n n

∞

=

−∑

When 1,x = you have the divergent harmonic series 1 .n Therefore,

1

n

n

xn

∞

=∑ converges conditionally for 1.x = −

85. (a) No, the series does not satisfy 1n na a+ ≤ for all n.

For example, 1 1.9 8<

(b) Yes, the series converges.

21 1 1 12 3 2 31 1 1 12 2 3 3

1 1 1 11 12 2 3 3

n n n

n n

n n

S = − + + −

⎛ ⎞ ⎛ ⎞= + + − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + + + − + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

As ,n → ∞

( ) ( )21 1 3 12 .

1 1 2 1 1 3 2 2nS → − = − =− −

86. (a) No, the series does not satisfy 1 :n na a+ ≤

( ) 1

1

1 1 11 18 643

nn

na

∞+

=

− = − + − +∑ and

1 1 .8 3<

(b) No, the series diverges because 1n∑ diverges.

87. 3 2 3 21 1

10 110 ,n nn n

∞ ∞

= =

=∑ ∑

convergent p-series

88. 21

35n n

∞

= +∑

converges by limit comparison to convergent p-series

21 .n∑

Section 9.6 The Ratio and Root Tests 315


89. Diverges by nth-Term Test lim n

na

→∞= ∞

90. Converges by limit comparison to convergent geometric

series 1 .2n∑

91. Convergent geometric series

( )78 1r = <

92. Diverges by nth-Term Test

32lim n

na

→∞=

93. Convergent geometric series ( )1r e= or Integral

Test

94. Converges (conditionally) by Alternating Series Test

95. Converges (absolutely) by Alternating Series Test

96. Diverges by comparison to Divergent Harmonic Series:

ln 1 for 3n nn n

> ≥

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

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

99. 1 1 112 3 41 1 1 1 1 1 1 113 2 5 7 4 9 11 6

s

S

= − + − +

= + − + + − + + − +

(i) 4

2

1 1 1 1 1 1 112 3 4 5 6 4 1 4

1 1 1 1 1 1 1 12 2 4 6 8 10 4 2 4

n

n

sn n

sn n

= − + − + − + + −−

= − + − + − + −−

Adding: 4 2 31 1 1 1 1 1 1 1 112 3 2 5 7 4 4 3 4 1 2n n ns s s

n n n+ = + − + + − + + + − =

− −

(ii) lim nn

s s→∞

= (In fact, ln 2.s = )

0s ≠ because 1.2

s >

3 4 21 1 3lim2 2 2n n n

nS S s s s s s

→∞= = + = + =

So, .S s≠

Section 9.6 The Ratio and Root Tests

1. ( )( )

( )( )( )( )( ) ( )( )( )1 ! 1 1 2 !

1 12 ! 2 !

n n n n nn n n

n n+ + − −

= = + −+ −

2. ( )( )

( )( )( )( ) ( )( )

2 2 ! 2 2 ! 12 ! 2 2 1 2 2 ! 2 2 1

k kk k k k k k− −

= =− − −



3. Use the Principle of Mathematical Induction. When 1,k = the formula is valid because ( )( )

1

2 1 !1 .

2 1!=

⋅Assume that

( ) ( )2 !1 3 5 2 1

2 !n

nn

n⋅ ⋅ − =

and show that

( )( ) ( )( )1

2 2 !1 3 5 2 1 2 1 .

2 1 !n

nn n

n+

+⋅ ⋅ − + =

+

To do this, note that:

( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( )

( ) ( )( )( )

( )( )

1

1

1 3 5 2 1 2 1 1 3 5 2 1 2 1

2 !2 1 Induction hypothesis

2 !2 ! 2 1 2 2

2 ! 2 1

2 ! 2 1 2 22 ! 1

2 2 !2 1 !

n

n

n

n

n n n n

nn

nn n n

n n

n n nn n

nn

+

+

⋅ ⋅ − + = ⎡ ⋅ ⋅ − ⎤ +⎣ ⎦

= ⋅ +

+ += ⋅

+

+ +=

+

+=

+

The formula is valid for all 1.n ≥

4. Use the Principle of Mathematical Induction. When 3,k = the formula is valid because ( )( )32 3! 3 51 1.1 6!= = Assume that

( )

( )( )( )

2 ! 2 3 2 111 3 5 2 5 2 !

n n n nn n

− −=

⋅ ⋅ −

and show that

( )( )

( ) ( )( )( )

12 1 ! 2 1 2 11 .1 3 5 2 5 2 3 2 2 !

n n n nn n n

+ + − +=

⋅ ⋅ − − +

To do this, note that:

( )( ) ( ) ( )( )

1 1 11 3 5 2 5 2 3 1 3 5 2 5 2 3

2 ! 2 3n

n n n n

n n

= ⋅⋅ ⋅ − − ⋅ ⋅ − −

−=

( )( ) ( )

2 1 12 ! 2 3

n

n n

−⋅

−

( )( )

( )( )( )( )

( )( ) ( )( )( ) ( )( )( ) ( )( )

( )1

2 ! 2 1 2 1 2 22 ! 2 1 2 2

2 2 1 ! 2 1 2 12 ! 2 1 2 2

2 1 ! 2 1 2 12 2 !

n

n

n

n n n nn n n

n n n nn n n

n n nn

+

− + += ⋅

+ +

+ − +=

+ +

+ − +=

+

The formula is valid for all 3.n ≥

5. ( ) ( ) ( )1

3 3 94 4 161 2

n

nn

∞

=

= + +∑

1 234, 1.875S S= ≈

Matches (d).

6. 1

3 1 3 9 14 ! 4 16 2

n

n n

∞

=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑

1 23, 1.034

S S= ≈

Matches (c).



7. ( ) 1 3

1

1

3 39! 2

9

n

n nS

+∞

=

−= − +

=

∑

Matches (f).

8. ( )( )

1

1

1

1 4 4 42 ! 2 24

2

n

n n

S

−∞

=

−= − +

=

∑

Matches (b).

9. 2

1

1 2

4 4 85 3 2 7

2, 3.31

n

n

nn

S S

∞

=

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠= =

∑

Matches (a).

10. 0

1

44 4

4

n

ne

eS

∞−

=

= + +

=

∑

Matches (e).

11. (a) Ratio Test: ( ) ( )( )

2 1 21

2

1 5 8 1 5 5lim lim lim 1.8 85 8

nn

nn n nn

na na nn

++

→∞ →∞ →∞

+ +⎛ ⎞= = = <⎜ ⎟⎝ ⎠

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 sum

of the series.

12. (a) Ratio Test:

( )( )

2

21

2 2

1 11 ! 2 2 1lim lim lim 0 1.1 1 1

!

n

n n nn

nna n nna n n

n

+

→∞ →∞ →∞

+ ++ ⎛ ⎞+ + ⎛ ⎞= = = <⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠

Converges

(b) (c)

(d) The sum is approximately 7.15485 (e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum

of the series.

13. 1

12n

n

∞

=∑

1

11

1 2 2 1lim lim lim 12 21 2

n nn

n nn n nn

aa

++

+→∞ →∞ →∞= = = <

Therefore, the series converges by the Ratio Test.

n 5 10 15 20 25

nS 9.2104 16.7598 18.8016 19.1878 19.2491

n 5 10 15 20 25

nS 7.0917 7.1548 7.1548 7.1548 7.1548

00

12

20

0 110

10



14. 1

1!n n

∞

=∑

( )

( )

1 1 1 !lim lim

1 !

! 1lim lim 01 ! 1

n

n nn

n n

naa n

nn n

+

→∞ →∞

→∞ →∞

+=

= = =+ +


15. 0

!3n

n

n∞

=∑

( )11

1 ! 3 1lim lim lim3 ! 3

nn

nn n nn

na na n+

+→∞ →∞ →∞

+ += ⋅ = = ∞

Therefore, by the Ratio Test, the series diverges.

16. 0

3!

n

n n

∞

=∑

( )

11 3 ! 3lim lim lim 0

1 ! 3 1

nn

nn n nn

a na n n

++

→∞ →∞ →∞= ⋅ = =

+ +

Therefore, by the Ratio Test, the series converges.

17. 1

65

n

nn

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

( )( )( )

11 1 6 5

lim lim6 5

1 6 6lim 15 5

nn

nn nn

n

naa n

nn

++

→∞ →∞

→∞

+=

+ ⎛ ⎞= = >⎜ ⎟⎝ ⎠

Therefore, the series diverges by the Ratio Test.

18. 1

109

n

nn

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

( )( )( )

11 1 10 9

lim lim10 9

1 10 10lim 19 9

nn

nn nn

n

naa n

nn

++

→∞ →∞

→∞

+=

+ ⎛ ⎞= = >⎜ ⎟⎝ ⎠


19. 1 4n

n

n∞

=∑

( ) 11 1 4 1lim lim lim 1 4 1

44

nn

nn n nn

na na nn

++

→∞ →∞ →∞

+ += = = <


20. 3

1 4nn

n∞

=∑

( )

( )

131

3

3

3

1 4lim lim

4

1 1lim 14 4

n

nnn nn

n

naa n

nn

+

+

→∞ →∞

→∞

+=

+= = <


21. 21

4n

n n

∞

=∑

( )

( )( )

211

2

2

2

4 1lim lim

4

lim 4 4 11

nn

nn nn

n

naa n

nn

++

→∞ →∞

→∞

+=

= = >+


22. ( ) ( )( )

( )( ) ( )

( )

1

1

1

1 21

3 21 2 1

2lim 01

n

n

n n

n

nn n

n na an n n n

nn n

+∞

=

+

→∞

− ++

+ += ≤ =

+ + +

+=

+

∑

Therefore, by Theorem 9.14, the series converges. Note: The Ratio Test is inconclusive because

1lim 1.n

n n

aa+

→∞=

The series converges conditionally.

23. ( )0

1 2!

n n

n n

∞

=

−∑

( )1

1 2 !lim lim1 ! 2

2lim 01

nn

nn nn

n

a na n

n

++

→∞ →∞

→∞

= ⋅+

= =+


24. ( ) ( )1

21

1 3 2n n

n n

−∞

=

−∑

( )( )

( )

1 21

2

2

2

3 2lim lim

2 1 3 2

3 3lim 122 2 1

nn

nn nn

n

a na n n

nn n

++

→∞ →∞

→∞

= ⋅+ +

= = >+ +




25. 1

!3n

n

nn

∞

=∑

( )( )

11

1 ! 3lim lim lim1 3 ! 3

nn

nn n nn

na n na n n+

+→∞ →∞ →∞

+= ⋅ = = ∞

+


26. ( )5

1

2 !

n

nn

∞

=∑

( )( ) ( )

( )( )( )

51

5

5

5

2 2 !lim lim

2 !1

2 2 2 1lim

1

n

n nn

n

na na nn

n n n

n

+

→∞ →∞

→∞

+= ⋅

+

+ += = ∞

+


27. 0 !

n

n

en

∞

=∑

( )

( )

11 1 !

lim lim!

!lim lim 01 ! 1

nn

nn nn

n n

e naa e n

n een n

++

→∞ →∞

→∞ →∞

+=

⎛ ⎞= = =⎜ ⎟⎜ ⎟+ +⎝ ⎠


28. 1

!n

n

nn

∞

=∑

( ) ( ) 11 1 ! 1

lim lim!

1lim1

nn

nn nn

n

n

n naa n n

nn e

++

→∞ →∞

→∞

+ +=

⎛ ⎞= =⎜ ⎟+⎝ ⎠


29. ( )0

61

n

nn n

∞

= +∑

( )( )

111 6 2 6 1 1lim lim lim 0 0.

2 26 1

n nnn

nnn n nn

na na n n en

+++

→∞ →∞ →∞

+ +⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠+

To find 1lim :2

n

n

nn→∞

+⎛ ⎞⎜ ⎟+⎝ ⎠

Let 12

nnyn+⎛ ⎞= ⎜ ⎟+⎝ ⎠

( ) ( )

[ ] ( ) ( ) ( ) ( )( )( )

2

2

ln 1 ln 21ln ln2 1

2 11 1 1 2lim ln lim lim 1

1 1 2n n n

n nny nn n

n n nn ny

n n n→∞ →∞ →∞

+ − ++⎛ ⎞= =⎜ ⎟+⎝ ⎠

⎡ ⎤− ⎡ + − + ⎤⎡ + − + ⎤ ⎣ ⎦= = = −⎢ ⎥⎢ ⎥− + +⎢ ⎥⎣ ⎦ ⎣ ⎦

by L'Hôpital's Rule. So, 1.ye

→


30. ( )( )

2

0

!3 !n

nn

∞

=∑

( )( )

( )( )

( )( )( )( )

2 21

2

1 ! 3 ! 1lim lim lim 0

3 3 ! 3 3 3 2 3 1!n

n n nn

n n naa n n n nn+

→∞ →∞ →∞

⎡ + ⎤ +⎣ ⎦= ⋅ = =+ + + +


31. 0

52 1

n

nn

∞

= +∑

( )( )

( )( )

( )1 11

1

5 2 1 5 2 1 5 1 1 2 5lim lim lim lim 12 1 2 25 2 1 2 1

n n n nn

nn n nn n n nn

aa

+ ++

+→∞ →∞ →∞ →∞

+ + += = = = >

++ +




32. ( )( )

( )( )

( )( )

4

0

4 4 41

4

1 22 1 !

2 1 !2 2lim lim lim 02 3 ! 2 2 3 2 2

n n

n

nn

nn n nn

n

naa n n n

∞

=

++

→∞ →∞ →∞

−+

+= ⋅ = =

+ + +

∑


33. ( )( )

( )( )( )

( )

1

0

1

1 !1 3 5 2 1

1 ! 1 3 5 2 1 1 1lim lim lim1 3 5 2 1 2 3 ! 2 3 2

n

n

n

n n nn

nn

n na na n n n n

+∞

=

+

→∞ →∞ →∞

−⋅ ⋅ +

+ ⋅ ⋅ + += ⋅ = =

⋅ ⋅ + + +

∑


Note: The first few terms of this series are 1 2! 3!1 .1 3 1 3 5 1 3 5 7

− + − + −⋅ ⋅ ⋅ ⋅ ⋅ ⋅

34. ( )( )1

1 2 4 6 22 5 8 3 1

n

n

nn

∞

=

− ⋅ ⋅⋅ ⋅ −∑

( )( )( )

( )1 2 4 2 2 2 2 5 3 1 2 2 2lim lim lim2 5 3 1 3 2 2 4 2 3 2 3

n

n n nn

n n na na n n n n+

→∞ →∞ →∞

⋅ + ⋅ − += ⋅ = =

⋅ − + ⋅ +


Note: The first few terms of this series are 2 2 4 2 4 62 2 5 2 5 8

⋅ ⋅ ⋅− + − +

⋅ ⋅ ⋅

35. 1

15n

n

∞

=∑

11 1lim lim 1

5 5

nn

n nn na

→∞ →∞

⎡ ⎤= = <⎢ ⎥⎣ ⎦

Therefore, by the Root Test, the series converges.

36. 1

1n

n n

∞

=∑

11 1lim lim lim 0

nn

n nn n na

n n→∞ →∞ →∞

⎡ ⎤= = =⎢ ⎥⎣ ⎦


37. 1 2 1

n

n

nn

∞

=

⎛ ⎞⎜ ⎟+⎝ ⎠

∑

1lim lim lim2 1 2 1 2

nn nn

n n n

n nan n→∞ →∞ →∞

⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠


38. 1

21

n

n

nn

∞

=

⎛ ⎞⎜ ⎟+⎝ ⎠

∑

2 2lim lim lim 21 1

nn nn

n n n

n nan n→∞ →∞ →∞

⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠

Therefore, by the Root Test, the series diverges.

39. 2

2 11

n

n

nn

∞

=

+⎛ ⎞⎜ ⎟−⎝ ⎠

∑

2 1 2 1lim lim lim 21 1

nn nn

n n n

n nan n→∞ →∞ →∞

+ +⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠


40. 1

4 32 1

n

n

nn

∞

=

+⎛ ⎞⎜ ⎟−⎝ ⎠

∑

4 3 4 3lim lim lim 22 1 2 1

nn nn

n n n

n nan n→∞ →∞ →∞

+ +⎛ ⎞= = =⎜ ⎟− −⎝ ⎠


41. ( )( )2

1

ln

n

nn n

∞

=

−∑

( )( )

1 1lim lim lim 0lnln

nn nn nn n n

ann→∞ →∞ →∞

−= = =




42. 3

1

32 1

n

n

nn

∞

=

−⎛ ⎞⎜ ⎟+⎝ ⎠

∑

3

3 3

3lim lim2 1

3 3 27lim2 1 2 8

nn nn

n n

n

nan

nn

→∞ →∞

→∞

−⎛ ⎞= ⎜ ⎟+⎝ ⎠

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠


43. ( )1

2 1nn

nn

∞

=

+∑

( ) ( )lim lim 2 1 lim 2 1nn nnn

nn n n

a n n→∞ →∞ →∞

= + = +

To find lim ,n

nn

→∞let lim .n

ny n

→∞= Then

( )ln lim ln

1 ln 1lim ln lim lim 0.1

nn

n n n

y n

n nnn n

→∞

→∞ →∞ →∞

=

= = = =

So, ln 0,y = so 0 1y e= = and

( ) ( )lim 2 1 2 1 1 3.n

nn

→∞+ = + =


44. 33

0 0

1nn

n ne

e

∞ ∞−

= =

=∑ ∑

1

31 1 1lim lim lim

3 3

nn nn n nn n n

ae→∞ →∞ →∞

⎛ ⎞= = =⎜ ⎟⎝ ⎠

Therefore, the series converges by the Root Test.

45. 1 3n

n

n∞

=∑

1 1 1lim lim lim

3 3 3

n nn

n nn n n

n na→∞ →∞ →∞

⎛ ⎞= = =⎜ ⎟⎝ ⎠

Therefore, the series converges by the Root Test. Note: You can use L'Hôpital's Rule to show

1lim 1:nn

n→∞

=

Let 1 1 ln, ln lnn ny n y nn n

= = =

ln 1lim lim 0 11n n

n n yn→∞ →∞

= = ⇒ →

46. 1 500

n

n

n∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

lim lim lim500 500

nn nn

n n n

n na→∞ →∞ →∞

⎛ ⎞ ⎛ ⎞= = = ∞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


47. 21

1 n

n n n

∞

=

1⎛ ⎞−⎜ ⎟⎝ ⎠

∑

2

2

1 1lim lim

1 1lim 0 0 0 1

nn nn

n n

n

an n

n n

→∞ →∞

→∞

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= − = − = <⎜ ⎟⎝ ⎠


48. 1

ln n

n

nn

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

ln lnlim lim lim 0 1n

n nnn n n

n nan n→∞ →∞ →∞

⎛ ⎞= = = <⎜ ⎟⎝ ⎠


49. ( )2 ln n

n

nn

∞

=∑

( )

1lim lim lim 0

lnln

nn nn nn n n

n nann→∞ →∞ →∞

= = =


50. ( )( )

( )( )2 21 1

! !n n

nnn n

n n

n n

∞ ∞

= =

=∑ ∑

( )( ) 22

! !lim lim limnn n nn n n

n nann→∞ →∞ →∞

= = = ∞


51. ( ) 1

1

1 5n

n n

+∞

=

−∑

1

5 51

5lim 0

n n

n

a an n

n

+

→∞

= < =+

=

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

52. 1 1

100 1100n nn n

∞ ∞

= =

=∑ ∑

This is the divergent harmonic series.

53. 3 21 1

3 13n n nn n

∞ ∞

= =

=∑ ∑

This is a convergent p-series.



54. 1

23

n

n

π∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Because 2 1,3

r π= > this is a divergent Geometric

Series.

55. 1

52 1n

nn

∞

= −∑

5 5lim2 1 2n

nn→∞

=−

Therefore, the series diverges by the nth-Term Test

56. 21 2 1n

nn

∞

= +∑

( )2 2

2

2 1 1lim lim 01 2 1 2n n

n n nn n→∞ →∞

+= = >

+

This series diverges by limit comparison to the divergent harmonic series

1

1.n n

∞

=∑

57. ( ) ( )2 2

1 1 1

1 3 1 3 3 1 32 2 9 2

n n nn n

n nn n n

− −∞ ∞ ∞

= = =

− − ⎛ ⎞= = −⎜ ⎟⎝ ⎠

∑ ∑ ∑

Because 3 1,2

r = > this is a divergent geometric series.

58. 3

1

103n n

∞

=∑

3 2

3 210 3 10lim

1 3n

nn→∞

=

Therefore, the series converges by a Limit Comparison Test with the p-series

3 21

1 .n n

∞

=∑

59. 1

10 32n

n

nn

∞

=

+∑

( )10 3 2 10 3lim lim 101 2

n

nn n

n n nn→∞ →∞

+ += =

Therefore, the series converges by a Limit Comparison Test with the geometric series

0

1 .2

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

60. 21

24 1

n

n n

∞

= −∑

( ) ( )2

2

ln 2 2 ln 2 22lim lim lim4 1 8 8

n nn

n n nn n→∞ →∞ →∞= = = ∞

−


61. cos 13 3n n

n≤

Therefore the series 1

cos3n

n

n∞

=∑ converges

by Direct comparison with the convergent geometric

series 1

1 .3n

n

∞

=∑ So, cos

3nn∑ converges.

62. ( )2

1ln

n

n n n

∞

=

−∑

( ) ( ) ( )1

1 11 ln 1 lnn na a

n n n n+ = ≤ =+ +

1lim 0ln ( )n n n→∞

=

Therefore, by the Alternating Series Test, the series converges.

63. 1

7!

n

n

nn

∞

=∑

( )( )

11 1 7 ! 7lim lim lim 0

1 ! 7

nn

nn n nn

na na n n n

++

→∞ →∞ →∞

+= ⋅ = =

+


64. ( )2

1

ln

n

nn

∞

=∑

( )2 3 2

ln 1nn n

≤

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

3 21

1 .n n

∞

=∑

65. ( ) 1

1

1 3!

n n

n n

−∞

=

−∑

( )

11

3 ! 3lim lim lim 01 ! 3 1

nn

nn n nn

a na n n+

−→∞ →∞ →∞= ⋅ = =

+ +

Therefore, by the Ratio Test, the series converges. (Absolutely)



66. ( )1

1 32

n n

nn n

∞

=

−∑

( ) ( )

11

13 2 3 3lim lim lim

1 2 3 2 1 2

n nn

n nn n nn

a n na n n

++

+→∞ →∞ →∞= ⋅ = =

+ +


67. ( )( )1

33 5 7 2 1

n

n n

∞

=

−⋅ ⋅ +∑

( )( )( )

( )( )

11 3 3 5 7 2 1 3lim lim lim 0

3 5 7 2 1 2 3 2 33

nn

nn n nn

naa n n n

++

→∞ →∞ →∞

− ⋅ ⋅ += ⋅ = =

⋅ ⋅ + + +−


68. ( )( )1

3 5 7 2 118 2 1 !n

n

nn n

∞

=

⋅ ⋅ +−∑

( )( )( )( )

( )( )

( )( )( )( )

11

3 5 7 2 1 2 3 18 2 1 ! 2 3 2 1 1lim lim lim18 2 1 2 1 ! 3 5 7 2 1 18 2 1 2 1 18

nn

nn n nn

n n n n n naa n n n n n n+

+→∞ →∞ →∞

⋅ ⋅ + + − + −= ⋅ = =

+ − ⋅ ⋅ + + −

Therefore, by the Ratio Test, the series converge.

69. (a) and (c) are the same.

( )( )

( )( ) ( )( ) ( )( )

1

1 0

2 3 4

1 55! 1 !

2 5 3 5 4 55

2! 3! 4!

nn

n n

nnn n

+∞ ∞

= =

+=

+

= + + + +

∑ ∑

70. (b) and (c) are the same.

( )( ) ( )

( ) ( ) ( )

1

0 1

2 3

3 34 4

3 3 34 4 4

1

1 2 3 4

n n

n nn n

∞ ∞ −

= =

+ =

= + + + +

∑ ∑

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

( )( )

( )( )

1

0 1

1 12 1 ! 2 1 !

1 113! 5!

n n

n nn n

−∞ ∞

= =

− −=

+ −

= − + −

∑ ∑

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

( )( )

( ) 1

12 1

2 3

1 11 2 2

1 1 12 2 2 3 2

n n

n nn nn n

+∞ ∞

−= =

− −=

−

= − + −⋅ ⋅

∑ ∑

73. Replace n with 1.n +

11 0

17 7n n

n n

n n∞ ∞

+= =

+=∑ ∑

74. Replace n with 2.n +

( )

2

2 0

9 92 ! !

n n

n nn n

+∞ ∞

= =

=−∑ ∑

75. (a) Because

10

5103 1.59 10 ,

2 10!−≈ ×

use 9 terms.

(b) ( )9

1

30.7769

2 !

k

kk k=

−≈ −∑

76. (a) Use 10 terms, 9,k = see Exercise 3.

(b) ( )( )

( )( ) ( )

( )( )

0 0

0

3 3 2 !1 3 5 2 1 2 ! 2 1

6 !0.40967

2 1 !

k k k

k k

k

k

kk k k

kk

∞ ∞

= =

∞

=

− −=

⋅ ⋅ + +

−= ≈

+

∑ ∑

∑

…

77. ( ) ( )1 4 1 3 2lim lim

4 1 4lim 13 2 3

nn

n nn n

n

n n aaa a

nn

+

→∞ →∞

→∞

− +=

−= = >

+

The series diverges by the Ratio Test.

78. ( ) ( )1 2 1 5 4lim lim

2 1 2lim 15 4 5

nn

n nn n

n

n n aaa a

nn

+

→∞ →∞

→∞

+ −=

+= = <

−

The series converges by the Ratio Test.



79. ( ) ( )1sin 1

lim lim

sin 1lim 0 1

nn

n nn n

n

n n aaa a

nn

+

→∞ →∞

→∞

+=

+= = <


80. ( ) ( )1 cos 1lim lim

cos 1lim 0 1

nn

n nn n

n

n n aaa a

nn

+

→∞ →∞

→∞

+=

+= = <


81. ( ) ( )( )1 1 1 1lim lim lim 1 1nn

n n nn n

n aaa a n+

→∞ →∞ →∞

+ ⎛ ⎞= = + =⎜ ⎟⎝ ⎠

The Ratio Test is inconclusive. But, lim 0,n

na

→∞≠ so the series diverges.

82. The series diverges because lim 0.nn

a→∞

≠

( )( )

1

1 22

1 33

14

1 14 2

12 0.7937

a

a

a

=

= =

= ≈

In general, 1 0.n na a+ > >

83.

( )( )( )

( )

1

1 2 11 3 2 1 2 1

lim lim 1 21 3 2 1

1 1lim 12 1 2

n

n nn

n

n nn na

nan

nn

+

→∞ →∞

→∞

⋅ +⋅ − +

=⋅

⋅ −

+= = <

+


84. 0

13n

n

n∞

=

+∑

( )( )

( )

1 1lim lim lim3 3

Let lim 1

ln lim ln 1

1lim ln 1

ln 1 1lim 0.1

nn nn nn n n

n

n

n

n

n

n

n na

y n

y n

nn

nn n

→∞ →∞ →∞

→∞

→∞

→∞

→∞

+ += =

= +

= +

= +

+= = =

+

Because 0ln 0, 1,y y e= = = so

1 1lim .3 3n

n→∞

+=


85. ( )3

1ln n

n n

∞

=∑

( )

1 1lim lim lim 0lnln

n nn nn n na

nn→∞ →∞ →∞= = =


86.

( )( )( )( )( )

( )( )

( )( )

1

1 3 5 2 1 2 11 2 3 2 1 2 2 1

lim lim1 3 5 2 11 2 3 2 1

2 1lim 0 12 2 1

n

n nn

n

n nn n na

nan

nn n

+

→∞ →∞

→∞

⋅ ⋅ − +⋅ ⋅ − +

=⋅ ⋅ −⋅ ⋅ −

+= = <

+


87. ( )( )

11 2 3

lim lim lim3 32 3

nn

nn n nn

xa x xa x

++

→∞ →∞ →∞= = =

For the series to converge: 1 3 3.3x x< ⇒ − < <

For 3,x = the series diverges.

For 3,x = − the series diverges.

88. 1 1 1lim lim lim4 4 4

nn nn

n n n

x x xa→∞ →∞ →∞

+ + += = =

For the series to converge,

1 1 4 1 4

45 3.

x x

x

+< ⇒ − < + <

⇒ − < <

For ( )0

3, 1 ,n

nx

∞

=

= ∑ diverges.

For ( )0

5, 1 ,n

nx

∞

=

= − −∑ diverges.

89. ( ) ( )

( )

11 1 1

lim lim

lim 1 11

nn

nn nn

n

x naa x n

n x xn

++

→∞ →∞

→∞

+ +=

= + = ++


1 1 1 1 1

2 0.

x x

x

+ < ⇒ − < + <

⇒ − < <

For ( )1

10, ,

n

nx

n

∞

=

−= ∑ converges.

For ( ) ( )1 1

1 1 12, ,n n

n nx

n n

∞ ∞

= =

− −= − =∑ ∑ diverges.



90. 1

1 2 1lim lim lim 1 1

2 1

nn

nn n nn

xa x xa x

++

→∞ →∞ →∞

−= = − = −

−


1 1 1 1 1

0 2.

x x

x

− < ⇒ − < − <

⇒ < <

For ( )0

2, 2 1 ,n

nx

∞

=

= ∑ diverges.

For ( )0

0, 2 1 ,n

nx

∞

=

= −∑ diverges.

91. ( )

( )

1

11 !

2lim lim!

2

lim 12

n

nnn nn

n

xnaa xn

xn

+

+

→∞ →∞

→∞

+=

= + = ∞

The series converges only at 0.x =

92. ( )

11 11 1

lim lim lim 01 ! ! 1

n nn

n n nn

xx xan na n

++

→∞ →∞ →∞

++ += = =

+ +

The series converges for all x.

93. See Theorem 9.17, page 641.


95. No. Let 1 .10,000na

n=

+

The series 1

110,000n n

∞

= +∑ diverges.

96. One example is 1

1100 .n n

∞

=

⎛ ⎞− +⎜ ⎟⎝ ⎠

∑

97. The series converges absolutely. See Theorem 9.17.

98. (a) Converges (Ratio Test) (b) Inconclusive (See Ratio Test) (c) Diverges (Ratio Test) (d) Diverges (Root Test) (e) Inconclusive (See Root Test) (f) Diverges (Root Test, 1e > )

99. Assume that

1lim 1n nn

a a L+→∞

= > or that 1lim .n nn

a a+→∞

= ∞

Then there exists 0N > such that 1 1n na a+ > for all

.n N> Therefore,

1 , lim 0n n n nn

a a n N a a+→∞

> > ⇒ ≠ ⇒ ∑ diverges.

100. First, let

lim 1nn

na r

→∞= <

and choose R such that 0 1.r R≤ < < There must

exist some 0N > such that nna R< for all

.n N> So, for nnn N a R> < and because the

geometric series

0

n

nR

∞

=∑

converges, you can apply the Comparison Test to conclude that

1

nn

a∞

=∑

converges which in turn implies that 1

nn

a∞

=∑ converges.

Second, let

lim 1.nn

na r R

→∞= > >

Then there must exist some 0M > such that n

na R> for infinitely many .n M> So, for

infinitely many ,n M> you have 1nna R> > which

implies that lim 0nn

a→∞

≠ which in turn implies that

1

nn

a∞

=∑ diverges.

101. 3 21

1

n n

∞

=∑

( )

3 23 21

3 21lim lim lim 1

1 11n

n n nn

a n na nn+

→∞ →∞ →∞

⎛ ⎞= ⋅ = =⎜ ⎟+⎝ ⎠+

102. 1 21

1

n n

∞

=∑

( )

1 21

1 2

1 2

1lim lim11

lim 11

n

n nn

n

a na n

nn

+

→∞ →∞

→∞

= ⋅+

⎛ ⎞= =⎜ ⎟+⎝ ⎠

103. 41

1

n n

∞

=∑

( )

441

41lim lim lim 1

1 11n

n n nn

a n na nn+

→∞ →∞ →∞

⎛ ⎞= ⋅ = =⎜ ⎟+⎝ ⎠+

104. 1

1p

n n

∞

=∑

( )

1 1lim lim lim 11 11

ppn

pn n nn

a n na nn+

→∞ →∞ →∞

⎛ ⎞= ⋅ = =⎜ ⎟+⎝ ⎠+



105. 1

1 ,pn n

∞

=∑ p-series

1 1lim lim lim 1n nn p p nn n na

n n→∞ →∞ →∞= = =

So, the Root Test is inconclusive.

Note: lim 1p nn

n→∞

= because if ,p ny n= then

ln lnpy nn

= and ln 0p nn

→ as .n → ∞

So 1y → as .n → ∞

106. Ratio Test:

( )( ) ( )( )

1 lnlim lim 1,

1 ln 1

pn

pn nn

n naa n n+

→∞ →∞= =

+ +inconclusive.

Root Test:

( ) ( )1

1 1lim lim limln ln

n nn p p nnn n na

n n n n→∞ →∞ →∞= =

1lim 1.nn

n→∞

= Furthermore, let ( )ln p ny n= ⇒

( )ln ln ln .py nn

=

( )( )( ) ( )ln ln

lim ln lim lim 0 lim ln 1.ln 1

p n

n n n n

p n py nn n n→∞ →∞ →∞ →∞

= = = ⇒ =

So, ( )1

1lim 1,ln p nnn n n→∞

= inconclusive.

107. ( )( )

2

1

!,

!n

nxn

∞

=∑ x positive integer

(a) ( )2!1: !,

!n

x nn

= =∑ ∑ diverges

(b) ( )( )

2!2:

2 !n

xn

= ∑ converges by the Ratio Test:

( )( )

( )( )

( )( )( )

2 221 ! 1! 1lim lim 12 2 ! 2 ! 2 2 2 1 4n n

n nnn n n n→∞ →∞

⎡ + ⎤ +⎣ ⎦ = = <+ + +

(c) ( )( )

2!3:

3 !n

xn

= ∑ converges by the Ratio Test:

( )( )

( )( )

( )( )( )( )

2 221 ! 1!lim lim 0 1

3 3 ! 3 ! 3 3 3 2 3 1n n

n nnn n n n n→∞ →∞

⎡ + ⎤ +⎣ ⎦ = = <+ + + +

(d) Use the Ratio Test:

( )( )

( )( ) ( ) ( )

( )

2 221 ! !!

lim lim 1!1 ! !n n

n xnnn

xnx n xn x→∞ →∞

⎡ + ⎤⎣ ⎦ = +⎡ + ⎤ +⎣ ⎦

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



108. For 1 1 1

1, 2, 3, , .k k k

n n n n n nn n n

n a a a a a a= = =

= − ≤ ≤ ⇒ − ≤ ≤∑ ∑ ∑…

Taking limits as 1 1 1 1 1

, .n n n n nn n n n n

k a a a a a∞ ∞ ∞ ∞ ∞

= = = = =

→ ∞ − ≤ ≤ ⇒ ≤∑ ∑ ∑ ∑ ∑

109. The differentiation test states that if

1

nn

U∞

=∑

is an infinite series with real terms and ( )f x is a real function such that ( )1 nf n U= for all positive integers n and 2 2d f dx exists at 0,x = then

1

nn

U∞

=∑

converges absolutely if ( ) ( )0 0 0f f ′= = and diverges otherwise. Below are some examples.

Convergent Series Divergent Series

( ) 33

1 , f x xn

=∑ ( )1, f x xn

=∑

( )11 cos , 1 cosf x xn

⎛ ⎞− = −⎜ ⎟⎝ ⎠

∑ ( )1sin , sinf x xn

=∑

110. Using the Ratio Test,

( )

( )( ) ( )( )

1 11

11 ! 19 1 19 19 1! 19 19lim lim lim lim 1

7 7 7 7 71 1 1 1

n n nn

n n nnn n n nn

na n nnna en n n

− −+

−→∞ →∞ →∞ →∞

⎡ ⎤⎡ ⎤⎡ ⎤− ⋅ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎢ ⎥= = ⎢ ⎥ = = ⋅ <⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ ⎢ + ⎥⎢ ⎥ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

So, the series converges.

111. First prove Abel's Summation Theorem:

If the partial sums of na∑ are bounded and if { }nb decreases to zero, then n na b∑ converges.

Let 1

.k

k ii

S a=

= ∑ Let M be a bound for { }.kS

( ) ( )( ) ( ) ( )

( )

1 1 2 2 1 1 2 1 2 1

1 1 2 2 2 3 1 1

1

11

n n n n n

n n n n n

n

i i i n ni

a b a b a b S b S S b S S b

S b b S b b S b b S b

S b b S b

−

− −

−

+=

+ + + = + − + + −

= − + − + + − +

= − +∑

The series ( )11

i i ii

S b b∞

+=

−∑ is absolutely convergent because ( ) ( )1 1i i i i iS b b M b b+ +− ≤ − and ( )11

i ii

b b∞

+=

−∑ converges

to 1.b

Also, lim 0n nn

S b→∞

= because { }nS bounded and 0.nb → Thus, 1 1

limn

n n i inn i

a b a b∞

→∞= =

=∑ ∑ converges.

Now let 1nb

n= to finish the problem.



Section 9.7 Taylor Polynomials and Approximations

1. 212 1y x= − +

Parabola Matches (d)

2. 4 21 18 2 1y x x= − +

y-axis symmetry Three relative extrema Matches (c)

3. ( )1 2 1 1y e x−= ⎡ + + ⎤⎣ ⎦

Linear Matches (a)

4. ( ) ( )31 2 13 1 1 1y e x x− ⎡ ⎤= − − − +⎣ ⎦

Cubic Matches (b)

5. ( ) ( )

( ) ( )

1 2

3 2

8 8 4 4

14 42

f x x fx

f x x f

−

−

= = =

′ ′= − = −

( ) ( ) ( )( )

( )

1 4 4 4

1 14 4 62 2

P x f f x

x x

′= + −

= − − = − +

1P is the first degree Taylor polynomial for f at 4.

6. ( ) ( )

( ) ( )

1 33

4 3

6 6 8 3

12 88

f x x fx

f x x f

−

−

= = =

′ ′= − = −

( ) ( ) ( )( )

( )

1 8 8 8

1 13 8 48 8

P x f f x

x x

′= + −

= − − = − +

1P is the first degree Taylor polynomial for f at 8.

7. ( )

( )

sec 24

sec tan 24

f x x f

f x x x f

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= =⎜ ⎟⎝ ⎠

( )

( )

1

1

4 4 4

2 24

P x f f x

P x x

π π π

π

⎛ ⎞ ⎛ ⎞⎛ ⎞′= + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= + −⎜ ⎟⎝ ⎠

1P is called the first degree Taylor polynomial for f at .4π

8. ( )

( ) 2

tan 14

sec 24

f x x f

f x x f

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= =⎜ ⎟⎝ ⎠

( )

1

1

1 24 4 4 4

2 12

P f f x x

P x x

π π π π

π

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞′= + − = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= + −

1P is called the first degree Taylor polynomial for f at .4π

f

P1 (4, 4)

−1 110

8

f

P1

02

12

4

(8, 3)

−

−1

5

P1

f

, 2

�4

�2

4π ))

�2

�2

−

−3

3

Section 9.7 Taylor Polynomials and Approximations 329


9. ( ) ( )

( ) ( )( ) ( )

1 2

3 2

5 2

4 4 1 4

2 1 2

3 1 3

f x x fx

f x x f

f x x f

−

−

−

= = =

′ ′= − = −

′′ ′′= =

( ) ( )( ) ( )( )

( ) ( )

22

2

11 1 1 1

234 2 1 12

fP f f x x

x x

′′′= + − + −

= − − + −

10. ( )

( )

( ) 3 2

sec 24

sec tan 24

sec sec tan 3 24

f x x f

f x x x f

f x x x x f

π

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= =⎜ ⎟⎝ ⎠

⎛ ⎞′′ ′′= + =⎜ ⎟⎝ ⎠

( ) ( )

( )

2

2

2

2

44 4 4 2 4

32 2 24 2 4

fP x f f x x

P x x x

ππ π π π

π π

′′⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞′= + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x 0 0.8 0.9 1.0 1.1 1.2 2

( )f x Error 4.4721 4.2164 4.0 3.8139 3.6515 2.8284

( )2P x 7.5 4.46 4.215 4.0 3.815 3.66 3.5

x –2.15 0.585 0.685 4π 0.885 0.985 1.785

( )f x –1.8270 1.1995 1.2913 1.4142 1.5791 1.8088 –4.7043

( )2P x 15.5414 1.2160 1.2936 1.4142 1.5761 1.7810 4.9475

−2 6

−2

10

P2

f

(1, 4)

−30

3

4



11. ( )( )( )( )

22

2 44

2 4 66

121 12 241 1 12 24 720

cos

1

1

1

f x x

P x x

P x x x

P x x x x

=

= −

= − +

= − + −

(a)

(b) ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2

2

2

4

4 44

4 44

5 56

6 6

6 66

sin

cos 1

0 0 1

sin

cos 1

0 1 0

sin

cos 1

0 1 0

f x x P x x

f x x P x

f P

f x x P x x

f x x P x

f P

f x x P x x

f x x P x

f P

′′ = − = −

′′′′ = − = −

′′′′ = = −

′′′′′′ = =

= =

= =

= − = −

= − = −

= − =

(c) In general, ( ) ( ) ( ) ( )0 0n nnf P= for all n.

12. ( ) ( )2 , 0 0xf x x e f= =

(a) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( ) ( )

2

2

2

4 42

2 0 0

4 2 0 2

6 6 0 6

8 12 0 12

x

x

x

x

f x x x e f

f x x x e f

f x x x e f

f x x x e f

′ ′= + =

′′ ′′= + + =

′′′ ′′′= + + =

= + + =

( )

( )

( )

22

2

32 2 3

3

4 42 3 2 3

4

22!

63!

124! 2

xP x x

xP x x x x

x xP x x x x x

= =

= + = +

= + + = + +

(b)

(c) ( ) ( )

( ) ( )( ) ( ) ( ) ( )

2

3

4 44

0 2 0

0 6 0

0 12 0

f P

f P

f P

′′′′ = =

′′′′′′ = =

= =

(d) ( ) ( ) ( ) ( )0 0n nnf P=

13. ( ) ( )

( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

3

3

3

3

4 43

0 13 0 39 0 927 0 27

81 0 81

x

x

x

x

x

f x e ff x e ff x e ff x e f

f x e f

= =

′ ′= =

′′ ′′= =

′′′ ′′′= =

= =

( ) ( ) ( ) ( ) ( ) ( ) ( )42 3 4

4

2 3 4

0 0 00 0

2! 3! 4!9 9 271 32 2 8

f f fP x f f x x x x

x x x x

′′ ′′′′= + + + +

= + + + +

14. ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

4 4

5 5

0 10 10 10 1

0 1

0 1

x

x

x

x

x

x


f x e f

f x e f

−

−

−

−

−

−

= =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

= =

= − = −

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

42 3 4

5

5 2 3 4 55

0 0 00 0

2! 3! 4!0

15! 2 6 24 120


f x x x xx x

′ ′′′′= + + + +

+ = − + − + −

3−3

−2

2

P6

P2

P4

f

−1

−3 3

f

P3

P2P4

2



15. ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

2

2

2

2

4 42

0 11 102 2

1 104 4

1 108 8

1 1016 16

x

x

x

x

x

f x e f

f x e f

f x e f

f x e f

f x e f

−

−

−

−

−

= =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

= =

( ) ( ) ( ) ( ) ( ) ( ) ( )42 3 4

4

2 3 4

0 0 00 0

2! 3! 4!1 1 1 112 8 48 384


x x x x

′′ ′′′′= + + + +

= − + − +

16. ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

3

3

3

3

4 43

0 11 103 31 109 91 1027 271 181 81

x

x

x

x

x

f x e f

f x e f

f x e f

f x e f

f x e f x

= =

′ ′= =

′′ ′′= =

′′′ ′′′= =

= =

( ) 2 3 44

2 3 4

1 1 9 1 27 1 8113 2! 3! 4!1 1 1 113 18 162 1944

P x x x x x

x x x x

= + + + +

= + + + +

17. ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

4 4

5 5

sin 0 0cos 0 1

sin 0 0cos 0 1

sin 0 0

cos 0 1

f x x ff x x ff x x ff x x f

f x x f

f x x f

= =

′ ′= =

′′ ′′= − =

′′′ ′′′= − = −

= =

= =

( ) ( ) 2 3 4 55

3 5

0 1 0 10 12! 3! 4! 5!

1 16 120

P x x x x x x

x x x

−= + + + + +

= − +

18. ( ) ( )( ) ( )( ) ( )( ) ( )3 3

sin 0 0cos 0

sin 0 0cos 0


ππ π ππ ππ π π

2

= =

′ ′= =

′′ ′′= − =

′′′ ′′′= − = −

( )3 3

2 3 33

002! 3! 6

P x x x x x xπ ππ π−= + + + = −

19. ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )4 4

0 00 1

2 0 23 0 3

4 0 4

x

x x

x x

x x

x x

f x xe ff x xe e ff x xe e ff x xe e f

f x xe e f

= =

′ ′= + =

′′ ′′= + =

′′′ ′′′= + =

= + =

( ) 2 3 44

2 3 4

2 3 402! 3! 4!1 12 6

P x x x x x

x x x x

= + + + +

= + + +

20. ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

2

2

2

2

4 42

0 02 0 02 4 0 2

6 6 0 6

12 8 0 12

x

x x

x x x

x x x

x x x

f x x e ff x xe x e ff x e xe x e ff x e xe x e f

f x e xe x e f

−

− −

− − −

− − −

− − −

= =

′ ′= − =

′′ ′′= − + =

′′′ ′′′= − + − = −

= − + =

( ) 2 3 44

2 3 4

2 6 120 02! 3! 4!12

P x x x x x

x x x

−= + + + +

= − +

21. ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

1

2

3

4

54 4

65 5

1 1 0 11

1 0 1

2 1 0 2

6 1 0 6

24 1 0 24

120 1 0 120

f x x fx

f x x f

f x x f

f x x f

f x x f

f x x f

−

−

−

−

−

−

= = + =+

′ ′= − + = −

′′ ′′= + =

′′′ ′′′= − + = −

= + =

= − + = −

( )2 3 4 5

5

2 3 4 5

2 6 24 12012! 3! 4! 5!

1

x x x xP x x

x x x x x

= − + − + −

= − + − + −



22. ( ) ( )

( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1

2

3

4

54 4

1 1 0 01 1

1 1

1 0 1

2 1 0 2

6 1 0 6

24 1 0 24

x xf x fx x

x

f x x f

f x x f

f x x f

f x x f

−

−

−

−

−

+ −= = =

+ +

= − +

′ ′= + =

′′ ′′= − + = −

′′′ ′′′= + =

= − + = −

( ) ( ) 2 3 44

2 3 4

2 6 240 12 6 24

P x x x x x

x x x x

= + − + −

= − + −

23. ( ) ( )( ) ( )( ) ( )3 2

sec 0 1sec tan 0 0sec sec tan 0 1

f x x ff x x x ff x x x x f

= =

′ ′= =

′′ ′′= + =

( ) 2 22

1 11 0 12! 2

P x x x x= + + = +

24. ( ) ( )( ) ( )( ) ( )( ) ( )

2

2

2 2 4

tan 0 0sec 0 12 sec tan 0 04 sec tan 2 sec 0 2

f x x ff x x ff x x x ff x x x x f

= =

′ ′= =

′′ ′′= =

′′′ ′′′= + =

( ) ( ) 3 33

2 16 30 1 0P x x x x x= + + + = +

25. ( ) ( )

( ) ( )( ) ( )( ) ( )

1

2

3

4

2 2 1 2

2 1 24 1 4

12 1 12

f x x fx

f x x ff x x ff x x f

−

−

−

−

= = =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

( ) ( ) ( ) ( )

( ) ( ) ( )

2 33

2 3

4 122 2 1 1 12! 3!

2 2 1 2 1 2 1

P x x x x

x x x

= − − + − − −

= − − + − − −

26. ( ) ( )

( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

22

3

4

5

4 46

1 2 1 4

2 2 1 46 2 3 8

24 2 3 4

120 2 15 8

f x x fx

f x x ff x x ff x x f

f x x f

−

−

−

−

−

= = =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

= =

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 3 44

2 3 4

1 1 3 8 3 4 15 82 2 2 24 4 2! 3! 4!1 1 3 1 52 2 2 24 4 16 8 64

P x x x x x

x x x x

= − − + − − − + −

= − − + − − − + −

27. ( ) ( )

( ) ( )

( ) ( )

( ) ( )

1 2

1 2

3 2

5 2

4 21 142 4

1 144 32

3 348 256

f x x x f

f x x f

f x x f

f x x f

−

−

−

= = =

′ ′= =

′′ ′′= − = −

′′′ ′′′= =

( ) ( ) ( ) ( )

( ) ( ) ( )

2 33

2 3

1 1 32 3 2562 4 4 44 2! 3!1 1 12 4 4 44 64 512

P x x x x

x x x

= + − − − + −

= + − − − + −

28. ( ) ( )

( ) ( )

( ) ( )

( ) ( )

1 3

2 3

5 3

8 38

8 21 183 12

2 189 144

10 10 1 5827 27 2 3456

f x x f

f x x f

f x x f

f x x f

−

−

−

= =

′ ′= =

′′ ′′= − = −

′′′ ′′′= = ⋅ =

( ) ( ) ( ) ( )2 33

1 1 52 8 8 812 288 20,736

P x x x x= + − − − + −



29. ( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( ) ( ) ( )

1

2

3

4 44

ln 2 ln 21 2 1 2

2 1 42 2 1 4

6 2 3 8

f x x f

f x x fx

f x x ff x x f

f x x f

−

−

−

−

= =

′ ′= = =

′′ ′′= − = −

′′′ ′′′= =

= − = −

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 3 44

2 3 4

1 1 4 1 4 3 8ln 2 2 2 2 22 2! 3! 4!1 1 1 1ln 2 2 2 2 22 8 24 64

P x x x x x

x x x x

= + − − − + − − −

= + − − − + − − −

30. ( ) ( )( ) ( )( ) ( )

2 2

2

2 2

coscos sin 22 cos 4 sin cos 2

f x x x ff x x x x ff x x x x x x f

π ππ ππ π

= = −

′ ′= − = −

′′ ′′= − − = − +

( ) ( ) ( )( )2

222

22

2P x x x

ππ π π π

−= − − − + −

31. (a) ( )3

33 3

P x x xππ= +

(b) ( )2 3

2 33

1 1 8 11 2 24 4 3 4

Q x x x xπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

32. (a) ( ) 2 3 4 2 44

2 0 241 0 12! 3! 4!

P x x x x x x x−= + + + + = − +

(b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 4 2 44

1 1 1 2 0 3 1 1 1 11 1 1 1 1 1 12 2 2! 3! 4! 2 2 4 8

Q x x x x x x x x−⎛ ⎞= + − − + − + − + − = − − + − − −⎜ ⎟⎝ ⎠

33. ( )( )( )

( )

1

33

3 55

161 16 120

sinf x x

P x x

P x x x

P x x x x

=

=

= −

= − +

(a)

x 0.00 0.25 0.50 0.75 1.00

sin x 0.0000 0.2474 0.4794 0.6816 0.8415

( )1P x 0.0000 0.2500 0.5000 0.7500 1.0000

( )3P x 0.0000 0.2474 0.4792 0.6797 0.8333

( )5P x 0.0000 0.2474 0.4794 0.6817 0.8417

f

P3Q3

−0.5 0.5

−4

4

−3 3

−2

f

Q

P4

4P2

2



(b)

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

34. (a) ( ) ( )( ) ( )

11

x

x

f x e f ef x e f e

= =

′ ′= =

( ) ( ) ( ) ( )4 xf x f x f x e′′ ′′′= = = and ( ) ( ) ( ) ( )41 1 1f f f e′′ ′′′= = =

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1

22

2 3 44

1

1 12

1 1 1 12 6 24

P x e e x

eP x e e x x

e e eP x e e x x x x

= + −

= + − + −

= + − + − + − + −

(b)

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

35. ( ) arcsinf x x=

(a) ( )3

3 6xP x x= +

(b) (c)

x 1.00 1.25 1.50 1.75 2.00 xe e 3.4903 4.4817 5.7546 7.3891

( )1P x e 3.3979 4.0774 4.7570 5.4366

( )2P x e 3.4828 4.4172 5.5215 6.7957

( )4P x e 3.4903 4.4809 5.7485 7.3620

x –0.75 –0.50 –0.25 0 0.25 0.50 0.75

( )f x –0.848 –0.524 –0.253 0 0.253 0.524 0.848

( )3P x –0.820 –0.521 –0.253 0 0.253 0.521 0.820

−3

P5

P3 P1f

2�−2�

3

−6 6

−1

7

y = ex

P1P2P4

1−1

2f

x

P3

π

2π

y

−



36. (a) ( ) arctanf x x=

( )3

3 3xP x x= −

(b) (c)

37. ( ) cosf x x=

38. ( ) arctanf x x=

39. ( ) ( )2ln 1f x x= +

40. ( ) 2 44 xf x xe−=

41. ( )

( )

3 2 3 49 9 272 2 8

12

1 3

4.3984

xf x e x x x x

f

= ≈ + + + +

≈

42. ( )

( )

2 2 3 412

15 0.0328

xf x x e x x x

f

−= ≈ − +

≈

43. ( ) ( ) ( ) ( ) ( ) ( )( )

2 3 41 1 1 12 8 24 64ln ln 2 2 2 2 2

2.1 0.7419

f x x x x x x

f

= ≈ + − − − + − − −

≈

44. ( ) ( ) ( )2

22 2 2cos 22

7 6.79548

f x x x x x

f

ππ π π π

π

⎛ ⎞−= ≈ − − − + −⎜ ⎟

⎝ ⎠

⎛ ⎞ ≈ −⎜ ⎟⎝ ⎠

x –0.75 –0.50 –0.25 0 0.25 0.50 0.75

( )f x –0.6435 –0.4636 –0.2450 0 0.2450 0.4636 0.6435

( )3P x –0.6094 –0.4583 –0.2448 0 0.2448 0.4583 0.6094

x

f

1−1 12

π

π

4

4

π2

π2

−

−

P3

y

6

6 8

4

2

−4

−6

−6x

P6 P2

P8 P4y

f (x) = cos x

−2

2

1

x

f x x( ) = arctan

1 3−3 −2

PP33PP11PP13P9

yPP7 PP1P5

3

2

−2 2 3 4−3−4

−3

x

P6P2

P8P4

y

f (x) = ln (x2 + 1)

x

2

4

−4 4

yy = 4xe(−x2/4)



45. ( ) ( ) ( )5cos ; sin Maxf x x f x x= = − ⇒ on [0, 0.3] is 1.

( ) ( )5 54

1 0.3 2.025 105!

R x −≤ = ×

Note: you could use ( ) ( ) ( )65 : cos ,R x f x x= − max on [0, 0.3] is 1.

( ) ( )6 65

1 0.3 1.0125 106!

R x −≤ = ×

Exact error: 60.000001 1.0 10−= ×

46. ( ) ( ) ( )6; Maxx xf x e f x e= = ⇒ on [0, 1] is 1.e

( ) ( )1

6 35 1 0.00378 3.78 10

6!eR x −≤ ≈ = ×

47. ( ) ( ) ( ) ( )( )

24

7 22

6 9arcsin ; Max

1

x xf x x f x

x

+= = ⇒

−on [0, 0.4] is ( ) ( )4 0.4 7.3340.f ≈

( ) ( )4 33

7.3340 0.4 0.00782 7.82 10 .4!

R x −≤ ≈ = × The exact error is 48.5 10 .−× [Note: You could use 4.]R

48. ( ) ( ) ( ) ( )( )

( ) ( )

24

42

4

24 1arctan ;

1

Max on [0, 0.4] is 0.4 22.3672.

x xf x x f x

x

f

+= =

−

⇒ ≈

( ) ( )43

22.3672 0.4 0.02394!

R x ≤ ≈

49. ( ) sing x x=

( ) ( )1 1ng x+ ≤ for all x.

( ) ( ) ( ) 11 0.3 0.0011 !

nnR x

n+≤ <

+

By trial and error, 3.n =

50. ( ) cosf x x=

( ) ( )1 1nf x+ ≤ for all x and all n.

( )( ) ( )( )

( )( )

1 1

1

1 !

0.10.001

1 !

n n

n

n

f z xR x

n

n

+ +

+

=+

≤ <+


51. ( )( ) ( )1

x

n x

f x e

f x e+

=

=

Max on [0, 0.6] is 0.6 1.8221.e ≈

( ) ( ) 11.8221 0.6 0.001

1 !n

nRn

+≤ <+


52. ( ) ( ) ( )( ) ( ) ( )

1 2

11

ln , , ,

!1 nnn

f x x f x x f x x

nf xx

− −

++

′ ′′= = = −

= −

…

The maximum value of ( ) ( )1nf x+ on [1, 1.25] is n!

( ) ( )

( )

1

1

! 0.25 0.0011 !

0.250.001

1

nn

n

nRn

n

+

+

≤ <+

<+

By trial and error, 3n =

53. ( ) ( )ln 1f x x= +

( ) ( ) ( )( )

11

1 !Max

1

nn

n

nf x

x+

+

−= ⇒

+on [0, 0.5] is n!.

( ) ( ) ( ) 1

1 0.5! 0.5 0.00011 ! 1

nn

nnR

n n

++≤ = <

+ +

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



54. ( ) ( )2cosf x xπ=

( )

( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( )

2 4 6

2

2 4 62 2 2

2 4 4 8 6 12

2 4 64 8 12

cos 12! 4! 6!

12! 4! 6!

12! 4! 6!

0.6 1 0.6 0.6 0.62! 4! 6!

x x xg x x

f x g x

x x x

x x x

f

π

π π π

π π π

π π π

= = − + − +

=

= − + − +

= − + − +

= − + − +

Because this is an alternating series,

( ) ( )

24

1 0.6 0.0001.2 !

nn

n nR an

π+≤ = <

By trial and error, 4.n = Using 4 terms ( )0.6 0.4257.f ≈

55. ( ) ( ), 1.3xf x e fπ−=

( ) ( ) xf x e ππ −′ = −

( ) ( ) ( ) ( )1 11 n nn xf x e ππ π+ ++ −= − ≤ − on [0, 1.3]

( )( ) ( )

111.3 0.0001

1 !

nn

nRnπ +

+≤ <+

By trial and error, 16.n = Using 16 terms, ( )1.3 0.01684.e π− ≈

56. ( ) xf x e−=

( ) xf x e−′ = −

( ) ( ) ( ) 11 1 1nn xf x e++ −= − ≤ on [0, 1]

( )

11 1 0.00011 !

nnR

n+≤ ≤

+


57. ( )

( )

2 3

43

4

4

4

1 , 02 6

0.0014!0.024

0.3936

0.3936 0.3936, 0

0.3936 0

x

z

z

z

z

x xf x e x x

eR x x

e x

xe

x ze

x

= ≈ + + + <

= <

<

<

< < <

− < <

58. ( )

( )

3

44

3

4

sin3!

sin 0.0014! 4!

0.024

0.3936

0.3936 0.3936

xf x x x

xzR x x

x

x

x

= ≈ −

= ≤ <

<

<

− < <

59. ( )2 4

cos 1 ,2! 4!x xf x x= ≈ − + fifth degree polynomial

( ) ( )1 1nf x+ ≤ for all x and all n.

( ) 65

6

1 0.0016!

0.72

0.9467

0.9467 0.9467

R x x

x

x

x

≤ <

<

<

− < <

Note: Use a graphing utility to graph ( )2 4cos 1 2 24y x x x= − − + in the viewing

window [ ] [ ]0.9467, 0.9467 0.001, 0.001− × − to verify

the answer.

60. ( ) 2 2 341 2 23

xf x e x x x−= ≈ − + −

( ) ( )2 22 , 4 ,x xf x e f x e− −′ ′′= − =

( ) ( ) ( )42 28 , 16x xf x e f x e− −′′′ = − =

( ) ( )( )4 2

4 4 2 43

2 4

16 20 0.0014! 24 3

0.0015

zz

z

f z eR x x x e x

e x

−−

−

= − = = <

<

1 4

22

0.0015 0.1970 0.1970,zzx e

e−⎛ ⎞< ≈ <⎜ ⎟⎝ ⎠

for 0.z <

So, 0 0.1970.x< <

In fact, by graphing ( ) 2xf x e−= and

2 341 2 2 ,3

y x x x= − + − you can verify that

( ) 0.001f x y− < on ( )0.19294, 0.20068 .−

61. The graph of the approximating polynomial P and the elementary function f both pass through the point

( )( ),c f c and the slopes of P and f agree at

( )( ), .c f c Depending on the degree of P, the nth

derivatives of P and f agree at ( )( ), .c f c

62. ( ) ( ) ( ) ( )2 2, ,f c P c f c P c′′= = and ( ) ( )2f c P c′′′′ =

63. See definition on page 652.




65. As the degree of the polynomial increases, the graph of the Taylor polynomial becomes a better and better approximation of the function within the interval of convergence. Therefore, the accuracy is increased.

66.

67. (a) ( )

( )

( )

( )

( ) ( )

2 3 44

2 3 4 55

5 4

1 1 112 6 24

1 1 12 6 24

x

x

f x e

P x x x x x

g x xe

Q x x x x x x

Q x x P x

=

= + + + +

=

= + + + +

=

(b) ( )

( )

( )

( ) ( )

3 5

5

4 62

6 5

sin

3! 5!sin

3! 5!

f x x

x xP x x

g x x x

x xQ x x P x x

=

= − +

=

= = − +

(c) ( ) ( )2 4

5sin 1 1

3! 5!x x xg x P x

x x= = = − +

68. (a) ( )3 5

5 3! 5!x xP x x= − + for ( ) sinf x x=

( )2 4

5 12! 4!x xP x′ = − +

This is the Maclaurin polynomial of degree 4 for ( ) cos .g x x=

(b) ( )2 4 6

6 12 4! 6!x x xQ x = − + − for cos x

( ) ( )3 5

6 53! 5!x xQ x x P x′ = − + − = −

(c) ( )2 3 4

12! 3! 4!x x xR x x= + + + +

( )2 3

12! 3!x xR x x′ = + + +

The first four terms are the same!

69. (a) ( ) ( )22

22

132

xQ x

π += − +

(b) ( ) ( )22

26

132

xR x

π −= − +

(c) No. The polynomial will be linear. Horizontal translations of the result in part (a) are possible only at 2 8x n= − + (where n is an integer) because the period of f is 8.

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

( ) ( ) ( )

( ) ( )

( )( ) ( ) ( )

0

0

0

lim

lim

lim .

h

h

h

f x h f xf x

hf x h f x

hf x h f x

f xh

→

→

→

− + − −′ − =

− − +=

+ − −′= =

−

Similarly, f ′′ is odd, f ′′′ is even, etc. Therefore, ( )4, , ,f f f′′ etc. are all odd functions, which implies that

( ) ( )0 0 0.f f ′′= = = So, in the formula

( ) ( ) ( ) ( ) 200 0

2!nf x

P x f f x′′

′= + + + all the

coefficients of the even power of x are zero.

71. Let f be an even function and nP be the nth Maclaurin polynomial for f. Because f is even, f ′ is odd, f ′′ is even, f ′′′ is odd, etc. All of the odd derivatives of f are odd and so, all of the odd powers of x will have coefficients of zero. nP will only have terms with even powers of x.

72. Let

( ) ( ) ( ) ( )20 1 2

nn nP x a a x c a x c a x c= + − + − + + −

where ( ) ( ).

!

i

if c

ai

=

( ) ( )0nP c a f c= =

For

( ) ( )( ) ( ) ( ) ( )1 , ! ! .

!

kk k

n nf c

k n P c a k k f ck

⎛ ⎞≤ ≤ = = =⎜ ⎟⎜ ⎟

⎝ ⎠

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

x20−20 10

2

−2

−4

4

6

8

10 f

y

P1

P3

P2

Section 9.8 Power Series 339


Section 9.8 Power Series

1. Centered at 0

2. Centered at 0

3. Centered at 2

4. Centered at π

5. ( )0

11

nn

n

xn

∞

=

−+∑

( )( )

1 11 1 1lim lim

2 1

1lim2

n nn

n nn nn

n

xu nLu n x

n x xn

+ ++

→∞ →∞

→∞

− += = ⋅

+ −

+= =

+

1 1x R< ⇒ =

6. ( )0

4 n

nx

∞

=∑

( )( )

11 4

lim lim lim 4 44

nn

nn n nn

xuL x xu x

++

→∞ →∞ →∞= = = =

14 14

x R< ⇒ =

7. ( )2

1

4 n

n

xn

∞

=∑

( ) ( )( ) ( )

( )

1

1 2 2

22

lim

4 1lim lim 4 4

4 1

n

n n

n

nn n

uLu

x n n x xx n n

+

→∞

+

→∞ →∞

=

+= = =

+

14 14

x R< ⇒ =

8. ( )0

15

n n

nn

x∞

=

−∑

( )( )

1 1 11 1 5

lim lim lim5 51 5

n n nn

n n nn n nn

x xxuLu x

+ + ++

→∞ →∞ →∞

−= = = =

−

1 55x

R< ⇒ =

9. ( )

2

0 2 !

n

n

xn

∞

=∑

( ) ( )( )

( )( )

1

2 2

2

2

lim

2 2 !lim

2 !

lim 02 2 2 1

n

n n

n

nn

n

uLu

x nx n

xn n

+

→∞

+

→∞

→∞

=

+=

= =+ +

So, the series converges for all .x R⇒ = ∞

10. ( ) 2

0

2 !!

n

n

n xn

∞

=∑

( ) ( )( )

( )( )( )

2 2

2

2

2 2 ! 1 !1lim lim2 ! !

2 2 2 1lim

1

nn

nn nn

n

n x nuLu n x n

n n xn

+

→∞ →∞

→∞

+ ++= =

+ += = ∞

+

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

11. 0 4

n

n

x∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Because the series is geometric, it converges only if

1,4x

< or 4 4.x− < <

12. 0 7

n

n

x∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Because the series is geometric, it converges only if

1,7x

< or 7 7.x− < <

13. ( )1

1 n n

n

xn

∞

=

−∑

( )( )

1 11 1

lim lim1 1

lim1

n nn

n nn nn

n

xu nu n x

nx xn

+ ++

→∞ →∞

→∞

−= ⋅

+ −

= =+

Interval: 1 1x− < <

When 1,x = the alternating series ( )1

1 n

n n

∞

=

−∑ converges.

When 1,x = − the p-series 1

1

n n

∞

=∑ diverges.

Therefore, the interval of convergence is ( ]1, 1 .−



14. ( ) ( )1

01 1n n

nn x

∞+

=

− +∑

( ) ( )( ) ( )

( )

2 11 21lim lim1 1

2lim

1

n nn

n nn nn

n

n xuu n x

n xx

n

+ +

→∞ →∞

→∞

− ++=

− +

+= =

+


When 1,x = the series ( ) ( )1

01 1n

nn

∞+

=

− +∑ diverges.

When 1,x = − the series ( )0

1n

n∞

=

− +∑ diverges.

Therefore, the interval of convergence is ( )1, 1 .−

15. 5

0 !

n

n

xn

∞

=∑

( ) ( )5 1 5

1 1 !lim lim lim 0

5 ! 1

nn

nn n nn

x nu xu n n

++

→∞ →∞ →∞

+= = =

+

The series converges for all x. The interval of convergence is ( ), .−∞ ∞

16. ( )( )0

32 !

n

n

xn

∞

=∑

( )( )

( )( )

( )( )

11 3 2 !

lim lim2 1 ! 3

3lim 02 2 2 1

nn

nn nn

n

x nuu n x

xn n

++

→∞ →∞

→∞

= ⋅+

= =+ +

Therefore, the interval of convergence is ( ), .−∞ ∞

17. ( )0

2 !3

n

n

xn∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

( ) ( )( ) ( )

( )( )

11 2 2 ! 3

lim lim2 ! 3

2 2 2 13

nn

nn nn

n xuu n x

n n x

++

→∞ →∞

+=

+ += = ∞

The series converges only for 0.x =

18. ( )( )( )0

11 2

n n

n

xn n

∞

=

−+ +∑

( )( )( )

( )( )( )

( )1 11 1 1 2 1

lim lim lim2 3 31

n nn

n nn n nn

x n n n xu xu n n nx

+ ++

→∞ →∞ →∞

− + + += ⋅ = =

+ + +−


When 1,x = the alternating series ( )( )( )0

11 2

n

n n n

∞

=

−+ +∑ converges.

When 1,x = − the series ( )( )0

11 2n n n

∞

= + +∑ converges by limit comparison to 21

1 .n n

∞

=∑

Therefore, the interval of convergence is [ ]1, 1 .−

19. ( ) 1

1

14

n n

nn

x+∞

=

−∑

Because the series is geometric, it converges only if 4 1x < or 4 4.x− < <

20. ( ) ( )0

1 ! 53

n n

nn

n x∞

=

− −∑

( ) ( ) ( )( ) ( )

( )( )1 1 11 1 1 ! 5 3 1 5

lim lim lim31 ! 5 3

n n nn

n n nn n nn

n x n xuu n x

+ + ++

→∞ →∞ →∞

− + − + −= = = ∞

− −

The series converges only for 5.x =



21. ( ) ( )1

1

1 49

n n

nn

xn

+∞

=

− −∑

( ) ( ) ( )( )

( ) ( ) ( )( )

2 1 11 1 4 1 9

lim lim1 4 9

4 1lim 41 9 9

n n nn

n n nn nn

n

x nuu x n

xn xn

+ + ++

→∞ →∞

→∞

− − +=

− −

−= = −

+

9R =


When ( ) ( )1 1

1 1

1 9 113,

9

n nn

nn n

xn n

+ +∞ ∞

= =

− −= =∑ ∑ converges.

When ( ) ( )1

1 1

1 9 15,9

n n

nn n

xn n

+∞ ∞

= =

− − −= − =∑ ∑ diverges.

Therefore, the interval of convergence is ( ]5, 13 .−

22. ( )( )

1

10

31 4

n

nn

xn

+∞

+=

−+∑

( ) ( )( ) ( )

( )( )( )

2 21

1 1

3 2 4lim lim

3 1 4

3 1 3lim4 2 4

n nn

n nn nn

n

x nuu x n

x n xn

+ ++

+ +→∞ →∞

→∞

⎡ ⎤− +⎣ ⎦=⎡ ⎤− +⎣ ⎦

− + −= =

+

4R = Interval: 1 7x− < <

When ( )

1

10 0

4 17,1 4 1

n

nn n

xn n

+∞ ∞

+= =

= =+ +∑ ∑ diverges.

When ( )( )

( )( )

1 1

10 0

4 11,

1 4 1

n n

nn n

xn n

+ +∞ ∞

+= =

− −= − =

+ +∑ ∑ converges.

Therefore, the interval of convergence is [ )1, 7 .−

23. ( ) ( )

( ) ( )( ) ( )

( )( )

1 1

0

2 21

1 1

1 11

1 1 1 11lim lim lim 12 21 1

1

n n

n

n nn

n nn n nn

xn

x n xu n xu n nx

R

+ +∞

=

+ ++

+ +→∞ →∞ →∞

− −+

− − + −+= ⋅ = = −

+ +− −

=

∑

Center: 1x =

Interval: 1 1 1x− < − < or 0 2x< <

When 0,x = the series 0

11n n

∞

= +∑ diverges by the integral test.

When 2,x = the alternating series ( ) 1

0

11

n

n n

+∞

=

−+∑ converges.

Therefore, the interval of convergence is ( ]0, 2 .

24. ( ) ( )1

1

1 22

n n

nn

xn

+∞

=

− −∑

( ) ( )( )

( ) ( )2 1 11

1

1 2 1 2 2 2lim lim lim1 2 2 2 1 2

n n n nn

n nn n nn

x xa x n xn na n

+ + ++

+→∞ →∞ →∞

− − − − − −= = =

+ +

2 1 2 2 2 0 42

x x x−< ⇒ − < − < ⇒ < <

when 0,x =

( ) ( ) ( )( ) ( )1

1 1 1

1 21 2 1diverges.

2 2

n n n

n nn n nn n n

+∞ ∞ ∞

= = =

−− − −= =∑ ∑ ∑

when 4,x =

( ) ( )1 1

1 1

1 2 1converges.

2

n n

nn n

nn n

+ +∞ ∞

= =

− −=∑ ∑

Therefore the interval of convergence is ( ]0, 4 .



25. 1

1

33

n

n

x −∞

=

−⎛ ⎞⎜ ⎟⎝ ⎠

∑ is geometric. It converges if

3 1 3 3 0 6.3

x x x−< ⇒ − < ⇒ < <

Therefore, the interval of convergence is ( )0, 6 .

26. ( ) 2 1

0

12 1

n n

n

xn

+∞

=

−+∑

( )( )

( )( )

( )( )

1 2 31

2 1

2 2

1 2 1lim lim

2 3 1

2 1lim

2 3

n nn

n nn nn

n

x nuu n x

nx x

n

+ ++

+→∞ →∞

→∞

− += ⋅

+ −

+= =

+

1R = Interval: 1 1x− < <

When ( )0

11,

2 1

n

nx

n

∞

=

−=

+∑ converges.

When ( ) 1

0

11,

2 1

n

nx

n

+∞

=

−= −

+∑ converges.

Therefore, the interval of convergence is [ ]1, 1 .−

27. ( ) 1

12

1n

n

n xn

∞−

=

−+∑

( )( )( )

( )( )( )

11

2

1 2 1lim lim2 2

2 1lim 2

2

nn

nn nn

n

n xu nu n n x

x nx

n n

+−→∞ →∞

→∞

+ − += ⋅

+ −

− += =

+

12

R =

Interval: 1 12 2

x− < <

When 1,2

x = − the series 1 1n

nn

∞

= +∑ diverges by the nth

Term Test.

When 1,2

x = the alternating series

( ) 1

1

11

n

n

nn

−∞

=

−+∑ diverges.

Therefore, the interval of convergence is 1 1, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

28. ( ) 2

0

1!

n n

n

xn

∞

=

−∑

( )( ) ( )

1 2 21

2

2

1 !lim lim1 ! 1

lim 01

n nn

n nn nn

n

xu nu n x

xn

+ ++

→∞ →∞

→∞

−= ⋅

+ −

= =+


29. ( )

3 1

0 3 1 !

n

n

xn

+∞

= +∑

( )( )

( )( )( )

3 41

3 1

3

3 4 !lim lim

3 1 !

lim 03 4 3 3 3 2

nn

nn nn

n

x nuu x n

xn n n

++

+→∞ →∞

→∞

+=

+

= =+ + +


30. ( )1

!2 !

n

n

n xn

∞

=∑

( )( )

( )

( )( )( )

11 1 ! 2 !

lim lim2 2 ! !

1lim 0

2 2 2 1

nn

nn nn

n

n x nuu n n x

n xn n

++

→∞ →∞

→∞

+= ⋅

+

+= =

+ +


31. ( ) ( )1 1

2 3 4 11

!

nn

n n

n xn x

n

∞ ∞

= =

⋅ ⋅ += +∑ ∑

( )( )

11 2 2lim lim lim

1 1

nn

nn n nn

n xa n x xa n x n

++

→∞ →∞ →∞

+ += = =

+ +

Converges if 1 1 1.x x< ⇒ − < <

At 1,x = ± diverges.

Therefore the interval of convergence is ( )1, 1 .−



32. ( )( )( )2 1

1

2 4 6 23 5 7 2 1

n

n

nx

n

∞+

=

⋅ ⋅⋅ ⋅ +∑

( )( )( )( )

( )( )

( )( )

2 3 21 2

2 1

2 4 2 2 2 3 5 2 1 2 2lim lim lim

3 5 7 2 1 2 3 2 4 2 2 3

1

nn

nn n nn

n n x n n xu xu n n n x n

R

++

+→∞ →∞ →∞

⋅ + ⋅ + += ⋅ = =

⋅ ⋅ + + ⋅ +

=

When 1,x = ± the series diverges by comparing it to

1

12 1n n

∞

= +∑

which diverges. Therefore, the interval of convergence is ( )1, 1 .−

33. ( ) ( )( )1

1

1 3 7 11 4 1 34

n n

nn

n x+∞

=

− ⋅ ⋅ − −∑

( ) ( )( )( )( ) ( )( )

( )( )

2 11

11

1 3 7 11 4 1 4 3 3 4lim lim4 1 3 7 11 4 1 3

4 3 3lim

4

n n nn

n nnn nn

n

n n xuu n x

n x

+ ++

++→∞ →∞

→∞

− ⋅ ⋅ ⋅ − + −= ⋅

− ⋅ ⋅ ⋅ − −

+ −= = ∞

0R = Center: 3x = Therefore, the series converges only for 3.x =

34. ( )( )1

! 11 3 5 2 1

n

n

n xn

∞

=

+⋅ ⋅ −∑

( ) ( )( )( )

( ) ( )( )

( )( )11 1 11 ! 1 ! 1 1lim lim lim 1

1 3 5 2 1 2 1 1 3 5 2 1 2 1 2

n nn

n n nn

n xn x n xa xn n na n

++

→∞ →∞ →∞

+ ++ + += = = +

⋅ ⋅ − + ⋅ ⋅ − +

Converges if 1 1 1 2 1 2 3 1.2

x x x+ < ⇒ − < + < ⇒ − < <

At ( ) ( )

!2 2 4 6 21, 1,1 3 5 2 1 1 3 5 2 1

n

nn nx a

n n⋅ ⋅

= = = >⋅ ⋅ − ⋅ ⋅ −

diverges.

At ( )( ) ( ) ( )

! 2 2 4 23, 1 ,1 3 2 1 1 3 2 1

nn

nn nx a

n n− ⋅

= − = = −⋅ − ⋅ −

diverges.

Therefore, the interval of convergence is ( )3, 1 .−

35. ( )

( )( )

1

11

11

11lim lim

n

nn

n nn

nnn nn

x cc

x cu c x cu c cx c

−∞

−=

−+

−→∞ →∞

−

−= ⋅ = −

−

∑

R c= Center: x c= Interval: c x c c− < − < or 0 2x c< <

When 0,x = the series ( ) 1

11 n

n

∞−

=

−∑ diverges.

When 2 ,x c= the series 1

1n

∞

=∑ diverges.

Therefore, the interval of convergence is ( )0, 2 .c



36. ( )( )0

!,

!

k n

n

n xk

kn

∞

=∑ is a positive integer.

( )

( )( )( )

( )( )( ) ( )

11 1 ! 1!

lim lim lim!1 ! 1 1

k kkn nn

kn n nn

n x xn xn xaknk na k nk k nk nk k

++

→∞ →∞ →∞

⎡ + ⎤ +⎣ ⎦= = =⎡ + ⎤ + − + +⎣ ⎦

Converges if 1 .kk

xR k

k< ⇒ =

37. 0

n

n

xk

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Because the series is geometric, it converges only if 1x k < or .k x k− < <

Therefore, the interval of convergence is ( ), .k k−

38. ( ) ( )

( ) ( )( ) ( ) ( )

( )( )

1

1

2 11

11

1

1 1lim lim lim1 11

n n

nn

n n nn

n nnn n nn

x cnc

x c n x cu nc x cu n c c n cx c

R c

+∞

=

+ ++

++→∞ →∞ →∞

− −

− − −= ⋅ = = −

+ +− −

=

∑

Center: x c=

Interval: c x c c− < − < or 0 2x c< <

When 0,x = the p-series 1

1

n n

∞

=

−∑ diverges.

When 2 ,x c= the alternating series ( ) 1

1

1 n

n n

+∞

=

−∑ converges.

Therefore, the interval of convergence is ( )0, 2 .c

39. ( ) ( )

( ) ( )( )( ) ( ) ( )

( )1

11

1 1!

1 1 !lim lim lim1 ! 1 1 1

1

n

n

nn

nn n nn

k k k n xn

k k k n k n x k n xu n xu n k k k n x n

R

∞

=

++

→∞ →∞ →∞

+ + −

+ + − + += ⋅ = =

+ + + − +

=

∑

When 1,x = ± the series diverges and the interval of convergence is ( )1, 1 .−

( ) ( )1 11

1 2k k k n

n⎡ + + − ⎤

≥⎢ ⎥⋅⎣ ⎦

40. ( )( )

( ) ( )( )( )

( )( )

( )( )1

11

!1 3 5 2 1

1 ! 1 3 5 2 1 1 1lim lim lim1 3 5 2 1 2 1 ! 2 1 2

2

n

n

nn

n n nn

n x cn

n x c n n x cu x cu n n n x c n

R

∞

=

++

→∞ →∞ →∞

−⋅ ⋅ −

+ − ⋅ ⋅ − + −= ⋅ = = −

⋅ ⋅ − + − +

=

∑

Interval: 2 2x c− < − < or 2 2c x c− < < +

The series diverges at the endpoints. Therefore, the interval of convergence is ( )2, 2 .c c− +

( )( ) ( )

( )( )

2! 2 2 4 6 2!2 11 3 5 2 1 1 3 5 2 1 1 3 5 2 1

nn c c nnn n n

⎡ ⎤+ − ⋅ ⋅⎢ ⎥= = >

⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ −⎢ ⎥⎣ ⎦



41. ( )

2 1

0 11

! 1 2 1 !

n n

n n

x x x xn n

−∞ ∞

= =

= + + + =−∑ ∑

42. ( ) ( ) ( ) ( )1 1

0 11 1 1n nn n

n nn x n x

∞ ∞+ −

= =

− + = −∑ ∑

43. ( ) ( )

2 1 2 1

0 12 1 ! 2 1 !

n n

n n

x xn n

+ −∞ ∞

= =

=+ −∑ ∑

Replace n with 1.n −

44. ( ) ( ) 12 1 2 1

0 1

1 12 1 2 1

n nn n

n n

x xn n

−+ −∞ ∞

= =

− −=

+ −∑ ∑

Replace n with 1.n −

45. (a) ( ) ( ) ( )0

, 3, 3 Geometric3

n

n

xf x∞

=

⎛ ⎞= −⎜ ⎟⎝ ⎠

∑

(b) ( ) ( )1

1, 3, 3

3 3

n

n

n xf x−∞

=

⎛ ⎞′ = −⎜ ⎟⎝ ⎠

∑

(c) ( ) ( ) ( )2

2

1, 3, 3

9 3

n

n

n n xf x−∞

=

− ⎛ ⎞′′ = −⎜ ⎟⎝ ⎠

∑

(d) ( ) [ )

( )

1

0

11

3 , 3, 31 3

1 33 3 , converges1 3 1

n

n

nn

xf x dxn

n n

+∞

=

++

⎛ ⎞= −⎜ ⎟+ ⎝ ⎠

⎡ ⎤−−⎛ ⎞⎢ ⎥=⎜ ⎟+ +⎝ ⎠⎢ ⎥⎣ ⎦

∑∫

∑ ∑

46. (a) ( ) ( ) ( ) ( ]1

1

1 5, 0, 10

5

n n

nn

xf x

n

+∞

=

− −= ∑

(b) ( ) ( ) ( ) ( )1 1

1

1 5, 0, 10

5

n n

nn

xf x

+ −∞

=

− −′ = ∑

(c) ( ) ( ) ( )( ) ( )1 2

2

1 1 5, 0, 10

5

n n

nn

n xf x

+ −∞

=

− − −′′ = ∑

(d) ( ) ( ) ( )( ) [ ]

1 1

1

1 5, 0, 10

1 5

n n

nn

xf x dx

n n

+ +∞

=

− −=

+∑∫

47. (a) ( ) ( ) ( ) ( ]1 1

0

1 1, 0, 2

1

n n

n

xf x

n

+ +∞

=

− −=

+∑

(b) ( ) ( ) ( ) ( )1

01 1 , 0, 2n n

nf x x

∞+

=

′ = − −∑

(c) ( ) ( ) ( ) ( )1 1

11 1 , 0, 2n n

nf x n x

∞+ −

=

′′ = − −∑

(d) ( ) ( ) ( )( )( ) [ ]

1 2

1

1 1, 0, 2

1 2

n n

n

xf x dx

n n

+ +∞

=

− −=

+ +∑∫

48. (a) ( ) ( ) ( ) ( ]1

1

1 2, 1, 3

n n

n

xf x

n

+∞

=

− −= ∑

(b) ( ) ( ) ( ) ( )1 1

11 2 , 1, 3n n

nf x x

∞+ −

=

′ = − −∑

(c) ( ) ( ) ( )( ) ( )1 2

21 1 2 , 1, 3n n

nf x n x

∞+ −

=

′′ = − − −∑

(d) ( ) ( ) ( )( ) [ ]1 1

1

1 2, 1, 3

1

n n

n

xf x dx

n n

+ +∞

=

− −=

+∑∫

49. ( ) ( )0

1 2

1 1 13 3 91 1

1, 1.33. Matches (c).

n

ng

S S

∞

=

= = + + +

= =

∑

50. ( ) ( )0

1 2

2 2 43 3 92 1

1, 1.67. Matches (a).

n

ng

S S

∞

=

= = + + +

= =

∑

51. ( ) ( )0

3.133.1

n

ng

∞

=

= ∑ diverges. Matches (b).

52. ( ) ( )0

232

n

ng

∞

=

− = −∑ alternating. Matches (d).

53. ( ) ( ) ( )0 0

1 1 1 1 18 8 4 4 162 1 ,

n n

n ng

∞ ∞

= =

⎡ ⎤= = = + + + ⋅ ⋅ ⋅⎣ ⎦∑ ∑

converges

1 2 31, 1.25, 1.3125S S S= = = Matches (b).

54. ( ) ( ) ( )0 0

1 1 1 1 18 8 4 4 162 1 ,

n n

n ng

∞ ∞

= =

⎡ ⎤− = − = − = − + −⎣ ⎦∑ ∑

converges

1 2 31, 0.75, 0.8125S S S= = = Matches (c).

55. ( ) ( ) ( )0 0

9 9 916 16 82 ,

n n

n ng

∞ ∞

= =

⎡ ⎤= =⎣ ⎦∑ ∑ diverges

1 21781,S S= = Matches (d).

56. ( ) ( ) ( )0 0

3 3 38 8 42 ,

n n

n ng

∞ ∞

= =

⎡ ⎤= =⎣ ⎦∑ ∑ converges

1 21, 1.75S S= = Matches (a).

57. A series of the form

( ) ( ) ( )

( )

20 1 2

0

nn

n

nn

a x c a a x c a x c

a x c

∞

=

− = + − + − +

+ − +

∑

is called a power series centered at c, where c is constant.



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

0.R = Otherwise, according to Theorem 8.20, there exists a real number 0R > (radius of convergence) such that the series converges absolutely for x c R− < and

diverges for .x c R− >

59. A single point, an interval, or the entire real line.

60. You differentiate and integrate the power series term by term. The radius of convergence remains the same. However, the interval of convergence might change.

61. Answers will vary.

1

n

n

xn

∞

=∑ converges for 1 1.x− ≤ < At 1,x = − the

convergence is conditional because 1n∑ diverges.

21

n

n

xn

∞

=∑ converges for 1 1.x− ≤ ≤ At 1,x = ± the

convergence is absolute.

62. Many answers possible.

(a) 1 2

n

n

x∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ Geometric: 1 22x x< ⇒ <

(b) ( )1

1 n n

n

xn

∞

=

−∑ converges for 1 1x− < ≤

(c) ( )1

2 1 n

nx

∞

=

+∑ Geometric:

2 1 1 1 0x x+ < ⇒ − < <

(d) ( )1

24

n

nn

xn

∞

=

−∑ converges for 2 6x− ≤ <

63. (a) ( ) ( )2 1

0 2 1 !

n

n

xf xn

+∞

=

=+∑

( )

( )

( )( )

2 31

2 1

2

2 1 !lim lim

2 3 !

lim 02 2 2 3

nn

nn nn

n

nu xu n x

xn n

++

+→∞ →∞

→∞

+= ⋅

+

= =+ +

Therefore, the interval of convergence is ( ).−∞ ∞

( ) ( )( )

2

0

12 !

n n

n

xg x

n

∞

=

−= ∑

( )( )

( )( )

1 2 21

2

1 2 !lim lim

2 2 ! 1

1lim 02 2

n nn

n nn nn

n

x nuu n x

n

+ ++

→∞ →∞

→∞

−= ⋅

+ −

= =+

Therefore, the interval of convergence is ( ).−∞ ∞

(b) ( ) ( )( ) ( )

2

0

12 !

n n

n

xf x g x

n

∞

=

−′ = =∑

(c) ( ) ( )( )

( )( )

( )( ) ( )

12 1 2 1

1 0

2 1

0

1 12 1 ! 2 1 !

12 1 !

n nn n

n n

n n

n

x xg x

n n

xf x

n

+− +∞ ∞

= =

+∞

=

− −′ = =

− +

−= − = −

+

∑ ∑

∑

(d) ( ) sinf x x= and

( ) cosg x x=

64. (a) ( )0 !

n

n

xf xn

∞

=

= ∑

( )

11 !lim lim

1 !

lim 01

nn

nn nn

n

u x nu n x

xn

++

→∞ →∞

→∞

= ⋅+

= =+

The series converges for all x. Therefore, the interval of convergence is ( ), .−∞ ∞

(b) ( ) ( ) ( )1 1

1 1 0! 1 ! !

n n n

n n n

nx x xf x f xn n n

− −∞ ∞ ∞

= = =

′ = = = =−∑ ∑ ∑

(c) ( )

( )

2 3 4

11

! 2! 3! 4!

0 1

n

n

x x x xf x xn

f

∞

=

= = + + + + +

=

∑

(d) ( ) xf x e=



65. ( )( )

( )( )

( ) ( )( )

( )( )

( ) ( )( )

( )( )

( )( )

( )( )

12 1 2 1

0 1

2 2

0 0

2 1 2 1

1 1

12 1 2 1

1 1

1 12 1 ! 2 1 !

1 2 1 12 1 ! 2 !

1 2 12 ! 2 1 !

1 10

2 1 ! 2 1 !

n nn n

n n

n nn n

n n

n n n

n n

n nn n

n n

x xy

n n

n x xy

n n

n x xy

n n

x xy y

n n

−+ −∞ ∞

= =

∞ ∞

= =

− −∞ ∞

= =

−− −∞ ∞

= =

− −= =

+ −

− + −′ = =

+

− −′′ = =

−

− −′′ + = + =

− −

∑ ∑

∑ ∑

∑ ∑

∑ ∑

66. ( )( )

( )( )

( ) ( )( )

( )( )

( ) ( )( )

( )( )

( )( )

( )( )

12 2 2

0 1

2 1 2 1

1 1

2 2 2 2

1 1

12 2 2 2

1 1

1 12 ! 2 2 !

1 2 12 ! 2 1 !

1 2 1 12 1 ! 2 2 !

1 10

2 2 ! 2 2 !

n nn n

n n

n nn n

n n

n nn n

n n

n nn n

n n

x xy

n n

n x xy

n n

n x xy

n n

x xy y

n n

− −∞ ∞

= =

− −∞ ∞

= =

− −∞ ∞

= =

−− −∞ ∞

= =

− −= =

−

− −′ = =

−

− − −′′ = =

− −

− −′′ + = + =

− −

∑ ∑

∑ ∑

∑ ∑

∑ ∑

67. ( ) ( )( )( ) ( )( )( ) ( )

2 1 2 1

0 1

2 2

0 0

2 1 2 1

1 1

2 1 ! 2 1 !

2 12 1 ! 2 !

22 ! 2 1 !

0

n n

n n

n n

n n

n n

n n

x xyn n

n x xyn n

n x xy yn n

y y

+ −∞ ∞

= =

∞ ∞

= =

− −∞ ∞

= =

= =+ −

+′ = =

+

′′ = = =−

′′ − =

∑ ∑

∑ ∑

∑ ∑

68. ( ) ( )( )( ) ( )

( )( ) ( )

2 2 2

0 1

2 1 2 1

1 1

2 2 2 2

1 1

2 ! 2 2 !

2 22 2 ! 2 1 !

2 12 1 ! 2 2 !

0

n n

n n

n n

n n

n n

n n

x xyn n

n x xyn n

n x xy yn n

y y

−∞ ∞

= =

− −∞ ∞

= =

− −∞ ∞

= =

= =−

−′ = =

− −

−′′ = = =

− −

′′ − =

∑ ∑

∑ ∑

∑ ∑

69. ( ) 2 22 2 1

0 1 1

2 2 122 ! 2 ! 2 !

nn n

n n nn n n

n n xx nxy y yn n n

−−∞ ∞ ∞

= = =

−′ ′′= = =∑ ∑ ∑

( )

( ) ( )

( )( )( )

( ) ( )( )

( ) ( ) ( )( )

2 2 2 2

1 1 0

2 2 2

1 0

2 2

10

2

10

2 2 1 22 ! 2 ! 2 !

2 2 1 2 12 ! 2 !

2 2 2 1 2 1 2 12 1 ! 2 ! 2 1

2 1 2 1 2 10

2 1 !

n n n

n n nn n n

n n

n nn n

n n

n nn

n

nn

n n x nx xy xy yn n n

n n x n xn n

n n x n x nn n n

n x n nn

−∞ ∞ ∞

= = =

−∞ ∞

= =

∞

+=

∞

+=

−′′ ′− − = − −

− += −

⎡ ⎤+ + + += − ⋅⎢ ⎥

+ +⎢ ⎥⎣ ⎦

+ ⎡ + − + ⎤⎣ ⎦= =+

∑ ∑ ∑

∑ ∑

∑

∑

70. ( )( )

( )( )

( ) ( )

4

21

4 1

21

1

11

2 ! 3 7 11 4 1

1 42 ! 3 7 11 4 1

1 4 4 1

n n

nn

n n

nn

n

n

xy

n n

nxy

n n

n ny

∞

=

−∞

=

∞

=

−= +

⋅ ⋅ ⋅ −

−′ =

⋅ ⋅ ⋅ −

− −′′ =

∑

∑

∑ ( )

4 2

22 ! 3 7 11 4 1

n

n

x

n n

−

⋅ ⋅ ⋅ −( )

( )

( )( )

( )( )

( ) ( )( ) ( )

( )( )

( )( )

4 22

22

4 2 4 22 2 2

2 22 1

1 14 2 4 2 2

2 2 2 21 1

1 42 ! 3 7 11 4 5

1 4 12 ! 3 7 11 4 5 2 ! 3 7 11 4 1

1 4 1 1 2 12 1 ! 3 7 11 4 1 2 ! 3 7 11 4 1 2 1

n n

nn

n nn n

n nn n

n nn n

n nn n

nxx

n n

nx xy x y x x

n n n n

n x x nn n n n n

−∞

=

− +∞ ∞

= =

+ ++ +∞ ∞

+= =

−= − +

⋅ ⋅ ⋅ −

− −′′ + = − + + +

⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ −

− + − += − =

+ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ − +

∑

∑ ∑

∑ ∑ 0



71. ( ) ( )( )

2

0 220

1

2 !

k k

kk

xJ x

k

∞

=

−= ∑

(a) ( )( )

( )( )

( )( )

1 22 2 2 21

2 22 22 2

1 2 ! 1lim lim lim 0

1 2 12 1 !

k k kk

k kkk k kk

x k xuu x kk

+ ++

+→∞ →∞ →∞

− −= ⋅ = =

− +⎡ + ⎤⎣ ⎦

Therefore, the interval of convergence is .x−∞ < < ∞

(b) ( )( )

( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )( )

( ) ( )( ) ( )

2

0 20

2 12 11

0 2 211 0

2 2 21

0 2 211 0

2 21 12 2

0 0 0 10 0

14 !

2 221 14 ! 4 1 !

2 2 1 2 2 2 11 1

4 ! 4 1 !

2 2 1 21 14 1 ! !

kk

kk

kkk k

k kk k

k kk k

k kk k

kk k

kk k

xJk

k xkxJk k

k k x k k xJ

k k

k x xx J xJ x Jk k

∞

=

+−∞ ∞+

+= =

−∞ ∞+

+= =

+∞ ∞+ +

+= =

= −

+′ = − = −⎡ + ⎤⎣ ⎦

− + +′′ = − = −⎡ + ⎤⎣ ⎦

+′′ ′+ + = − + −+

∑

∑ ∑

∑ ∑

∑ ∑ ( ) ( )( )

( )( )

( ) ( )( ) ( ) ( )

( )( )

2 2 2 2

210

2 2

20

2 2

20

14 1 ! ! 4 !

1 2 2 1 21 1 14 1 4 14 !

1 4 2 2 4 4 04 4 4 4 4 44 !

k kk

k kk

k k

kk

k k

kk

xk k k

x kk kk

x k kk k kk

+ +∞

+=

+∞

=

+∞

=

+ −+

⎡ ⎤− += − + − +⎢ ⎥

+ +⎢ ⎥⎣ ⎦

− − − +⎡ ⎤= − + =⎢ ⎥+ + +⎣ ⎦

∑

∑

∑

(c) ( )2 4 6

6 14 64 2304x x xP x = − + −

(d) ( )( )

( )( ) ( )

( )( ) ( )

12 2 11 1

0 2 2 20 00 0 00

1 1 1 1 11 0.9212 3204 ! 4 ! 2 1 4 ! 2 1

k k kk k

k k kk k k

x xJ dx dx

k k k k k

+∞ ∞ ∞

= = =

⎡ ⎤− − −= = ⎢ ⎥ = = − + ≈

⎢ + ⎥ +⎣ ⎦∑ ∑ ∑∫ ∫

(integral is approximately 0.9197304101)

72. ( ) ( )( )

( )( )

2 2 1

1 2 1 2 10 0

1 12 ! 1 ! 2 ! 1 !

k kk k

k kk k

x xJ x x

k k k k

+∞ ∞

+ += =

− −= =

+ +∑ ∑

(a) ( )( ) ( )

( )( )

( )( )( )

1 2 3 2 1 21

2 3 22 1

1 2 ! 1 ! 1lim lim lim 0

2 1 ! 2 ! 2 2 11

k k kk

kk kk k kk

x k k xuu k k k kx

+ + ++

+ +→∞ →∞ →∞

− + −= ⋅ = =

+ + + +−

Therefore, the interval of convergence is .x−∞ < < ∞

−6 6

−5

3



(b) ( ) ( )( )

( ) ( ) ( )( )

( ) ( ) ( )( )( )

2 1

1 2 10

2

1 2 10

2 1

1 2 11

12 ! 1!

1 2 12 ! 1 !

1 2 1 22 ! 1 !

k k

kk

k k

kk

k k

kk

xJ x

k k

k xJ x

k k

k k xJ x

k k

+∞

+=

∞

+=

−∞

+=

−=

+

− +′ =+

− +′′ =+

∑

∑

∑

( ) ( ) ( )( )( )

( ) ( )( )

( )( )

( )( )

( ) ( )( )( )

( ) ( )( )

2 1 2 12 2

1 1 1 2 1 2 11 0

2 3 2 1

2 1 2 10 0

2 1 2 1

2 1 2 11 1 1

1 2 1 2 1 2 11

2 ! 1 ! 2 ! 1 !

1 12 ! 1 ! 2 ! 1 !

1 2 1 2 1 2 12 ! 1 ! 2 2 ! 1 ! 2

k kk k

k kk k

k kk k

k kk k

k kk k

k kk k k

k k x k xx J xJ x J

k k k k

x xk k k k

k k x k xx xk k k k

+ +∞ ∞

+ += =

+ +∞ ∞

+ += =

+ +∞ ∞ ∞

+ += = =

− + − +′′ ′+ + − = ++ +

− −+ −

+ +

− + − += + + − −

+ +

∑ ∑

∑ ∑

∑ ∑ ∑ ( )( )

( )( )

( ) ( )( ) ( )( )

( )( )

( ) ( )( )

( )( )

( )( )

2 1

2 1

2 3

2 10

2 1 2 3

2 1 2 11 0

2 1 2 3

2 1 2 11 0

2 1

2 11

12 ! 1 !

12 ! 1 !

1 2 1 2 2 1 1 12 ! 1 ! 2 ! 1 !

1 4 1 12 ! 1 ! 2 ! 1 !

12 1 ! !

k k

k

k k

kk

k kk k

k kk k

k kk k

k kk k

k k

kk

xk k

xk k

x k k k xk k k k

x k k xk k k k

xk k

+

+

+∞

+=

+ +∞ ∞

+ += =

+ +∞ ∞

+ += =

+∞

−=

⎡ ⎤−⎢ ⎥

+⎢ ⎥⎣ ⎦

−+

+

− ⎡ + + + − ⎤ −⎣ ⎦= ++ +

− + −= +

+ +

−= +

−

∑

∑ ∑

∑ ∑

∑ ( )( )

( )( )

( )( )

( ) ( )( )

2 3

2 10

1 2 3 2 3

2 1 2 10 0

2 3

2 10

12 ! 1 !

1 12 ! 1 ! 2 ! 1 !

1 1 10

2 ! 1 !

k k

kk

k kk k

k kk k

k k

kn

xk k

x xk k k k

xk k

+∞

+=

+ + +∞ ∞

+ += =

+∞

+=

−+

− −= +

+ +

− ⎡ − + ⎤⎣ ⎦= =+

∑

∑ ∑

∑

(c) ( ) 3 5 77

1 1 12 16 384 18,432xP x x x x= − + −

(d) ( ) ( ) ( )( ) ( )

( )( )

( ) ( )( )

( )( ) ( ) ( )

1 12 1 2 1

0 2 2 2 10 0

12 1 2 1

1 0 12 1 2 10 0

1 2 1 12 1 ! 1 ! 2 ! 1 !

1 12 ! 1 ! 2 ! 1 !

k kk k

k kk k

k kk k

k kk k

k x xJ x

k k k k

x xJ x J x J x

k k k k

+ ++ +∞ ∞

+ += =

++ +∞ ∞

+ += =

− + −′ = =+ + +

− − ′− = − = = −+ +

∑ ∑

∑ ∑ Note:

73. ( ) ( ) ( )2

01 cos

2 !

nn

n

xf x xn

∞

=

= − =∑

74. ( ) ( ) ( )2 1

01 sin

2 1 !

nn

n

xf x xn

∞ +

=

= − =+∑

−6 6

−4

4

−2

2

2�−2� −

−2

� �

2



75. ( ) ( ) ( )

( )

0 01 Geometric

1 1 for 1 11 1

n nn

n nf x x x

xx x

∞ ∞

= =

= − = −

= = − < <− − +

∑ ∑

76. ( ) ( )2 1

01 arctan , 1 1

2 1

nn

n

xf x x xn

+∞

=

= − = − ≤ ≤+∑

77. ( )0

, 4, 44

n

n

x∞

=

⎛ ⎞ −⎜ ⎟⎝ ⎠

∑

(a) ( )0 0

5 2 5 1 84 8 1 5 8 3

n n

n n

∞ ∞

= =

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ −⎝ ⎠⎝ ⎠∑ ∑

(b) ( )0 0

5 2 5 1 84 8 1 5 8 13

n n

n n

∞ ∞

= =

⎛ − ⎞ ⎛ ⎞= − = =⎜ ⎟ ⎜ ⎟ +⎝ ⎠⎝ ⎠∑ ∑

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

(d)

78. ( )0

3 n

nx

∞

=∑ converges on 1 1, .

3 3⎛ ⎞−⎜ ⎟⎝ ⎠

(a) ( )0 0

1 1 1 1: 3 26 6 2 1 1 2

n n

n nx

∞ ∞

= =

⎛ ⎞⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠⎝ ⎠∑ ∑

(b) ( )0 0

1 1 1 1 2: 36 6 2 1 1 2 3

n n

n nx

∞ ∞

= =

⎛ ⎞⎛ ⎞ ⎛ ⎞= − − = − = =⎜ ⎟⎜ ⎟ ⎜ ⎟ +⎝ ⎠ ⎝ ⎠⎝ ⎠∑ ∑

(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 2.3

y =

(d) 0 0

23 23

nN Nn

n nM

= =

⎛ ⎞⋅ = >⎜ ⎟⎝ ⎠

∑ ∑

79. False;

( )1

12

n n

nn

xn

∞

=

−∑

converges for 2x = but diverges for 2.x = −

80. False; it is not possible. See Theorem 9.20.

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

82. True

( )1 1

0 00

11

0 001 1

nn

n

nn n

n n

f x dx a x dx

a x an n

∞

=

∞ ∞+

= =

⎛ ⎞= ⎜ ⎟

⎝ ⎠

⎡ ⎤= =⎢ ⎥+ +⎣ ⎦

∑∫ ∫

∑ ∑

83. ( )( ) ( )

( )( )

( )( )( )

1 1 11 ! !lim lim lim 0

1 ! 1 ! ! ! 1 1n n n

n n nn

n p xn p n pa x xn n q n n qa n n q

+ +

→∞ →∞ →∞

+ ++ + += = =

+ + + + + + +

So, the series converges for all : .x R = ∞

M 10 100 1000 10,000

N 5 14 24 35

M 10 100 1000 10,000

N 3 6 9 13

1−10

3

−2.5 2.5

− �2

�2

0 60

4

0 60

1

3

60

−1

1.5

6

−0.5

−1



84. (a) ( )( ) ( )

2 3 4

2 3 4 2 2 1 2 4 2 2 4 22

1 2 2

1 2 2 2 1 2 1n n n nn

g x x x x x

S x x x x x x x x x x x x x+

= + + + + +

= + + + + + + + = + + + + + + + + +

2 22

0 0lim 2n n

nn n n

S x x x∞ ∞

→∞ = =

= +∑ ∑

Each series is geometric, 1,R = and the interval of convergence is ( )1, 1 .−

(b) For ( ) 2 2 21 1 1 21, 2 .

1 1 1xx g x x

x x x+

< = + =− − −

85. (a) ( )

( ) ( ) ( )

30

2 3 4 5 60 1 2 0 1 2 0

3 3 3 3 2 3 33 0 1 2

,

1 1 1

nn n n

n

n n nn

f x c x c c

c c x c x c x c x c x c x

S c x x c x x x c x x x

∞

+=

= =

= + + + + + + +

= + + + + + + + + + + +

∑

3 3 2 33 0 1 2

0 0 0lim n n n

nn n n n

S c x c x x c x x∞ ∞ ∞

→∞ = = =

= + +∑ ∑ ∑

Each series is geometric, 1,R = and the interval of convergence is ( )1, 1 .−

(b) For ( )2

0 1 220 1 23 3 3 3

1 1 11, .1 1 1 1

c c x c xx f x c c x c xx x x x

+ +< = + + =

− − − −

86. For the series ,nnc x∑

1

1 1 1

1lim lim lim 1

nn n n n

nn n nn n n n

a c x c cx x Ra c x c c

++ + +

→∞ →∞ →∞ +

= = < ⇒ < =

For the series 2 ,nnc x∑

2 2

1 1 1 2 22

1lim lim lim 1 .

nn n n n

nn n nn n n n

b c x c cx x R x Rb c x c c

++ + +

→∞ →∞ →∞ +

= = < ⇒ < = ⇒ <

87. At ( )0 00 0

, ,n nn n

n nx x R c x x c R

∞ ∞

= =

= + − =∑ ∑ diverges.

At ( ) ( )0 00 0

, ,n nn n

n nx x R c x x c R

∞ ∞

= =

= − − = −∑ ∑ converges.

Furthermore, at 0 ,x x R= −

( )00 0

, diverges.n nn n

n nc x x C R

∞ ∞

= =

− =∑ ∑

So, the series converges conditionally at 0 .x R−



Section 9.9 Representation of Functions by Power Series

1. (a) ( )

10 0

1 1 44 1 4

11 4 4 4

n n

nn n

x x

a x xr

∞ ∞

+= =

=− −

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠∑ ∑

This series converges on ( )4, 4 .−

(b)

2

2

2

2 3

14 16 64

4 1

14

4

4 16

16

16 64

x x

xx

x

x x

x

x x

+ + +

−

−

−

−

2. (a) ( )

( )1

0 0

1 1 22 1 2 1

112 2 2

nn n

nn n

ax x r

xx∞ ∞

+= =

= =+ − − −

−⎛ ⎞= − =⎜ ⎟⎝ ⎠

∑ ∑


(b)

2 3

2

2

2 3

3

3 4

12 4 8 16

2 1

12

2

2 4

4

4 8

8

8 16

x x x

xx

x

x x

x

x x

x

x x

− + − +

+

+

−

− −

+

−

− −

3. (a) ( )

( )1

0 0

3 3 44 1 4 1

3 134 4 4

nn n

nn n

ax x r

xx∞ ∞

+= =

= =+ − − −

−⎛ ⎞= − =⎜ ⎟⎝ ⎠

∑ ∑


(b)

2

2

2

2 3

3 3 34 16 64

4 3334343 34 16

3163 316 64

x x

x

x

x

xx

x

x x

− + −

+

+

−

− −

+

4. (a) ( )

10 0

2 2 55 1 5 1

2 25 5 5

n n

nn n

ax x r

x x∞ ∞

+= =

= =− − −

⎛ ⎞= =⎜ ⎟⎝ ⎠

∑ ∑


(b)

2

2

2

2 2 2 5 25 125

5 2225252 25 25

225

x x

x

x

x

x x

x

+ + +

−

−

−

5. ( )

( )1

0 0

1 1 1 213 2 1 11

2

11 12 2 2

nn

nn n

axx x r

xx∞ ∞

+= =

= = =−− − − −⎛ ⎞− ⎜ ⎟

⎝ ⎠

−−⎛ ⎞= =⎜ ⎟⎝ ⎠

∑ ∑

Interval of convergence: ( )1 1 1, 32

x −< ⇒ −

Section 9.9 Representation of Functions by Power Series 353


6. ( )

( )3 1

0 0

4 4 1 235 8 3 11

8

31 32 8 2

nn

nn n

axx x r

xx∞ ∞

+= =

= = =+− − + −⎛ ⎞− ⎜ ⎟

⎝ ⎠

++⎛ ⎞= =⎜ ⎟⎝ ⎠

∑ ∑

Interval of convergence: ( )3 1 11, 58

x +< ⇒ −

7. ( )0

1 31 3 1

n

n

a xx r

∞

=

= =− − ∑

Interval of convergence: 1 13 1 , 3 3

x ⎛ ⎞< ⇒ ⎜ ⎟⎝ ⎠

8. ( )0

1 51 5 1

n

n

a xx r

∞

=

= =− − ∑

Interval of convergence: 1 15 1 , 5 5

x ⎛ ⎞< ⇒ ⎜ ⎟⎝ ⎠

9. ( ) ( )

( ) ( )

( )

0

10

5 5 5 922 3 9 2 3 11 39

5 2 23 , 3 19 9 9

25 39

n

n

nn

nn

ax x rx

x x

x

∞

=

∞

+=

−= = =

− − + + −− +

⎛ ⎞= − + + <⎜ ⎟⎝ ⎠

= − +

∑

∑

Interval of convergence: ( )2 15 33 1 , 9 2 2

x ⎛ ⎞+ < ⇒ −⎜ ⎟⎝ ⎠

10. ( ) ( )( )

( )

( ) ( )0

0

3 3 12 1 3 2 2 1 2 3 2 1

2 23

2 23

n

n

n n

nn

ax x x r

x

x

∞

=

∞

=

= = =− + − + − −

⎡ ⎤= − −⎢ ⎥⎣ ⎦

− −=

∑

∑

Interval of convergence: 3 1 72 , 2 2 2

x ⎛ ⎞− < ⇒ ⎜ ⎟⎝ ⎠

11.

( )0

1

10

2 2 3 2 32 22 3 11 13 3

2 23 3

1 23

n

n

n nn

nn

axx rx

x

x

∞

=

+∞

+=

= = =+ −⎛ ⎞+ − −⎜ ⎟

⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

−=

∑

∑

Interval of convergence: 2 3 31 , 3 2 2

x− ⎛ ⎞< ⇒ −⎜ ⎟⎝ ⎠

12. ( ) ( )( )

( )

( ) ( )0

10

4 4 4 113 2 11 3 3 1 3 11 3 1

3 3411 11

3 34

11

n

n

n n

nn

ax x x r

x

x

∞

=

∞

+=

= = =+ + − − − − −

⎡− − ⎤= ⎢ ⎥

⎣ ⎦

− −=

∑

∑

Interval of convergence: ( )3 203 1 3, 11 3

x ⎛ ⎞− − < ⇒ ⎜ ⎟⎝ ⎠

13.

( )

( )

2

0 0 0

4 3 12 3 3 1

1 11 3 1

1 13 3

nn n

nn n n

xx x x x

x x

x x x∞ ∞ ∞

= = =

= ++ − + −

−= +

− − −

⎡ ⎤⎛ ⎞ ⎢ ⎥= − − = −⎜ ⎟⎝ ⎠ ⎢ ⎥−⎣ ⎦

∑ ∑ ∑

Interval of convergence: 13x

< and ( )1 1, 1x < ⇒ −

14.

( )

( )

2

0 0

1

0

3 8 2 33 5 2 2 3 1

1 31 2 1 3

3 32

1 32

nn

n n

nn n

n

xx x x x

x x

x x

x

∞ ∞

= =

∞+

=

−= −

+ − + −

= +− − −

⎛ ⎞= − +⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

∑ ∑

∑

Interval of convergence: 12x

< and

1 13 1 , 3 3

x ⎛ ⎞< ⇒ −⎜ ⎟⎝ ⎠

15.

( )( )2

2

0 0

2 1 11 1 1

1 1 2n n n

n n

x x x

x x∞ ∞

= =

= +− − +

= + − =∑ ∑

Interval of convergence: 2 1x < or ( )1, 1 because 2 2

1 22

2lim lim2

nn

nn nn

u x xu x

++

→∞ →∞= =



16. 2

22 2

0 0

5 1 15 1 5 5

15

n nn

n n

a x xx rx

∞ ∞

= =

⎛ ⎞ −⎛ ⎞= = = − =⎜ ⎟ ⎜ ⎟+ −⎛ ⎞− ⎝ ⎠⎝ ⎠− ⎜ ⎟⎝ ⎠

∑ ∑

Interval of convergence: ( )2

21 5 5 5, 55x x< ⇒ − < < ⇒ −

17. ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

0

2

0 0 0

20 0 0

2 3 4 5 6 2

0

1 11

1 1 11

2 1 1 1 1 11 1 1

2 0 2 0 2 0 2 2 , 1, 1 See Exercise 15.

n n

n

n n n n n

n n n

n nn n n

n n n

n

n

xx

x x xx

h x x x xx x x

x x x x x x x

∞

=

∞ ∞ ∞

= = =

∞ ∞ ∞

= = =

∞

=

= −+

= − − = − =−

− ⎡ ⎤= = + = − + = − +⎣ ⎦− + −

= + + + + + + + = −

∑

∑ ∑ ∑

∑ ∑ ∑

∑

18. ( ) ( ) ( ) ( ) ( )

( )

( ) ( )

20 0

2 3 4 5

0

2 1 2 1

0 0

1 1 1 11 See Exercise 17.1 2 1 2 1 2 2

1 11 1 0 2 0 2 0 22 2

1 2 , 1, 12

n n n

n n

n n

n

n n

n n

xh x x xx x x

x x x x x x

x x

∞ ∞

= =

∞

=

∞ ∞+ +

= =

= = − = − −− + −

⎡ ⎤ ⎡ ⎤= − − = − + − + − +⎣ ⎦⎣ ⎦

= − = − −

∑ ∑

∑

∑ ∑

19. By taking the first derivative, you have ( )2

1 1 .1 1

ddx x x

−⎡ ⎤ =⎢ ⎥+⎣ ⎦ +Therefore,

( )

( ) ( ) ( ) ( ) ( )112

0 1 0

1 1 1 1 1 , 1, 1 .1

n n nn n n

n n n

d x nx n xdxx

∞ ∞ ∞+−

= = =

⎡ ⎤−= − = − = − + −⎢ ⎥

+ ⎣ ⎦∑ ∑ ∑

20. By taking the second derivative, you have ( )

2

321 2 .

1 1ddx x x

⎡ ⎤ =⎢ ⎥+⎣ ⎦ +Therefore,

( )

( ) ( ) ( ) ( ) ( ) ( )( ) ( )2

1 23 2

0 1 2 0

2 1 1 1 1 1 2 1 , 1, 1 .1

n n n nn n n n

n n n n

d dx nx n n x n n xdx dxx

∞ ∞ ∞ ∞− −

= = = =

⎡ ⎤ ⎡ ⎤= − = − = − − = − + + −⎢ ⎥ ⎢ ⎥

+ ⎣ ⎦ ⎣ ⎦∑ ∑ ∑ ∑

21. By integrating, you have ( )1 ln 1 .1

dx xx

= ++∫ Therefore,

( ) ( ) ( ) 1

0 0

1ln 1 1 , 1 1.

1

n nn n

n n

xx x dx C x

n

+∞ ∞

= =

−⎡ ⎤+ = − = + − < ≤⎢ ⎥ +⎣ ⎦

∑ ∑∫

To solve for C, let 0x = and conclude that 0.C = Therefore,

( ) ( ) ( ]1

0

1ln 1 , 1, 1 .

1

n n

n

xx

n

+∞

=

−+ = −

+∑



22. By integrating, you have

( ) 11 ln 1

1dx x C

x= + +

+∫ and ( ) 21 ln 1 .

1dx x C

x= − − +

−∫

( ) ( ) ( ) ( )2ln 1 ln 1 ln 1 .f x x x x= − = + − ⎡− − ⎤⎣ ⎦ Therefore,

( )

( ) ( )

( ) ( )

2

1 1

1 20 0 0 0

1 2 22 2

0 0 0

1 1ln 11 1

11

1 1

1 1 121 2 2 1

n n nn n n

n n n n

n n nn

n n n

x dx dxx x

x xx dx x dx C Cn n

x xxC C Cn n n

+ +∞ ∞ ∞ ∞

= = = =

+ ++∞ ∞ ∞

= = =

− = −+ −

⎡ ⎤− ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥= − − = + − +⎢ ⎥⎢ ⎥ ⎢ ⎥ + +⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤− − −−⎣ ⎦= + = + = ++ + +

∫ ∫

∑ ∑ ∑ ∑∫ ∫

∑ ∑ ∑


( ) ( )2 2

2

0ln 1 , 1, 1 .

1

n

n

xxn

+∞

=

− = − −+∑

23. ( ) ( ) ( ) ( )2 22

0 0

1 1 1 , 1, 11

nn n n

n nx x

x

∞ ∞

= =

= − = − −+ ∑ ∑

24. ( ) ( )

( )

22

0

2 1

0

2 2 1 See Exercise 23.1

1 2

n n

n

n n

n

x x xx

x

∞

=

∞+

=

= −+

= −

∑

∑

Because ( )( )222ln 1 ,

1d xxdx x

+ =+

you have

( ) ( ) ( ) 2 22 2 1

0 0

1ln 1 1 2 , 1 1.

1

n nn n

n n

xx x dx C x

n

+∞ ∞+

= =

−⎡ ⎤+ = − = + − ≤ ≤⎢ ⎥ +⎣ ⎦

∑ ∑∫


( ) ( ) [ ]2 2

2

0

1ln 1 , 1, 1 .

1

n n

n

xx

n

+∞

=

−+ = −

+∑

25. Because, ( )0

1 1 ,1

n n

nx

x

∞

=

= −+ ∑ you have ( ) ( ) ( ) ( ) ( )22 2

20 0 0

1 1 11 4 1 4 1 2 , , .4 1 2 2

nn n n nn n

n n nx x x

x

∞ ∞ ∞

= = =

⎛ ⎞= − = − = − −⎜ ⎟+ ⎝ ⎠∑ ∑ ∑

26. Because ( )21 1 arctan 2 ,

4 1 2dx x

x=

+∫ you can use the result of Exercise 25 to obtain

( ) ( ) ( ) 2 12

20 0

1 41 1 1arctan 2 2 2 1 4 2 , .4 1 2 1 2 2

n n nn n n

n n

xx dx x dx C x

x n

+∞ ∞

= =

−⎡ ⎤= = − = + − < <⎢ ⎥+ +⎣ ⎦

∑ ∑∫ ∫


( ) ( ) 2 1

0

1 4 1 1arctan 2 2 , , .2 1 2 2

n n n

n

xx

n

+∞

=

− ⎛ ⎤= −⎜ ⎥+ ⎝ ⎦∑



27. ( )2 2 3

ln 12 2 3x x xx x x− ≤ + ≤ − +

28. ( )2 3 4 2 3 4 5

ln 12 3 4 2 3 4 5x x x x x x xx x x− + − ≤ + ≤ − + − +

29. ( ) ( ) ( ) ( ) ( )1 2 3

1

1 1 1 1 11 2 3

n n

n

x x x xn

+∞

=

− − − − −= − + −∑

(a)

(b) From Example 4,

( ) ( ) ( ) ( )1 1

1 0

1 1 1 1ln , 0 2, 1.

1

n n n n

n n

x xx x R

n n

+ +∞ ∞

= =

− − − −= = < ≤ =

+∑ ∑

(c) 0.5:x =

( ) ( ) ( )1

1 1

1 1 2 1 20.693147

n n n

n nn n

+∞ ∞

= =

− − −= ≈ −∑ ∑

(d) This is an approximation of 1ln .2

⎛ ⎞⎜ ⎟⎝ ⎠

The error is approximately 0. [The error is less than the first omitted term,

( )51 181 51 2 8.7 10 .−⋅ ≈ × ]

x 0.0 0.2 0.4 0.6 0.8 1.0 2

2 2xS x= − 0.000 0.180 0.320 0.420 0.480 0.500

( )ln 1x + 0.000 0.182 0.336 0.470 0.588 0.693 2 3

3 2 3x xS x= − + 0.000 0.183 0.341 0.492 0.651 0.833

x 0.0 0.2 0.4 0.6 0.8 1.0 2 3 4

4 2 3 4x x xS x= − + − 0.0 0.18227 0.33493 0.45960 0.54827 0.58333

( )ln 1x + 0.0 0.18232 0.33647 0.47000 0.58779 0.69315 2 3 4 5

5 2 3 4 5x x x xS x= − + − + 0.0 0.18233 0.33698 0.47515 0.61380 0.78333

−4 8

−3

S3f

S2

5

−4 8

−3

f

S5

S4

5

−3

0 4

n = 1n = 3

n = 6n = 2

3



30. ( )( )

2 1 3 5

0

12 1 ! 3! 5!

n n

n

x x xxn

+∞

=

−= − + −

+∑

(a)

(b) ( )( )

2 1

0

1sin ,

2 1 !

n n

n

xx R

n

+∞

=

−= = ∞

+∑

(c) ( ) ( )( )

2 1

0

1 1 20.4794255386

2 1 !

n n

n n

+∞

=

−≈

+∑

(d) This is an approximation of 1sin .2

⎛ ⎞⎜ ⎟⎝ ⎠

The error is

approximately 0.

31. ( ) lineg x x=

Matches (c)

32. ( )3,

3xg x x= − cubic with 3 zeros.

Matches (d)

33. ( )3 5

3 5x xg x x= − +

Matches (a)

34. ( )3 5 7

,3 5 7x x xg x x= − + −

Matches (b)

In Exercises 35–38, arctan ( )2 +1

=0= 1 .

2 + 1

nn

n

xxn

∞

−∑

35. ( ) ( ) ( )( )

2 1

2 10 0

1 4 11 1 1 1arctan 14 2 1 2 1 4 4 192 5120

n nn

nn nn n

+∞ ∞

+= =

−= − = = − + +

+ +∑ ∑

Because 1 0.001,5120

< you can approximate the series by its first two terms: 1 1 1arctan 0.245.4 4 192≈ − ≈

36. ( )

( ) ( )( )

( ) ( )( )( ) ( ) ( )( )

4 22

0

4 32

0

4 3 4 33 4 24 30

0 0

arctan 12 1

arctan 1 , 04 3 2 1

3 4 3arctan 1 14 3 2 1 4 3 2 1 4

27 2187 177,147192 344,064 230,686,720

nn

n

nn

n

n nn n

nn n

xxn

xx dx C Cn n

x dxn n n n

+∞

=

+∞

=

+ +∞ ∞

+= =

= −+

= − + =+ +

= − = −+ + + +

= − +

∑

∑∫

∑ ∑∫

Because 177,147 230,686,720 0.001,< you can approximate the series by its first two terms: 0.134.

37. ( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )( )

2 122 4 12

0 0

2 4 2

0

21 2

4 200

arctan 1 1 12 1 2 1

arctan 1 Note: 04 2 2 1

arctan 1 1 114 2 2 1 2 8 1152

nn

n

n n

nn

n

nn

n

xx xx x n n

x xdx C Cx n n

x dxx n n

++∞ ∞

= =

+∞

=

∞

+=

= − = −+ +

= − + =+ +

= − = − ++ +

∑ ∑

∑∫

∑∫

Because 1 0.001,1152

< you can approximate the series by its first term: 21 2

0

arctan 0.125.x dxx

≈∫

4

6

−4

−6



38. ( )

( ) ( )( )

( ) ( )( )

2 32

0

2 42

0

1 2 22 40

0

arctan 12 1

arctan 12 4 2 1

1 1 1arctan 12 4 2 1 2 64 1152

nn

n

nn

n

nn

n

xx xn

xx x dxn n

x x dxn n

+∞

=

+∞

=

∞

+=

= −+

= −+ +

= − = − ++ +

∑

∑∫

∑∫

Because 1 0.001,1152

< you can approximate the series by its first term: 1 2 20

arctan 0.016.x x dx ≈∫

In Exercises 39–42, use =0

1 = , < 1.1

n

nx x

x

∞

− ∑

39. ( )

12

0 1

1 1 , 111

n n

n n

d d x nx xdx x dxx

∞ ∞−

= =

⎡ ⎤⎡ ⎤= = = <⎢ ⎥⎢ ⎥−⎣ ⎦− ⎣ ⎦∑ ∑

40. ( )

12

1 1, 1

1n n

n n

x x nx nx xx

∞ ∞−

= =

= = <−

∑ ∑

41. ( ) ( ) ( )

( )

( )

2 2 2

1

1

0

1 11 1 1

, 1

2 1 , 1

n n

n

n

n

x xx x x

n x x x

n x x

∞−

=

∞

=

+= +

− − −

= + <

= + <

∑

∑

42. ( )( )

( ) ( ) 12

0 0

12 1 2 1 , 1

1n n

n n

x xx n x n x x

x

∞ ∞+

= =

+= + = + <

−∑ ∑

(See Exercise 41.)

43. ( )

( ) ( )

( )

1

1 1 1

2

12

1 1 12 2 2

1 1 22 1 1 2

n

n n

n n n

P n

E n nP n n n−∞ ∞ ∞

= = =

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= =⎡ − ⎤⎣ ⎦

∑ ∑ ∑

Because the probability of obtaining a head on a single

toss is 1,2

it is expected that, on average, a head will be

obtained in two tosses.

44. (a) ( )( )

1

21 1

23

1 2 2 2 1 23 9 3 9 1 2 3

nn

n nn n

−∞ ∞

= =

⎛ ⎞= = =⎜ ⎟⎝ ⎠ ⎡ − ⎤⎣ ⎦

∑ ∑

(b)

( )

1

1 1

2

1 9 9 910 10 100 10

9 1 9100 1 9 10

n n

n nn n

−∞ ∞

= =

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ⋅ =⎡ − ⎤⎣ ⎦

∑ ∑

45. Because( )

1 1 ,1 1x x

=+ − −

substitute ( )x− into the

geometric series.

46. Because( )2 2

1 1 ,1 1x x

=− −

substitute ( )2x into the

geometric series.

47. Because( )

1 15 ,1 1x x

⎛ ⎞= ⎜ ⎟⎜ ⎟+ − −⎝ ⎠

substitute ( )x− into the

geometric series and then multiply the series by 5.

48. Because ( ) 1ln 1 ,1

x dxx

− = −−∫ integrate the series

and then multiply by ( )1 .−

49. Let arctan arctan .x y θ+ = Then,

( )( ) ( )

( ) ( )

tan arctan arctan tan

tan arctan tan arctantan

1 tan arctan tan arctan

tan1

arctan .1

x y

x yx y

x yxy

x yxy

θ

θ

θ

θ

+ =

+=

−

+=

−

⎛ ⎞+=⎜ ⎟−⎝ ⎠

Therefore,

arctan arctan arctan for 1.1x yx y xy

xy⎛ ⎞+

+ = ≠⎜ ⎟−⎝ ⎠

50. (a) From Exercise 49, you have

( ) ( )( )( )

120 119 1 239120 1 120 1arctan arctan arctan arctan arctan119 239 119 239 1 120 119 1 239

28,561arctan arctan 128,561 4

π

⎡ ⎤+ −⎛ ⎞− = + − = ⎢ ⎥⎜ ⎟ − −⎝ ⎠ ⎢ ⎥⎣ ⎦⎛ ⎞

= = =⎜ ⎟⎝ ⎠



(b) ( )( )

( )( )

2

2

2 1 51 1 1 10 52 arctan arctan arctan arctan arctan arctan5 5 5 24 121 1 5

2 5 121 1 1 5 5 1204 arctan 2 arctan 2 arctan arctan arctan arctan arctan5 5 5 12 12 1191 5 12

1 1 120 14 arctan arctan arctan arctan5 239 119 2

⎡ ⎤⎢ ⎥= + = = =⎢ ⎥−⎣ ⎦

⎡ ⎤⎢ ⎥= + = + = =⎢ ⎥−⎣ ⎦

− = − ( )( )see part a .39 4

π=

51. (a) ( )

( ) ( )( )( )

2

1 11 1 1 42 22 arctan arctan arctan arctan arctan2 2 2 31 1 2

4 3 1 71 1 4 1 252 arctan arctan arctan arctan arctan arctan arctan 12 7 3 7 1 4 3 1 7 25 4

π

⎡ ⎤+⎢ ⎥= + = =⎢ ⎥

−⎢ ⎥⎣ ⎦

⎡ ⎤−⎛ ⎞− = + − = = = =⎢ ⎥⎜ ⎟ +⎝ ⎠ ⎢ ⎥⎣ ⎦

(b) ( ) ( ) ( ) ( ) ( ) ( )3 5 7 3 5 70.5 0.5 0.5 1 7 1 7 1 71 1 1 18 arctan 4 arctan 8 4 3.142 7 2 3 5 7 7 3 5 7

π⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − ≈ − + − − − + − ≈⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

52. (a) ( ) ( )( )( )

1 2 1 31 1 5 6arctan arctan arctan arctan2 3 1 1 2 1 3 5 6 4

π⎡ ⎤+ ⎛ ⎞+ = = =⎢ ⎥ ⎜ ⎟−⎢ ⎥ ⎝ ⎠⎣ ⎦

(b)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 5 7 3 5 7

1 14 arctan arctan2 3

1 2 1 2 1 2 1 3 1 3 1 31 14 4 4 0.4635 4 0.3217 3.142 3 5 7 3 3 5 7

π ⎡ ⎤= +⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − + − + − + − ≈ + ≈⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

53. From Exercise 21, you have

( ) ( ) ( )

( )

11

0 1

1

1

1 1ln 1

1

1.

n nn n

n n

n n

n

x xx

n n

xn

−+∞ ∞

= =

+∞

=

− −+ = =

+

−=

∑ ∑

∑

So, ( ) ( ) ( )11

1 1

1 1 2112

1 3ln 1 ln 0.4055.2 2

n nn

nn nn n

+∞ ∞+

= =

−− =

⎛ ⎞= + = ≈⎜ ⎟⎝ ⎠

∑ ∑


( ) ( ) ( )11

1 1

1 1 3113

1 4ln 1 ln 0.2877.3 3

n nn

nn nn n

+∞ ∞+

= =

−− =

⎛ ⎞= + = ≈⎜ ⎟⎝ ⎠

∑ ∑


( ) ( ) ( )11

1 1

1 2 5215

2 7ln 1 ln 0.3365.5 5

n nnn

nn nn n

+∞ ∞+

= =

−− =

⎛ ⎞= + = ≈⎜ ⎟⎝ ⎠

∑ ∑

56. From Example 5, you have ( )2 1

0arctan 1 .

2 1

nn

n

xxn

+∞

=

= −+∑

( ) ( ) ( )2 1

0 0

111 12 1 2 1

arctan 1 0.78544

nn n

n nn nπ

+∞ ∞

= =

− = −+ +

= = ≈

∑ ∑


( ) ( ) ( ) ( )2 1

2 10 0

1 211 12 2 1 2 1

1arctan 0.4636.2

nn n

nn nn n

+∞ ∞

+= =

− = −+ +

= ≈

∑ ∑


( ) ( ) ( ) ( )

( ) ( )

12 1 2 1

1 0

2 1

0

1 11 13 2 1 3 2 1

1 31

2 11arctan 0.3218.3

n nn n

n n

nn

n

n n

n

∞ ∞+

− += =

+∞

=

− = −− +

= −+

= ≈

∑ ∑

∑

59. ( ) arctanf x x= is an odd function (symmetric to the

origin).



60. The approximations of degree ( )3, 7, 11, , 4 1, 1, 2,n n− =… … have relative extrema.

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

62. Because 1

0 1,n n

n nn n

d a x na xdx

∞ ∞−

= =

⎡ ⎤=⎢ ⎥

⎣ ⎦∑ ∑ the radius of

convergence is the same, 3.

63. Because the first series is the derivative of the second series, the second series converges for 1 4x + < (and

perhaps at the endpoints, 3x = and 5.x = − )

64. You can verify that the statement is incorrect by calculating the constant terms of each side:

( )

0 0

0

1 15 5

1 1 11 1 15 5 5

nn

n n

n

n

x xx x

x x

∞ ∞

= =

∞

=

⎛ ⎞ ⎛ ⎞+ = + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑ ∑

∑

The formula should be

0 0 0

11 .5 5

n nn n

n n n

xx x∞ ∞ ∞

= = =

⎡ ⎤⎛ ⎞ ⎛ ⎞+ = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

∑ ∑ ∑

65. ( )( )0

13 2 1

n

nn n

∞

=

−+∑

From Example 5 you have ( )2 1

0arctan 1 .

2 1

nn

n

xxn

+∞

=

= −+∑

( )( )

( )( ) ( )

( ) ( )

20 0

2 1

0

1 1 33 2 1 33 2 1

1 1 33

2 1

13 arctan3

3 0.90689976

n n

nnn n

nn

n

n n

n

π

∞ ∞

= =

+∞

=

− −=

+ +

−=

+

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

∑ ∑

∑

66. ( )( ) ( ) ( )

( )

2 12 1

2 10 0

1 31

3 2 1 ! 2 1 !

3sin 0.8660253 2

n nnn

nn nn n

π π

π

++∞ ∞

+= =

−= −

+ +

⎛ ⎞= = ≈⎜ ⎟⎝ ⎠

∑ ∑

67. Using a graphing utility, you obtain the following partial sums for the left hand side. Note that 1 0.3183098862.π ≈

0

1

0: 0.31830987841: 0.3183098862

n Sn S= ≈

= ≈

Section 9.10 Taylor and Maclaurin Series

1. For 0,c = you have:

( )( ) ( ) ( ) ( )

( )

2

2

2 3 42

0

2 0 2

24 8 161 2 .2! 3! 4! !

x

n nn x n

nx

n

f x e

f x e f

xx x xe xn

∞

=

=

= ⇒ =

= + + + + + = ∑


( )( ) ( ) ( ) ( )

( )

3

3

2 33

0

3 0 3

39 271 3 .2! 3! !

x

n nn x n

nx

n

f x e

f x e f

xx xe xn

∞

=

=

= ⇒ =

= + + + + = ∑

Section 9.10 Taylor and Maclaurin Series 361


3. For 4,c π= you have:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )4 4

2cos4 2

2sin4 2

2cos4 2

2sin4 2

2cos4 2

f x x f

f x x f

f x x f

f x x f

f x x f

π

π

π

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= − = −⎜ ⎟⎝ ⎠

⎛ ⎞′′ ′′= − = −⎜ ⎟⎝ ⎠

⎛ ⎞′′′ ′′′= =⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

and so on. Therefore, you have:

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

0

2 3 4

1 2

0

4 4cos

!

4 4 42 12 4 2! 3! 4!

1 42 .2 !

nn

n

nn n

n

f xx

n

x x xx

xn

π π

π π ππ

π

∞

=

+∞

=

⎡ − ⎤⎣ ⎦=

⎡ ⎤⎡ − ⎤ ⎡ − ⎤ ⎡ − ⎤⎛ ⎞ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥= − − − + + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

− ⎡ − ⎤⎣ ⎦=

∑

∑

[Note: ( ) ( )1 21 1, 1, 1, 1, 1, 1, 1, 1,n n+− = − − − − … ]

4. For 4,c π= you have:

( )

( )

( )

( )

( ) ( ) ( )4 4

2sin4 2

2cos4 2

2sin4 2

2cos4 2

2sin4 2

f x x f

f x x f

f x x f

f x x f

f x x f

π

π

π

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= =⎜ ⎟⎝ ⎠

⎛ ⎞′′ ′′= − = −⎜ ⎟⎝ ⎠

⎛ ⎞′′′ ′′′= − = −⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

and so on. Therefore you have:

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )( )

0

2 3 4

11 2

0

4 4sin

!

4 4 42 12 4 2! 3! 4!

1 42 1 .2 1 !

nn

n

nn n

n

f xx

n

x x xx

xn

π π

π π ππ

π

∞

=

+−∞

=

⎡ − ⎤⎣ ⎦=

⎡ ⎤⎡ − ⎤ ⎡ − ⎤ ⎡ − ⎤⎛ ⎞ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥= + − − − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎧ ⎫− ⎡ − ⎤⎪ ⎪⎣ ⎦= +⎨ ⎬+⎪ ⎪⎩ ⎭

∑

∑



5. For 1,c = you have

( ) ( )

( ) ( )( ) ( )( ) ( )

1

2

3

4

1 1 1

1 1

2 1 2

6 1 6

f x x fx

f x x f

f x x f

f x x f

−

−

−

−

= = =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

and so on. Therefore, you have

( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

0

2 3

2 3

0

1 11!

2 1 6 11 1

2! 3!

1 1 1 1

1 1

nn

n

n n

n

f xx n

x xx

x x x

x

∞

=

∞

=

−=

− −= − − + − +

= − − + − − − +

= − −

∑

∑


( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1

2

3

4

1 1 2 11

1 2 1

2 1 2 2

6 1 2 6

f x x fx

f x x f

f x x f

f x x f

−

−

−

−

= = − = −−

′ ′= − =

′′ ′′= − = −

′′′ ′′′= − =

and so on. Therefore you have

( ) ( )( )

( ) ( ) ( )( ) ( )

02 3

1

0

2 211 !

1 2 2 2

1 2

nn

n

n n

n

f xx n

x x x

x

∞

=

∞+

=

−=

−= − + − − − + − −

= − −

∑

∑

7. For 1,c = you have,

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

3

4 44

5 55

ln 1 0

1 1 1

1 1 1

2 1 2

6 1 6

24 1 24

f x x f

f x fx

f x fx

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= − = −

′′′ ′′′= =

= − = −

= =


( ) ( )( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )

0

2 3 4 5

2 3 4 5

1

0

1 1ln

!

1 2 1 6 1 24 10 1

2! 3! 4! 5!

1 1 1 11

2 3 4 5

11 .

1

nn

n

nn

n

f xx

n

x x x xx

x x x xx

xn

∞

=

+∞

=

−=

− − − −= + − − + − + −

− − − −= − − + − + −

−= −

+

∑

∑


( )( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( )2 3 4

0 0

1

1 1 1 1 1 11 1 .

! 2! 3! 4! !

x

n nx

n nnx

n n

f x e

f x e f e

f x x x x xe e x e

n n

∞ ∞

= =

=

= ⇒ =

⎡ ⎤− − − − −⎢ ⎥= = + − + + + + =⎢ ⎥⎣ ⎦

∑ ∑




( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )4 4

sin 3 0 0

3 cos 3 0 3

9 sin 3 0 0

27 cos 3 0 27

81 sin 3 0 0

f x x f

f x x f

f x x f

f x x f

f x x f

= =

′ ′= =

′′ ′′= − =

′′′ ′′′= − = −

= =

and so on. Therefore you have

( ) ( ) ( ) ( )

( )

2 13

0 0

0 1 327sin 3 0 3 0 0! 3! 2 1 !

n nn n

n n

f x xxx xn n

+∞ ∞

= =

−= = + + − + + =

+∑ ∑

10. For 0,c = you have.

( ) ( ) ( )

( ) ( )

( )( )

( )

( ) ( )( )

( )

( ) ( ) ( )( )

( ) ( )

( ) ( ) ( )( )

( ) ( )

( ) ( ) ( )( )

( ) ( )

2

2

2

22

2

32

4 24 4

42

4 25 5

52

6 4 26 6

62

ln 1 0 0

2 0 01

2 2 0 21

4 30 0

1

12 6 10 12

1

48 10 50 0

1

240 5 15 15 10 240

1

f x x f

xf x fx

xf x fx

x xf x f

x

x xf x f

x

x x xf x f

x

x x xf x f

x

= + =

′ ′= =+−′′ ′′= =+

−′′′ ′′′= =

+

− + −= = −

+

− += =

+

− − + −= =

+


( )

( ) ( )

( )

2 3 4 5 62

0

2 24 62

0

0 2 0 12 0 240ln 1 0 0! 2! 3! 4! 5! 6!

1.

2 3 1

n n

n

n n

n

f x x x x x xx xn

xx xxn

∞

=

+∞

=

+ = = + + + − + + +

−= − + − =

+

∑

∑


( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( )

3 2

3 3

4 45 3 2 4

2 4

0

sec 0 1

sec tan 0 0

sec sec tan 0 1

5 sec tan sec tan 0 0

5 sec 18 sec tan sec tan 0 5

0 5sec 1 .! 2! 4!

n n

n

f x x f

f x x x f

f x x x x f

f x x x x x f

f x x x x x x f

f x x xxn

∞

=

= =

′ ′= =

′′ ′′= + =

′′′ ′′′= + =

= + + =

= = + + +∑



12. For 0,c = you have;

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( )

2

2

4 2 2

4 44 2 3

5 56 4 2 2 4

3 5 3

0

tan 0 0

sec 0 1

2 sec tan 0 0

2 sec 2 sec tan 0 2

8 sec tan sec tan 0 0

8 2 sec 11 sec tan 2 sec tan 0 16

0 2 16tan! 3! 5!

n n

n

f x x f

f x x f

f x x x f

f x x x x f

f x x x x x f

f x x x x x x f

f x x x xx x xn

∞

=

= =

′ ′= =

′′ ′′= =

′′′ ′′′⎡ ⎤= + =⎣ ⎦

⎡ ⎤= + =⎣ ⎦

⎡ ⎤= + + =⎣ ⎦

= = + + + = +∑ 52 .3 15

x+ +

13. The Maclaurin series for ( ) cosf x x= is ( )( )

2

0

1.

2 !

n n

n

xn

∞

=

−∑

Because ( ) ( )1 sinnf x x+ = ± or cos ,x± you have ( ) ( )1 1nf z+ ≤ for all z. So by Taylor's Theorem,

( )( ) ( )( ) ( )

1110 .

1 ! 1 !

nnn

nxf z

R x xn n

+++≤ = ≤

+ +

Because ( )

1

lim 0,1 !

n

n

xn

+

→∞=

+it follows that ( ) 0nR x → as .n → ∞ So, the Maclaurin series for cos x converges to

cos x for all x.

14. The Maclaurin series for ( ) 2xf x e−= is ( )0

2.

!

n

n

xn

∞

=

−∑

( ) ( ) ( ) 11 22 .nn xf x e++ −= − So, by Taylor's Theorem,

( )( ) ( )( )

( )( )

11 21 12

0 .1 ! 1 !

nn zn nf z e

Rn x x xn n

++ −+ +−

≤ = =+ +

Because ( )( )

( )( )

1 112 2lim lim 0,

1 ! 1 !

n nn

n n

x xn n

+ ++

→∞ →∞

−= =

+ +it follows that ( ) 0nR x → as .n → ∞

So, the Maclaurin Series for 2xe− converges to 2xe− for all x.

15. The Maclaurin series for ( ) sinhf x x= is ( )

2 1

0.

2 1 !

n

n

xn

+∞

= +∑

( ) ( )1 sinhnf x x+ = (or cosh x). For fixed x,

( )( ) ( )( )

( )( )

11 1sinh

0 0 as .1 ! 1 !

nn n

nf z z

R x x x nn n

++ +≤ = = → → ∞

+ +

(The argument is the same if ( ) ( )1 coshnf x x+ = ). So, the Maclaurin series for sinh x converges to sinh x for all x.

16. The Maclaurin series for ( ) coshf x x= is ( )

2

0.

2 !

n

n

xn

∞

=∑

( ) ( )1 sinhnf x x+ = (or cosh x). For fixed x,

( )( ) ( )( )

( )( )

11 1sinh

0 0 as .1 ! 1 !

nn n

nf z z

R x x x nn n

++ +≤ = = → → ∞

+ +

(The argument is the same if ( ) ( )1 coshnf x x+ = ). So, the Maclaurin series for cosh x converges to cosh x for all x.



17. Because ( ) ( ) ( )( )2 31 1 21 1 ,

2! 3!k k k x k k k x

x kx− + + ++ = − + − + you have

( ) ( ) ( )( ) ( )( )( )

( ) ( )

2 3 42 2 3 4

0

2 3 2 3 4 2 3 4 51 1 2 1 2 3 4 5

2! 3! 4!

1 1 .n n

n

x x xx x x x x x

n x

−

∞

=

+ = − + − + − = − + − + −

= − +∑

18. Because ( ) ( ) ( )( )2 31 1 21 1


x kx− + + ++ = − + − +

you have

( ) ( ) ( )( ) ( )( )( ) ( ) ( )4 2 3 4 2 3 4

0

4 5 4 5 6 4 5 6 7 3 !1 1 4 1 4 10 20 35 1

2! 3! 4! 3! !n n

n

nx x x x x x x x x x

n

∞−

=

++ = − + − + = − + − + − = −∑

19. Because ( ) ( ) ( )( )2 31 1 21 1 ,


x kx− + + ++ = − + − + you have

( ) ( )( ) ( )( )( )

( )( ) ( )( )( )

( )

2 31 2

2 3

2 3

1

1 2 3 2 1 2 3 2 5 211 12 2! 3!

1 3 1 3 51

2 2 2! 2 3!

1 3 5 2 11 .

2 !

n

nn

x xx x

x xx

n xn

−

∞

=

⎛ ⎞⎡ + − ⎤ = + + + +⎜ ⎟⎣ ⎦ ⎝ ⎠

= + + + +

⋅ ⋅ −= + ∑

20. Because ( ) ( ) ( )( )2 31 1 21 1


x kx− + + ++ = − + − + you have

( ) ( )( ) ( )( )( )

( )( ) ( )( )( )

( )

61 22 2 4

2 4 62 3

2

1

1 2 3 2 1 2 3 2 5 211 12 2! 3!

1 3 1 3 5112 2 2! 2 3!

1 3 5 2 11

2 !n

nn

xx x x

x x x

nx

n

−

∞

=

⎡ ⎤+ − = − + − +⎣ ⎦

= − + − +

⋅ ⋅ −= + ∑

21. 1 22

2

1 1 12 24

xx

−⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠+ ⎢ ⎥⎣ ⎦and because ( ) ( ) ( )1 2

1

1 1 3 5 2 11 1 ,

2 !

n n

nn

n xx

n

∞−

=

− ⋅ ⋅ −+ = + ∑ you have

( ) ( )( ) ( ) ( )2 2

3 121 1

1 1 3 5 2 1 2 1 1 3 5 2 11 1 11 .2 2 ! 2 2 !4

n n n n

n nn n

n x n xn nx

∞ ∞

+= =

⎡ ⎤− ⋅ ⋅ − − ⋅ ⋅ −⎢ ⎥= + = +⎢ ⎥+ ⎣ ⎦

∑ ∑

22. ( )

3

31 1 1 , 3

8 22x k

x

−⎛ ⎞= + = −⎜ ⎟⎝ ⎠+

( )

( ) ( )( ) ( ) ( )2 3

3 11

3 4 3 4 5 2 !1 1 11 3 1 18 2 2! 2 3! 2 8 2 !2

n nn

n

nx x x xnx

∞

+=

⎧ ⎫ ⎡ + ⎤⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + = + −⎨ ⎬ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ ⎣ ⎦⎪ ⎪⎩ ⎭

∑

23. ( )( ) ( )( ) ( ) ( )

1 2

12 3

2

1 1 , 1 2

1 2 1 2 1 2 1 2 3 2 1 3 5 2 31 11 1 1 12 2! 3! 2 2 !

n nn

n

x x k

nx x x x x x

n

∞+

=

+ = + =

− − − ⋅ ⋅ −+ = + + + + = + + −∑



24. ( ) ( )( ) ( )( )( )

( ) ( )

1 4 2 3

2 3 42 3 4

1

2

1 4 3 4 1 4 3 4 7 411 14 2! 3!

1 3 7 3 7 1114 4 2! 4 3! 4 4!

1 3 7 11 4 5114 4 !

nn

nn

x x x x

x x x x

nx x

n

+∞

=

− − −+ = + + + +

3 ⋅ ⋅ ⋅= + − + − +

− ⋅ ⋅ −= + + ∑

25. Because ( ) ( ) ( )11 2

2

1 1 3 5 2 31 1

2 2 !

n n

nn

n xxxn

+∞

=

− ⋅ ⋅ −+ = + + ∑

you have ( ) ( ) ( )1 221 22

2

1 1 3 5 2 31 1 .

2 2 !

n n

nn

n xxxn

+∞

=

− ⋅ ⋅ −+ = + + ∑

26. Because ( ) ( ) ( )11 2

2

1 1 3 5 2 31 1

2 2 !

n n

nn

n xxxn

+∞

=

− ⋅ ⋅ −+ = + + ∑

you have ( ) ( ) ( )1 331 23

2

1 1 3 5 2 31 1 .

2 2 !

n n

nn

n xxxn

+∞

=

− ⋅ ⋅ −+ = + + ∑

27.

( )

2 3 4 5

0

2 2 2 4 6 82 22 3 4

0 0

1! 2! 3! 4! 5!

21

! 2 ! 2 2 2! 2 3! 2 4!

nx

n

nn

xn

n n

x x x x xe xn

x x x x x xen n

∞

=

∞ ∞

= =

= = + + + + + +

= = = + + + + +

∑

∑ ∑

28.

( ) ( )

2 3 4 5

0

2 3 4 53

0 0

1! 2! 3! 4! 5!

3 1 3 9 27 81 2431 3! ! 2! 3! 4! 5!

nx

n

n n n nx

n n

x x x x xe xn

x x x x x xe xn n

∞

=

∞ ∞−

= =

= = + + + + + +

− −= = = − + − + − +

∑

∑ ∑

29. ( ) ( )1

1

1ln 1 , 0 2

nn

n

xx x

n

∞−

=

−= − < ≤∑

( ) ( ) 1

1

1ln 1 , 1 1

n n

n

xx x

n

−∞

=

−+ = − < ≤∑

30. ( ) ( )

( ) ( )

1

1

1 22

1

1 1ln , 0 2

1ln 1 , 1 1

n n

n

n n

n

xx x

n

xx x

n

−∞

=

−∞

=

− −= < ≤

−+ = − < ≤

∑

∑

31. ( )( )

( ) ( )( )

2 1

0

2 1

0

1sin

2 1 !

1 3sin 3

2 1 !

n n

n

n n

n

xx

n

xx

n

+∞

=

+∞

=

−=

+

−=

+

∑

∑

32. ( )( )

( ) ( )( )

2 1

0

2 1

0

1sin

2 1 !

1sin

2 1 !

n n

n

n n

n

xx

n

xx

nπ

π

+∞

=

+∞

=

−=

+

−=

+

∑

∑

33. ( )( )

( ) ( )( )

( )( )

2 2 4 6

0

2 2 2

0 0

2 4

1cos 1

2 ! 2! 4! 6!

1 4 1 4cos 4

2 ! 2 !

16 25612! 4!

n n

n

n n n n n

n n

x x x xxn

x xx

n n

x x

∞

=

∞ ∞

= =

−= = − + − +

− −= =

= − + −

∑

∑ ∑

34. ( )( )

( ) ( )( )

2

0

2

0

1cos

2 !

1cos

2 !

n n

n

n n

n

xx

n

xx

nπ

π

∞

=

∞

=

−=

−=

∑

∑



35. ( )( )

( ) ( )( )

( )( )

2 2 4

0

23 23 2

0

3

0

3 6

1cos 1

2 ! 2! 4!

1cos

2 !

12 !

12! 4!

n n

n

nn

n

n n

n

x x xxn

xx

n

xn

x x

∞

=

∞

=

∞

=

−= = − + −

−=

−=

= − + −

∑

∑

∑

36. ( )( )

( ) ( )( )

2 1

0

2 133

0

9 153

9 153

1sin

2 1 !

12 sin 2

2 1 !

23! 5!

2 223! 5!

n n

n

nn

n

xx

n

xx

n

x xx

x xx

+∞

=

+∞

=

−=

+

−=

+

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

= − + −

∑

∑

37.

( ) ( )

( )

2 3 4 5

2 3 4 5

3 5 7

3 5 7 2 1

0

12! 3! 4! 5!

12! 3! 4! 5!

2 2 223! 5! 7!

1sinh2

3! 5! 7! 2 1 !

x

x

x x

x x

n

n

x x x xe x

x x x xe x

x x xe e x

x e e

x x x xxn

−

−

−

+∞

=

= + + + + + +

= − + − + − +

− = + + + +

= −

= + + + + =+∑

38.

( ) ( )

2 3

2 3

2 4

2

0

12! 3!

12! 3!

2 222! 4!

2 cos 22 !

x

x

x x

nx x

n

x xe x

x xe x

x xe e

xh x e en

−

−

∞−

=

= + + + +

= − + − +

+ = + + +

= + = ∑

39. ( ) ( )

( ) ( ) ( ) ( ) ( )( )

2

2 4 6 2

0

1cos 1 cos 22

2 2 2 1 21 11 1 12 2! 4! 6! 2 2 !

n n

n

x x

x x x xn

∞

=

= ⎡ + ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥= + − + − − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∑

40. The formula for the binomial series gives ( ) ( ) ( )1 2

1

1 1 3 5 2 11 1 ,

2 !

n n

nn

n xx

n

∞−

=

− ⋅ ⋅ −+ = + ∑ which implies that

( ) ( ) ( )

( )( ) ( )

( )

21 22

1

22

2 1

1

3 5 7

1 1 3 5 2 11 1

2 !1ln 1

1

1 1 3 5 2 12 2 1 !

1 3 1 3 5 .2 3 2 4 5 2 4 6 7

n n

nn

n n

nn

n xx

n

x x dxx

n xx

n n

x x xx

∞−

=

+∞

=

− ⋅ ⋅ −+ = +

+ + =+

− ⋅ ⋅ −= +

+

⋅ ⋅ ⋅= − + − +

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

∑

∫

∑

41. ( )( )

2 23 5 4 62

0

1sin

3! 5! 3! 5! 2 1 !

n n

n

xx x x xx x x x xn

+∞

=

−⎛ ⎞= − + − = − + − =⎜ ⎟ +⎝ ⎠

∑

42. ( )( )

2 12 4 3 5

0

1cos 1

2! 4! 2! 4! 2 !

n n

n

xx x x xx x x xn

+∞

=

−⎛ ⎞= − + − = − + − =⎜ ⎟

⎝ ⎠∑

43. ( ) ( ) ( )

( )

3 5 22 4

0

3! 5! 1sin 1 , 02! 4! 2 1 !

1, 0

n n

n

x x x xx x x xx x n

x

∞

=

− + − −= = − + − = ≠

+

= =

∑



44. ( )( ) ( )

( )( ) ( )

2 1 2

2 20 0

2 ! 2 !arcsin 1 , 02 ! 2 1 2 ! 2 1

1, 0

n n

n nn n

n x n xx xx xn n n n

x

+∞ ∞

= =

= ⋅ = ≠+ +

= =

∑ ∑

45. ( ) ( ) ( )

( ) ( ) ( )

( )

2 3 4 2 3 4 5 6

2 3 4 2 3 4 5 6

3 5 7

23 5 7

1 12! 3! 4! 2! 3! 4! 5! 6!

1 12! 3! 4! 2! 3! 4! 5! 6!

2 2 223! 5! 7!

12 3! 5! 7!

ix

ix

ix ix

n nix ix

ix ix ix x ix x ix xe ix ix

ix ix ix x ix x ix xe ix ix

ix ix ixe e ix

xe e x x xxi

−

−

+−

= + + + + + = + − − + + − −

− − −= − + + + + = − − + + − − +

− = − + − +

−−= − + − + =

( ) ( )1

0sin

2 1 !nx

n

∞

=

=+∑

46. ( )

( )( ) ( )

2 4 6

22 4 6

0

2 2 22 See Exercise 45.2! 4! 6!

11 cos

2 2! 4! 6! 2 !

ix ix

n nix ix

n

x x xe e

xe e x x x xn

−

− ∞

=

+ = − + − +

−+= − + − + = =∑

47. ( )

( )

2 3 4 3 5

3 3 4 4 5 5 52

3 52

3 52

5

sin

12 6 24 6 120

2 6 6 6 120 12 24

3 30

3 30

xf x e x

x x x x xx x

x x x x x x xx x

x xx x

x xP x x x

=

⎛ ⎞⎛ ⎞= + + + + + − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + − + − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + + − +

= + + −

48. ( )

( )

2 4 4 2 4

2 2 3 3 4 4 4

3 4

3 4

4

cos

1 12 6 24 2 24

12 2 6 2 24 4 24

13 6

13 6

xg x e x

x x x x xx

x x x x x x xx

x xx

x xP x x

=

⎛ ⎞⎛ ⎞= + + + + + − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + − + − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + − − +

= + − −

49. ( ) ( )

( )

2 4 2 3 4 5

2 3 3 4 4 5 5 5

2 3 5

2 3 5

5

cos ln 1

12 24 2 3 4 5

2 3 2 4 4 5 6 24

32 6 40

32 6 40

h x x x

x x x x x xx

x x x x x x x xx

x x xx

x x xP x x

= +

⎛ ⎞⎛ ⎞= − + + − + − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − − + +

= − − +

−6 6

−2

P5

f

14

−6 6

−40

g

P4

8

−3 9

−4

h

P5

4



50. ( ) ( )

( )

2 3 4 2 3 4 5

2 3 3 3 4 4 4 4 5 5 5 5 52

2 3 5

2 3 5

5

ln 1

12 6 24 2 3 4 5

2 3 2 2 4 3 4 6 5 4 6 12 24

32 3 40

32 3 40

xf x e x

x x x x x x xx x

x x x x x x x x x x x x xx x

x x xx

x x xP x x

= +

⎛ ⎞⎛ ⎞= + + + + + − + − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + − + + − + − + + − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + + + +

= + + +

51. ( ) sin .1

xg xx

=+

Divide the series for sin x by ( )1 .x+

3 42

3 52 4

2

32

2 3

34

3 4

4 5

4 5

5 56 6

1 0 06 120

6

5 06

5 56 6

56 120

5 56 6

x xx x

x xx x x x

x xxx

x xx x

x x

x x

x x

− + − +

+ + − + + +

+

− −

− −

+

+

− +

− −

( )

( )

3 42

3 42

4

5 56 6

6 6

x xg x x x

x xP x x x

= − + − +

5 5= − + −

52. ( ) .1

xef xx

=+

Divide the series for xe by ( )1 .x+

2 3 4

2 3 4 5

2 3

2 3

3 4

3 4

4 5

4 5

312 3 8

1 12 6 24 120

1

02 6

2 2

3 24

3 338 120

3 38 8

x x x

x x x xx x

xx x

x x

x x

x x

x x

x x

+ − + +

+ + + + + + +

+

+ +

+

− +

− −

+

+

( )

( )

2 3 4

2 3 4

4

312 3 8

312 3 8

x x xf x

x x xP x

= + − + −

= + − +

53.

( )

4 32

3! 3!

sin

x xy x x x

f x x x

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠=

Matches (c)

−3 3

−3

f P5

3

−6 6

−4

g

4

P4

−6 6

−6

fP4

6



54.

( )

3 5 2 41

2! 4! 2! 4!

cos

x x x xy x x

f x x x

⎛ ⎞= − + = − +⎜ ⎟

⎝ ⎠=

Matches (d)

55.

( )

3 22 1

2! 2!x

x xy x x x x

f x xe

⎛ ⎞= + + = + +⎜ ⎟

⎝ ⎠

=

Matches (a)

56. ( )( )

2 3 4 2 2

2

1

11

y x x x x x x

f x xx

= − + = − +

=+

Matches (b)

57. ( ) ( )

( )( )

( )( )( )

( )( )( )

22

0 00

1 2 2

00

1 2 3

0 0

1 2 3

0

11 1

!

11 !

12 3 1 !

12 3 1 !

n nx xt

n

n nx

n

xn n

n

n n

n

te dt dt

n

tdt

n

tn n

xn n

∞−

=

+ +∞

=

+ +∞

=

+ +∞

=

⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟− = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤−= ⎢ ⎥

+⎢ ⎥⎣ ⎦

⎡ ⎤−= ⎢ ⎥

+ +⎢ ⎥⎣ ⎦

−=

+ +

∑∫ ∫

∑∫

∑

∑

58. ( ) ( )

( ) ( )( )

( ) ( )( )

1 333

0 02

1 3 14

2 0

1 3 14

2

1 1 3 5 2 31 1

2 2 !

1 1 3 5 2 38 3 1 2 !

1 1 3 5 2 38 3 1 2 !

n nx x

nn

xn n

nn

n n

nn

n ttt dt dtn

n tttn n

n xxxn n

−∞

=

− +∞

=

− +∞

=

⎡ ⎤− ⋅ ⋅ ⋅ ⋅ ⋅ −⎢ ⎥+ = + +⎢ ⎥⎣ ⎦

⎡ ⎤− ⋅ ⋅ ⋅ ⋅ ⋅ −⎢ ⎥= + +

+⎢ ⎥⎣ ⎦

− ⋅ ⋅ ⋅ ⋅ ⋅ −= + +

+

∑∫ ∫

∑

∑

59. Because ( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 3 4

0

1 1 1 1 1ln 1 , 0 2

1 2 3 4

n n

n

x x x xx x x

n

+∞

=

− − − − −= = − − + − + < ≤

+∑

you have ( ) ( )1

1

1 1 1 1ln 2 1 1 0.6931. 10,001 terms2 3 4

n

n n

∞+

=

= − + − + = − ≈∑

60. Because ( ) ( )( )

2 1 3 5 7

0

1sin ,

2 1 ! 3! 5! 7!

n n

n

x x x xx xn

+∞

=

−= = − + − +

+∑ you have

( ) ( )( ) ( )

0

1 1 1 1sin 1 1 0.8415. 4 terms2 1 ! 3! 5! 7!

n

n n

∞

=

−= = − + − + ≈

+∑

61. Because 2 3

01 ,

! 2! 3!

nx

n

x x xe xn

∞

=

= = + + + +∑

you have ( )2 3

2

0

2 2 21 2 7.3891. 12 terms2! 3! !

n

ne

n

∞

=

= + + + + = ≈∑

62. Because 2 3 4 5

01 ,

! 2! 3! 4! 5!

nx

n

x x x x xe xn

∞

=

= = + + + + + +∑ you have 1 1 1 1 11 12! 3! 4! 5!

e− = − + − + − +

and ( ) ( )1

1

1

11 1 1 1 1 11 1 0.6321. 6 terms2! 3! 4! 5! 7! !

n

n

e ee n

−∞−

=

−−= − = − + − + − + = ≈∑



63. Because

( )( )

( )( )

( )( )

2 2 4 6 8

0

2 22 4 6 8

0

2 13 5 7

0

1cos 1

2 ! 2! 4! 6! 8!

11 cos

2! 4! 6! 8! 2 2 !

11 cos2! 4! 6! 8! 2 2 !

n n

nn n

nn n

n

x x x x xxn

xx x x xxn

xx x x xx n

∞

=

+∞

=

+∞

=

−= = − + − + −

−− = − + − + =

+

−−= − + − + =

+

∑

∑

∑

you have ( )( )

2 1

0 0 0

11 coslim lim 0.2 2 !

n

x x n

xxx n

+∞

→ → =

−−= =

+∑

64. Because

( )( )

( )( )

2 1 3 5 7

0

22 4 6

0

1sin

2 1 ! 3! 5! 7!

1sin 13! 5! 7! 2 1 !

n n

n

n n

n

x x x xx xn

xx x x xx n

+∞

=

∞

=

−= = − + − +

+

−= − + − + =

+

∑

∑

you have ( )( )

2

0 0 0

1sinlim lim 1.2 1 !

n n

x x n

xxx n

∞

→ → =

−= =

+∑

65. Because 2 3

12! 3!

x x xe x= + + + +

( )

2 3 1

01

2! 3! 1 !

nx

n

x x xe xn

+∞

=

− = + + ++∑

and ( )

2

0

1 12! 3! 1 !

x n

n

e x x xx n

∞

=

−= + + +

+∑

you have ( )0 0

1lim lim 1.1 !

x n

x x

e xx n→ →

−= =

+∑

66. Because ( )2 3

ln 12 3x xx x+ = − + −

(See Exercise 29.)

( ) ( )2

0

ln 1 11

2 3 1

n n

n

x xx xx n

∞

=

+ −= − + −

+∑

you have ( ) ( )0 0 0

ln 1 1lim lim 1.

1

n n

x x n

x xx n

∞

→ → =

+ −= =

+∑

67.

( )

( )

( )( )

( )( )

( ) ( )

31 13

0 00

31

00

13 1

0 0

0

!

1!

13 1 !

13 1 !

1 1 11 14 14 3 1 !

n

x

n

n n

n

n n

n

n

n

n

xe dx dx

n

xdx

n

xn n

n n

n n

∞−

=

∞

=

+∞

=

∞

=

⎡ ⎤−⎢ ⎥=⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥=⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥=

+⎢ ⎥⎣ ⎦

−=

+

= − + − + − ++

∑∫ ∫

∑∫

∑

∑

Because ( )

1 0.0001,3 6 1 6!

<⎡ + ⎤⎣ ⎦

you need 6 terms.

( )( )

51 2

00

10.8075

3 1 !

nx

ne dx

n n−

=

−≈ ≈

+∑∫

68. ( )1 43 4 5 3 4 5 61 4 1 4 2

0 00

ln 12 3 4 3 4 2 5 3 6 4x x x x x x xx x dx x dx

⎛ ⎞ ⎡ ⎤+ = − + − + = − + − +⎜ ⎟ ⎢ ⎥⋅ ⋅ ⋅⎝ ⎠ ⎣ ⎦

∫ ∫

Because ( ) ( ) ( ) ( )5 3 41 4

0

1 4 1 4 1 40.0001, ln 1 0.00472.

15 3 8x x dx< + ≈ − ≈∫

69. ( )( )

( )( )( )

( )( )( )

12 2 11 1

0 00 0 00

1 1 1sin2 1 ! 2 1 2 1 ! 2 1 2 1 !

n n nn n

n n n

x xx dx dxx n n n n n

+∞ ∞ ∞

= = =

⎡ ⎤ ⎡ ⎤− − −⎢ ⎥ ⎢ ⎥= = =

+ + + + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∑ ∑ ∑∫ ∫

Because ( )1 7 7! 0.0001,⋅ < you need three terms:

( )1

0

sin 1 11 0.9461. using three nonzero terms3 3! 5 5!

x dxx

= − + − ≈⋅ ⋅∫

Note: You are using 0

sinlim 1.x

xx+→

=



70. ( )( )

( )( )( )

( )( )( )

41 120 0

0

14 1

0 0

0

1cos

2 !

14 1 2 !

14 1 2 !

n n

n

n n

n

n

n

xx dx dx

n

xn n

n n

∞

=

+∞

=

∞

=

⎡ ⎤−⎢ ⎥=⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥=

+⎢ ⎥⎣ ⎦

−=

+

∑∫ ∫

∑

∑

( )( )( )

31 20

0

1cos 0.904523

4 1 2 !

n

nx dx

n n=

−≈ ≈

+∑∫

Because ( ) ( )

1 0.0001,4 4 1 2 4 !

<⎡ + ⎤⎡ ⎤⎣ ⎦⎣ ⎦

you need 4 terms.

71. 2 4 61 2 1 2

0 0

1 23 5 7

2 2 20

arctan 13 5 7

3 5 7

x x x xdx dxx

x x xx

⎛ ⎞= − + − +⎜ ⎟

⎝ ⎠

⎡ ⎤= − + − +⎢ ⎥⎣ ⎦

∫ ∫

Because ( )2 91 9 2 0.0001,< you have

1 2

2 3 2 5 2 7 2 90

arctan 1 1 1 1 12 3 2 5 2 7 2 9 2

0.4872.

x dxx

⎛ ⎞≈ − + − +⎜ ⎟⎝ ⎠

≈

∫

Note: You are using 0

arctanlim 1.x

xx+→

=

72. ( ) ( )

( )( )( )

( )( )( )

4 21 2 120 0

0

1 24 3

0 0

4 30

1arctan

2 1

14 3 2 1

14 3 2 1 2

n n

n

n n

n

n

nn

xx dx dx

n

xn n

n n

+∞

=

+∞

=

∞

+=

⎡ ⎤−⎢ ⎥=

+⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥=

+ +⎢ ⎥⎣ ⎦

−=

+ +

∑∫ ∫

∑

∑

Because ( )( ) 4 3

1 0.00014 3 2 1 2 nn n + <

+ +

when 2,n = you need 2 terms.

( ) ( ) ( )1 2 2

3 70

1 1arctan 0.0412953 1 2 7 3 2

x dx ≈ − ≈⋅∫

73. 0.33 6 9 12 4 7 10 130.3 0.33

0.1 0.10.1

5 51 12 8 16 128 8 56 160 1664x x x x x x x xx dx dx x

⎛ ⎞ ⎡ ⎤+ = + − + − + = + − + − +⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦∫ ∫

Because ( )7 71 0.3 0.1 0.0001,56

− < you need two terms.

( ) ( )0.3 3 4 40.1

11 0.3 0.1 0.3 0.1 0.201.8

x dx ⎡ ⎤+ = − + − ≈⎢ ⎥⎣ ⎦∫

74. ( )4 6

1 22 2 2 2 4 6

0.23 5 70.2 0.22 2 4 60 0

0

1 1 1 1 31 1 1 12 2 2 2 21 1 1 12 2! 3! 2 8 16

1 1 11 12 8 16 6 40 112

x xx x x x x x

x x xx dx x x x dx x

⎛ ⎞ ⎛ ⎞⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠+ = + = + + + + = + − + −

⎡ ⎤⎡ ⎤+ = + − + − = + − + −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫

Because ( )50.20.0001,

40< you need 2 terms.

( )30.2 20

0.21 0.2 0.201333

6x dx+ ≈ + ≈∫



75. ( ) ( )

( )( ) ( )

( )

( ) ( )

( )( )

22

4 1 2 4 3 2 4 3 22 2

0 00 0 0 0

0

1 1 1 2cos

4 32 ! 4 3 2 !2 !2

n n nn n n

n n n

x x xx x dx dx

nn n nn

ππ

π π+ + +∞ ∞ ∞

= = =

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤− − −⎢ ⎥⎢ ⎥ ⎢ ⎥= = =

+ +⎛ ⎞⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑ ∑ ∑∫ ∫

Because ( ) ( )23 22 2 23 10! 0.0001,π ⋅ < you need five terms.

( ) ( ) ( ) ( ) ( )3 2 7 2 11 2 15 2 19 21

0

2 2 2 2 2cos 2 0.7040.

3 14 264 10,800 766,080x x dx

π π π π π⎡ ⎤⎢ ⎥= − + − + ≈⎢ ⎥⎣ ⎦

∫

76. ( ) ( ) ( ) ( )

12 3 4 2 3 4 51 1

0.5 0.50.5

cos 12! 4! 6! 8! 2 2! 3 4! 4 6! 5 8!x x x x x x x xx dx dx x

⎡ ⎤⎛ ⎞= − + − + − = − + − + −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦∫ ∫

Because ( )51 1 0.5 0.0001,201,600

− < you have

( ) ( ) ( ) ( ) ( )1 2 3 4 50.5

1 1 1 1cos 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 0.3243.4 72 2880 201,600

x dx ⎡ ⎤≈ − − − + − − − + − ≈⎢ ⎥⎣ ⎦∫


( ) ( )( )

( )( )

12 2 11 12 2

0 00 0 00

2 3

1 1 11 1 1 12 ! 2 ! 2 1 2 ! 2 12 2 2 2

1 1 1 11 0.3412.2 1 3 2 2! 5 2 3! 72

n n nn nx

n n nn n n

x xe dx dx

n n n n nπ π π π

π

+∞ ∞ ∞−

= = =

⎡ ⎤− − −⎢ ⎥= = =

+ +⎢ ⎥⎣ ⎦

⎛ ⎞≈ − + − ≈⎜ ⎟⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠

∑ ∑ ∑∫ ∫


( ) ( )( )

( ) ( )( )

22 2 12 22 2

1 10 0 1

1

0

2 3 4 5

6 7 8

1 11 1 12 ! 2 ! 2 12 2 2

1 2 112 ! 2 12

1 7 31 127 511 204712 1 3 2 2! 5 2 3! 7 2 4! 9 2 5! 112

8191 32,767 131,0712 6! 13 2 7! 15 2

n nn nx

n nn n

n n

nn

x xe dx dx

n n n

n n

π π π

π

π

+∞ ∞−

= =

+∞

=

⎡ ⎤− −⎢ ⎥= =

+⎢ ⎥⎣ ⎦

− −=

+

⎛ ⎞≈ − + − + −⎜ ⎟⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎝ ⎠

+ − +⋅ ⋅ ⋅ ⋅ ⋅

∑ ∑∫ ∫

∑

9524,287 0.1359.

8! 17 2 9! 19− ≈

⋅ ⋅ ⋅

79. ( ) ( )( )

( )

2 1

0

53

5

1 4cos 2

2 !

223

n n n

n

xf x x x

n

xP x x x

+∞

=

−= =

= − +

∑

The polynomial is a reasonable approximation on the

interval 3 3, .4 4

⎡ ⎤−⎢ ⎥⎣ ⎦

80. ( ) ( )

( )2 3 4 5

5

sin ln 12

7 112 4 48 96

xf x x

x x x xP x

= +

= − + −

The polynomial is a reasonable approximation on the interval ( )0.60, 0.73 .−

3−3

−2

f

2

P5

−4 8

−4

f

P5

4



81. ( )

( ) ( ) ( ) ( ) ( )3 4 5

5

ln , 1

1 1 71 11

24 24 1920

f x x x c

x x xP x x

= =

− − −= − − + −

The polynomial is a reasonable approximation on the interval 1, 2 .4⎡ ⎤⎢ ⎥⎣ ⎦

82. ( )

( ) ( ) ( ) ( ) ( ) ( )

3

2 3 4 5

5

arctan , 1

1 1 1 10.7854 0.7618 1 0.3412 0.0424 1.3025 5.5913

2! 3! 4! 5!

f x x x c

x x x xP x x

= ⋅ =

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤− − − −≈ + − − ⎢ ⎥ − ⎢ ⎥ + ⎢ ⎥ − ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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

83. See Guidelines, page 682.

84. 2 1 0na + = (odd coefficients are zero)

85. The binomial series is ( ) ( ) ( )( )2 31 1 21 1 .

2! 3!k k k k k k

x kx x x− − −

+ = + + + + The radius of convergence is 1.R =

86. (a) Replace x with ( ).x−

(b) Replace x with 3 .x

(c) Multiply series by x. (d) Replace x with 2 ,x then replace x with 2 ,x− and add the two together.

87.

( )

( )

20 0

2 3 4

20 0 0 0 0

2 3 2 4

2 2 3 3 40 0 0 0 0

tan ln 1cos cos

1 1 1tancos cos 2 cos 3 cos 4 cos

tancos cos 2 cos 3 cos 4

g g kxy xkv k v

gx g kx kx kx kxxkv k v v v v

gx gx gx gkx gk xxkv kv v v v

θθ θ

θθ θ θ θ θ

θθ θ θ θ

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥= − − − − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= − + + + +

( )

4

2 3 2 4

2 2 3 3 4 40 0 0

cos

tan2 cos 3 cos 4 cos

gx kgx k gxxv v v

θ

θθ θ θ

+

= + + + +

−2 4

−1

f

P5

3

−2 4

−2

g

3

P5



88.

( ) ( )( )( )( ) ( )

( ) ( )( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

0

23 42

2 2 3 3 4 4

2 2 3 3 4 4

2 3 4 2

2 22 2

160 , 64, , 3216

1 16 32 1 16 323232 64 1 2 3 64 1 2 4 64 1 2

2 2 23 322 64 3 64 16 4 64 16

23 32 3 3264 16 32 16

n n n

n n n nn n

v k g

x xxy x

x x xx

x xx xn n

θ

∞ ∞

− −= =

= ° = = = −

= − − − −

⎡ ⎤⎢ ⎥= − + + +⎢ ⎥⎣ ⎦

= − = −∑ ∑

89. ( )21 , 0

0, 0

xe xf xx

−⎧ ≠⎪= ⎨=⎪⎩

(a)

(b) ( ) ( ) ( ) 21

0 0

0 00 lim lim0

x

x x

f x f efx x

−

→ →

− −′ = =−

21

0

21 2

2 20 0 0

Let lim . Then

1 1 lnln lim ln lim ln lim .

x

x

x

x x x

eyx

e x xy xx x x

−

→

−

+ +→ → →

=

⎛ ⎞ ⎡ ⎤− −⎡ ⎤⎜ ⎟= = − − = = −∞⎢ ⎥⎢ ⎥⎜ ⎟ ⎣ ⎦ ⎣ ⎦⎝ ⎠

So, 0y e−∞= = and you have ( )0 0.f ′ =

(c) ( ) ( ) ( ) ( ) ( ) ( )

2

0

0 0 00 0

! 1! 2!

nn

n

f f x f xx f f x

n

∞

=

′ ′′= + + + = ≠∑ This series converges to f at 0x = only.

90. (a) ( ) ( )2

2

ln 1.

xf x

x

+=

From Exercise 10, you obtain:

( ) ( )2 2 2

20 0

2 4 6 8

8

1 111 1

1 .2 3 4 5

n nn n

n n

x xP

x n n

x x x xP

+∞ ∞

= =

− −= =

+ +

= − + − +

∑ ∑

(b)

0 2

−0.5

1.5

21 3

2

1

−1−2−3x

y



(c) ( ) ( )

( ) ( )

2

20

80

ln 1x

x

tF x dt

t

G x P t dt

+=

=

∫

∫

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

91. By the Ratio Test: ( )

1 !lim lim 01 ! 1

n

nn n

xx nn x n

+

→∞ →∞⋅ = =

+ +which shows that

0 !

n

n

xn

∞

=∑ converges for all x.

92. ( ) ( )

2 3 2 3 3 5 2

0

1ln ln 1 ln 11

2 2 2 ) 2 , 12 3 2 3 3 5 2 1

n

n

x x xx

x x x x x x xx x x x Rn

∞

=

+⎛ ⎞ = + − −⎜ ⎟−⎝ ⎠

⎛ ⎞ ⎛ ⎞= − + − − − − − − = + + + = =⎜ ⎟ ⎜ ⎟ +⎝ ⎠ ⎝ ⎠

∑

( ) ( ) ( )

( )

2 4 61 2 1 2 1 21 1 2 1 1 1 1ln 3 ln 2 1 1 1.0980651 1 2 2 3 5 7 12 80 448

ln 3 1.098612

⎡ ⎤⎛ ⎞+ ⎛ ⎞ ⎢ ⎥= ≈ + + + = + + + ≈⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎣ ⎦≈

93. 5 4 3 60 10

3! 63⎛ ⎞ 5 ⋅ ⋅

= = =⎜ ⎟⎝ ⎠

94. ( )( )2 2 33

2!2− − −⎛ ⎞

= =⎜ ⎟⎝ ⎠

95. ( )( )( )( )0.5 0.5 0.5 1.5 2.5 50.03906254! 1284

⎛ ⎞ − − −= = − = −⎜ ⎟

⎝ ⎠

96. ( )( )( )( )( )1 3 1 3 4 3 7 3 10 3 13 35!5

91 0.12483729

− − − − − −⎛ ⎞=⎜ ⎟

⎝ ⎠−

= ≈ −

97. ( )0

1 k n

n

kx x

n

∞

=

⎛ ⎞+ = ⎜ ⎟

⎝ ⎠∑

Example: ( )2 2

0

21 1 2n

nx x x x

n

∞

=

⎛ ⎞+ = = + +⎜ ⎟

⎝ ⎠∑

98. Assume e p q= is rational. Let N q> and form the following.

( ) ( )

1 1 1 11 12! ! 1 ! 2 !

eN N N

⎡ ⎤− + + + + = + +⎢ ⎥ + +⎣ ⎦

( ) ( ) ( )( ) ( )

( )

2

2

1Set ! 1 1 , a positive integer. But,!

1 1 1 1 1 1!1 ! 2 ! 1 1 2 1 1

1 1 1 1 1 11 , a contradiction.11 1 11 1

1

a N eN

a NN N N N N N N

N N N NNN

⎡ ⎤⎛ ⎞= − + + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤= + + = + + < + +⎢ ⎥

+ + + + + + +⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥= + + + = =⎢ ⎥+ + + ⎛ ⎞⎢ ⎥+⎣ ⎦ ⎢ ⎥− ⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦

…

x 0.25 0.50 0.75 1.00 1.50 2.00

( )F x 0.2475 0.4810 0.6920 0.8776 1.1798 1.4096

( )G x 0.2475 0.4810 0.6924 0.8865 1.6878 9.6063

Review Exercises for Chapter 9 377


99. ( )

( )( )( ) ( ) ( )

20 1 22

2 20 1 2

2 30 1 0 2 1 0 3 2 1

11

xg x a a x a xx x

x x x a a x a x

x a a a x a a a x a a a x

= = + + +− −

= − − + + +

= + − + − − + − − +

Equating coefficients,

0

1 0 1

2 1 0 2

3 2 1 3

4 3 2

01 1

0 10 23, etc.

aa a aa a a aa a a aa a a

=

− = ⇒ =

− − = ⇒ =

− − = ⇒ =

= + =

In general, 1 2.n n na a a− −= + The coefficients are the Fibonacci numbers.

100. Assume the interval is [ ]1, 1 .− Let [ ]1, 1 ,x ∈ −

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

2

2

12

12

1 1 1 , , 1

1 1 1 , 1, .

f f x x f x x f c c x

f f x x f x x f d d x

′ ′′= + − + − ∈

′ ′′− = + − − + − − ∈ −

So, ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

1 12 2

1 12 2

1 1 2 1 1

2 1 1 1 1 .

f f f x x f c x f d

f x f f x f c x f d

′ ′′ ′′− − = + − − +

′ ′′ ′′= − − − − + +

Because ( ) 1f x ≤ and ( ) 1,f x′′ ≤

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 22 21 1 1 12 2 2 22 1 1 1 1 1 1 1 1 3 4.f x f f x f c x f d x x x′ ′′ ′′≤ + − + − + + ≤ + + − + + = + ≤

So, ( ) 2.f x′ ≤

Note: Let ( ) ( )212 1 1.f x x= + − Then ( ) ( )1, 1f x f x′ ′′≤ = and ( )1 2.f ′ =

Review Exercises for Chapter 9

1. 1! 1na

n=

+

2. 2 1nna

n=

+

3. 24 : 6, 5, 4.67,nan

= + …

Matches (a)

4. 4 : 3.5, 3,2nna = − …

Matches (c)

5. ( ) 110 0.3 : 10, 3,nna −= …

Matches (d)

6. ( ) 1236 : 6, 4,

nna

−= − − …

Matches (b)

7. 5 2n

nan+

=

The sequence seems to converge to 5.

5 2lim lim

2lim 5 5

nn n

n

nan

n

→∞ →∞

→∞

+=

⎛ ⎞= + =⎜ ⎟⎝ ⎠

1200

8



8. sin2n

na π=

The sequence seems to diverge (oscillates).

sin : 1, 0, 1, 0, 1, 0,2

nπ− …

9. ( )78lim 3 3

n

n→∞

⎡ ⎤+ =⎢ ⎥⎣ ⎦

Converges

10. 5lim 1 11n n→∞

⎡ ⎤+ =⎢ ⎥+⎣ ⎦

Converges

11. 3

21lim

n

nn→∞

+= ∞

Diverges

12. 1lim 0n n→∞

=

Converges

13. 2lim 01n

nn→∞

=+

Converges

14. ( )

1lim limln 1n n

nn n→∞ →∞

= = ∞

Diverges

15. ( ) ( ) 1lim 1 lim 11

1lim 01

n n

n

n nn n n nn n

n n

→∞ →∞

→∞

+ ++ − = + −

+ +

= =+ +

Converges

16. 1 2

1 21 1lim 1 lim 12

n k

n ke

n k→∞ →∞

⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + =⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

Converges; 2k n=

17. sinlim 0n

nn→∞

=

Converges

18. Let ( )( )

1

lnln

nn n

n n

y b c

b cy

n

= +

+=

( )1lim ln lim ln ln .n nn nn n

y b b c cb c→∞ →∞

= ++

Assume b c≥ and note that the terms

ln ln ln lnn n n n

n n n n n nb b c c b b c c

b c b c b c+

= ++ + +

converge as .n → ∞ Hence na converges.

19. (a)

1

1

2

3

4

5

6

7

8

0.058000 1 , 1, 2, 3,4

0.058000 1 $8100.004

$8201.25$8303.77$8407.56$8512.66$8619.07$8726.80$8835.89

n

nA n

A

AAAAAAA

⎛ ⎞= + =⎜ ⎟⎝ ⎠

⎛ ⎞= + =⎜ ⎟⎝ ⎠

=

=

=

=

=

=

=

…

(b) 40 $13,148.96A =

20. (a) ( )175,000 0.70 , 1, 2,nnV n= = …

(b) ( )55 175,000 0.70 $29,412.25V = ≈

21. (a)

The series diverges ( )32geometric 1 .r = >

(b)

22. (a)

The series converges by the Alternating Series Test. (b)

n 5 10 15 20 25

nS 13.2 113.3 873.8 6648.5 50,500.3

n 5 10 15 20 25

nS 0.3917 0.3228 0.3627 0.3344 0.3564

0 12

−2

2

120

−10

120

0 120

1



23. (a)

The series converges by the Alternating Series Test. (b)

24. (a)

The series converges, by the Limit Comparison Test with 21 .n∑

(b)

25. 0

23

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ Geometric series with 1a = and 2.3

r =

( )1 1 3

1 1 2 3 1 3aS

r= = = =

− −

26. ( )2

0 0

2 24 4 3 123 3

nn

nn n

+∞ ∞

= =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

∑ ∑

See Exercise 25.

27. ( ) ( ) ( ) ( ) ( ) ( )1 0 0

1 1 6 10 8 10 110.6 0.8 0.6 0.6 0.8 0.8 0.6 0.8 5.51 0.6 1 0.8 10 4 10 2 2

n n n n

n n n

∞ ∞ ∞

= = =

⎡ ⎤+ = + = + = ⋅ + ⋅ = =⎣ ⎦ − −∑ ∑ ∑

28. ( )( )

( )

0 0 0

2 1 2 1 13 1 2 3 1 2

1 1 1 1 1 11 3 1 21 2 3 2 2 3 3 4

n n

n n nn n n n

∞ ∞ ∞

= = =

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − + − + − + = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

∑ ∑ ∑

29. (a) ( ) ( )( )0

0.09 0.09 0.0009 0.000009 0.09 1 0.01 0.0001 0.09 0.01 n

n

∞

=

= + + + = + + + = ∑

(b) 0.09 10.091 0.01 11

= =−

30. (a) ( ) ( )0

0.64 0.64 0.0064 0.000064 0.64 1 0.01 0.0001 0.64 0.01 n

n

∞

=

= + + + = + + + = ∑

(b) 0.64 640.641 0.01 99

= =−

31. Diverges. Geometric series with 1a = and 1.67 1.r = >

n 5 10 15 20 25

nS 0.4597 0.4597 0.4597 0.4597 0.4597

n 5 10 15 20 25

nS 0.8333 0.9091 0.9375 0.9524 0.9615

120

−0.1

1

0 120

1



32. Converges. Geometric series with 1a = and 0.67 1.r = <

33. Diverges. nth-Term Test. lim 0.nn

a→∞

≠

34. Diverges. nth-Term Test, 23lim .n

na

→∞=

35.

( ) ( ) ( )

( ) ( ) ( )

( )

1

2

2

0

8

0.7 8 0.7 8 16 0.7

8 16 0.7 16 0.7 16 0.7

16 18 16 0.7 8 45 meters1 0.7 3

n

n

n

D

D

D∞

=

=

= + =

= + + + + +

= − + = − + =−∑

36. ( )

( )

39

0

40

42,000 1.055

42,000 1 1.055$5,737,435.79

1 1.055

n

nS

=

=

−= ≈

−

∑

37. (See Exercise 116 in Section 9.2)

( )

( )( )12

0.06 2

0.06 12

11

300 1$7630.70

1

rt

r

P eA

e

e

e

−=

−

−= ≈

−

38. (See Exercise 116 in Section 9.2)

( )

12

12 10

12 1 112

12 0.035125 1 1 $17,929.060.035 12

trA Pr

⎡ ⎤⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ≈⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

39. ( )43 31

1

ln 1 1 1ln lim 03 9 9 9

b

b

xx x dxx x

∞ −

→∞

⎡ ⎤= − − = + =⎢ ⎥⎣ ⎦∫

By the Integral Test, the series converges.

40. 3 44 31 1

1 1

n n nn

∞ ∞

= =

=∑ ∑

Divergent p-series, 3 14

p = <

41. 2 21 1 1

1 1 1 1

n n nn n n n

∞ ∞ ∞

= = =

⎛ ⎞− = −⎜ ⎟⎝ ⎠

∑ ∑ ∑

Because the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges.

42. 2 21 1 1

1 1 1 12 2n n

n n nn n

∞ ∞ ∞

= = =

⎛ ⎞− = −⎜ ⎟⎝ ⎠

∑ ∑ ∑

Because the first series is a convergent p-series while the second series is a convergent geometric series, the difference converges.

43.

( )1

65 1

6 5 1 6lim1 5

n

n

n

nn

∞

=

→∞

−

⎡ − ⎤=⎢ ⎥

⎣ ⎦

∑

By a limit comparison test with the divergent harmonic

series 1

1,n n

∞

=∑ the series diverges.

44. 3

1

3 3 2

3

3

3lim lim 11 3

n

n n

nn n

n n n nn n n

∞

=

→∞ →∞

+

+= =

+

∑

By a limit comparison test with the divergent p-series

1 21

1 ,n n

∞

=∑ the series diverges.

45.

( )

31

3 3 2

3 2 3

12

1 2lim lim 11 2

n

n n

n n

n n nn n n

∞

=

→∞ →∞

+

+= =

+

∑

By a limit comparison test with the convergent p-series

3 21

1 ,n n

∞

=∑ the series converges.

46. ( )( ) ( )

1

12

1 2 1lim lim 11 2

n

n n

nn n

n n n nn n

∞

=

→∞ →∞

++

+ + += =

+

∑

By a limit comparison test with 1

1,n n

∞

=∑ the series

diverges.

47. ( )( )1

1 3 5 2 12 4 6 2n

nn

∞

=

⋅ ⋅ −⋅ ⋅∑

( )( )

1 3 5 2 1 3 5 2 1 1 12 4 6 2 2 4 2 2 2 2n

n nan n n n

⋅ ⋅ − −⎛ ⎞= = ⋅ >⎜ ⎟⋅ ⋅ −⎝ ⎠

Because 1 1

1 1 12 2n nn n

∞ ∞

= =

=∑ ∑ diverges (harmonic series), so

does the original series.



48. Because 1

13n

n

∞

=∑ converges,

1

13 5n

n

∞

= −∑ converges by the

Limit Comparison Test.

49. ( )5

1

1 n

n n

∞

=

−∑ converges by the Alternating Series Test.

51lim 0

n n→∞= and

( )1 5 5

1 1 .1

n na ann

+ = < =+

50. ( ) ( )2

1

1 11

n

n

nn

∞

=

− ++∑ converges by the Alternating Series

Test. 21lim 01n

nn→∞

+=

+and if

( ) ( ) ( )( )

2

22 2

2 11 , 01 1

x xxf x f xx x

− + −+ ′= = < ⇒+ +

terms

are decreasing. So, 1 .n na a+ <

51. ( )2

2

13

n

n

nn

∞

=

− −−∑ converges by the Alternating Series Test.

2lim 03n

nn→∞

=−

and if

( ) ( ) ( )( )

2

22 2

3, 0

3 3

nnf x f xn n

− +′= = < ⇒

− −terms are

decreasing. So, 1 .n na a+ <

52. ( )1

11

n

n

nn

∞

=

−+∑

11

2 1

lim 01

n n

n

n na an n

nn

+

→∞

+= ≤ =

+ +

=+

By the Alternating Series Test, the series converges.

53. Diverges by the nth-Term Test.

lim 1 03n

nn→∞

= ≠−

54. Converges by the Alternating Series Test.

( )1

3 ln 1 3 ln 3 ln, lim 01n n

n

n n na an n n+

→∞

+= < = =

+

55. 3 1 3 1 3lim lim 12 5 2 5 2

nn

n n

n nn n→∞ →∞

− −⎛ ⎞ ⎛ ⎞= = >⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Diverges by Root Test.

56. 4 4 4lim lim 17 1 7 1 7

nn

n n

n nn n→∞ →∞

⎛ ⎞ ⎛ ⎞= = <⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

Converges by Root Test.

57. 21 nn

n

e

∞

=∑

( )

( )

( )( )

21

21

2

2 2 1

2 1

1lim lim

1lim

1 1lim

0 1 0 1

nn

n n nn

n

n nn

nn

a n ea ne

e n

e n

ne n

+

→∞ →∞ +

+ +→∞

+→∞

+= ⋅

+=

+⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= = <

By the Ratio Test, the series converges.

58. 1

!n

n

ne

∞

=∑

( )11

1 !lim lim

!

1lim

nn

nn nn

n

na ea e n

ne

++→∞ →∞

→∞

+= ⋅

+= = ∞

By the Ratio Test, the series diverges.

59. 31

2n

n n

∞

=∑

( ) ( )

1 3 31

3 32 2lim lim lim 2

21 1

nn

nn n nn

a n na n n

++

→∞ →∞ →∞= ⋅ = =

+ +

By the Ratio Test, the series diverges.

60. ( )( )1

1 3 5 2 12 5 8 3 1n

nn

∞

=

⋅ ⋅ −⋅ ⋅ −∑

( )( )( )( )

( )( )

1 1 3 2 1 2 1 2 5 3 1 2 1 2lim lim lim2 5 3 1 3 2 1 3 2 1 3 2 3

n

n n nn

n n na na n n n n+

→∞ →∞ →∞

⋅ − + ⋅ − += ⋅ = =

⋅ − + ⋅ − +

By the Ratio Test, the series converges.



61. (a) Ratio Test: ( )( )( )

11 1 3 5 1 3 3lim lim lim 1,

5 53 5

nn

nn n nn

na na nn

++

→∞ →∞ →∞

+ +⎛ ⎞⎛ ⎞= = = <⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Converges

(b)

(c)

(d) The sum is approximately 3.75.

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

(c)

(d) The sum is approximately 0.0714.

63. (a) 21 1 1

NN

dxx x N

∞∞ ⎡ ⎤= − =⎢ ⎥⎣ ⎦∫

(b) 5 4 41 1 1

4 4NN

dxx x N

∞∞ ⎡ ⎤= − =⎢ ⎥⎣ ⎦∫

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

64. No. Let 23937.5,na

n= then 75 0.7.a = The series 2

1

3937.5

n n

∞


n 5 10 15 20 25

nS 2.8752 3.6366 3.7377 3.7488 3.7499

n 5 10 15 20 25

nS 0.0871 0.0669 0.0734 0.0702 0.0721

N 5 10 20 30 40

21

1N

n n=∑ 1.4636 1.5498 1.5962 1.6122 1.6202

21

Ndx

x∞

∫ 0.2000 0.1000 0.0500 0.0333 0.0250

N 5 10 20 30 40

51

1N

n n=∑ 1.0367 1.0369 1.0369 1.0369 1.0369

51

Ndx

x∞

∫ 0.0004 0.0000 0.0000 0.0000 0.0000

−1

0 12

4

0 120

0.3



65. ( ) ( )( ) ( )( ) ( )( ) ( )

3

3

3

3

0 13 0 3

9 0 927 0 27

x

x

x

x


−

−

−

−

= =

′ ′= − = −

′′ ′′= =

′′′ ′′′= − = −

( )2 3

3

2 3

9 271 32! 3!

9 91 32 2

x xP x x

x x x

= − + −

= − + −

66. ( )

( )

( )

( )

2

2

2 2 4

tan 14

sec 24

2 sec tan 44

4 sec tan 2 sec 164

f x x f

f x x f

f x x x f

f x x x x f

π

π

π

π

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

⎛ ⎞′ ′= − =⎜ ⎟⎝ ⎠

⎛ ⎞′′ ′′= − = −⎜ ⎟⎝ ⎠

⎛ ⎞′′′ ′′′= + − =⎜ ⎟⎝ ⎠

( )2 3

381 2 2

4 4 3 4P x x x xπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

67. Because ( )99

950.001,

180 9!π

<⋅

use four terms.

( ) ( ) ( )3 5 7

3 5 7

95 95 9595 95sin 95 sin 0.99594180 180 180 3! 180 5! 180 7!

π π ππ π⎛ ⎞° = ≈ − + − ≈⎜ ⎟⎝ ⎠

68. Because ( )60.750.001,

6!< use three terms.

( ) ( ) ( )2 40.75 0.75cos 0.75 1 0.7319

2! 4!≈ − + ≈

69. Because ( )150.750.001,

15< use 14 terms.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 4 5 6 140.75 0.75 0.75 0.75 0.75 0.75ln 1.75 0.75 0.559062

2 3 4 5 6 14≈ − + − + − + − ≈

70. Because ( )50.250.001,

5!< use five terms.

( ) ( ) ( )2 3 40.25 0.25 0.25 0.25

1 0.25 0.7792! 3! 4!

e− ≈ − + − + ≈



71. ( ) cos , 0f x x c= =

( )( ) ( )( )

( ) ( ) ( ) ( )

11

11

1 !

11 !

nn

n

nn

n

f zR x x

n

xf z R xn

++

++

=+

≤ ⇒ ≤+

(a) ( ) ( )( )

10.50.001

1 !

n

nR xn

+

≤ <+

This inequality is true for 4.n =

(b) ( ) ( )( )

110.001

1 !

n

nR xn

+

≤ <+


(c) ( ) ( )( )

10.50.0001

1 !

n

nR xn

+

≤ <+


(d) ( ) ( )12 0.00011 !

n

nR xn

+

≤ <+


72. ( )

( )

( )

( )

2 4

4

2 4 6

6

2 4 6 8 10

10

cos

12! 4!

12! 4! 6!

12! 4! 6! 8! 10!

f x x

x xP x

x x xP x

x x x x xP x

=

= − +

= − + −

= − + − + −

73. 0 10

n

n

x∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Geometric series which converges only if 10 1x < or

10 10.x− < <

74. ( )0

2 n

nx

∞

=∑

Geometric series which converges only if 2 1x < or 1 12 2.x− < <

75. ( ) ( )( )2

0

1 2

1

n n

n

x

n

∞

=

− −

+∑

( ) ( )( )

( )( ) ( )

1 1 21

2

1 2 1lim lim

2 1 2

2

n nn

n nn nn

x nuu n x

x

+ ++

→∞ →∞

− − += ⋅

+ − −

= −

1R = Center: 2 Because the series converges when 1x = and when

3,x = the interval of convergence is [ ]1, 3 .

76. ( )1

3 2 nn

n

xn

∞

=

−∑

( )( )

111 3 2

lim lim1 3 2

3 2

nnn

nnn nn

xu nu n x

x

+++

→∞ →∞

−= ⋅

+ −

= −

13

R =

Center: 2

Because the series converges at 53

and diverges at 7,3

the

interval of convergence is 5 7, .3 3⎡ ⎞

⎟⎢⎣ ⎠

77. ( )0

! 2 n

nn x

∞

=

−∑

( ) ( )( )

11 1 ! 2

lim lim! 2

nn

nn nn

n xuu n x

++

→∞ →∞

+ −= = ∞

−

which implies that the series converges only at the center 2.x =

78. ( )0 0

2 22 2

n n

nn n

x x∞ ∞

= =

− −⎛ ⎞= ⎜ ⎟⎝ ⎠

∑ ∑

Geometric series which converges only if

2 1 or 0 4.2

x x−< < <

−10 10

−10

f

P4

P10P6

10



79. ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )( )

( ) ( )( )( )

( ) ( )( )

( )

2

20

12 1 2 1

2 211 0

1 2

210

1 12 2 2 22 2

2 21 10 0

14 !

1 2 1 2 2

4 ! 4 1 !

1 2 2 2 1

4 1 !

1 2 2 2 1 1 2 21

4 1 ! 4 1 !

nn

nn

n nn n

n nn n

n n

nn

n nn nn

n nn n

xyn

n x n xy

n n

n n xy

n

n n x n x xx y xy x yn n

∞

=

+− +∞ ∞

+= =

+∞

+=

+ ++ +∞ ∞

+ += =

= −

− − +′ = =

⎡ + ⎤⎣ ⎦

− + +′′ =

⎡ + ⎤⎣ ⎦

− + + − +′′ ′+ + = + + −

⎡ + ⎤ ⎡ + ⎤⎣ ⎦ ⎣ ⎦

∑

∑ ∑

∑

∑ ∑ ( )

( ) ( )( )( )

( ) ( )( )

( )( )

( ) ( )( )( )

( )( )

( ) ( )( )

( )( )

2 2

20

11 2 2

2 2 21 10

12 2

2 210

1 22 2

2 210

4 !

2 2 2 1 1 2 2 11

4 !4 1 ! 4 1 !

1 2 2 2 1 1 114 !4 1 !

1 4 1 114 !4 1 !

n

nn

n nn n

nn nn

nn n

nnn

nn n

nnn

n

n n nx

nn n

n nx

nn

nx

nn

+∞

=

+∞+ +

+ +=

+∞+

+=

+∞+

+=

⎡ ⎤+ + − + −⎢ ⎥= − + +⎢ ⎥⎡ + ⎤ ⎡ + ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤− + + +⎢ ⎥= + −⎢ ⎥⎡ + ⎤⎣ ⎦⎣ ⎦

⎡ ⎤− +⎢ ⎥= + −⎢ ⎥⎡ + ⎤⎣ ⎦⎣ ⎦

∑

∑

∑

∑

( )( )

( )( )

12 2

2 20

1 1 11 04 ! 4 !

nn n

n nn

xn n

+∞+

=

⎡ ⎤−⎢ ⎥= + − =⎢ ⎥⎣ ⎦

∑

80. ( )

( ) ( ) ( ) ( )( )

( ) ( )( )( )

( ) ( )( )( )

( ) ( )( )

( )

( )

2

0

12 1 2 1

11 0

1 2

10

1 12 2 2 2 1 2

1 10 0 0

1

32 !

3 2 3 2 22 ! 2 1 !

3 2 2 2 12 1 !

3 2 2 2 1 1 3 2 2 1 33 3

2 1 ! 2 1 ! 2 !

1 3

n n

nn

n nn n

n nn n

n n

nn

n n nn n n n n

n n nn n n

n

xy

n

n x n xy

n n

n n xy

n

n n x n x xy xy y

n n n

∞

=

+− +∞ ∞

+= =

+∞

+=

+ + + + +∞ ∞ ∞

+ += = =

+

−=

− − +′ = =

+

− + +′′ =

+

− + + − + −′′ ′+ + = + +

+ +

−=

∑

∑ ∑

∑

∑ ∑ ∑

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

11 2 2 2 2 1 2

0 0 0

11 2 2 2 2

0 0

11 2 2 2 2

0 0

1 1 2 1 2

1

2 2 1 3 1 32 ! 2 ! 2 !

1 3 1 32 1 1

2 ! 2 !

1 3 1 32

2 ! 2 !

1 3 1 32

2 !

n nn n n n n n

n n nn n n

n nn n n n

n nn n

n nn n n n

n nn n

n nn n n n

nn

n x x xn n n

x xn

n n

x xn

n n

x xn

n

++ + + +∞ ∞ ∞

= = =

++ + +∞ ∞

= =

++ + +∞ ∞

= =

+ + +∞

=

+ − −+ +

− −= ⎡− + + ⎤ +⎣ ⎦

− −= − +

− −= +

∑ ∑ ∑

∑ ∑

∑ ∑

∑ ( )

( ) [ ]

11

1 2

1

22 1 ! 2

1 32 2 0

2 !

nn

n n n

nn

nn n

xn n

n

∞

−=

+∞

=

⋅−

−= − + =

∑

∑

81. ( )

10 0

2 2 33 1 3 1

2 23 3 3

n n

nn n

ax x r

x x∞ ∞

+= =

= =− − −

⎛ ⎞ =⎜ ⎟⎝ ⎠

∑ ∑



82. ( ) ( )

( )1

0 0

3 3 2 3 22 1 2 1 2 1

1 332 2 2

nn n

nn n

ax x x r

xx∞ ∞

+= =

= = =+ + − − −

−⎛ ⎞− =⎜ ⎟⎝ ⎠

∑ ∑

83. ( ) 2 .3

g xx

=−

Power series 0

23 3

n

n

x∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

Derivative: ( )1 1

1 1 0

2 1 2 2 13 3 3 9 3 9 3

n n n

n n n

x x xn n n− −∞ ∞ ∞

= = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

84. Integral: ( )( )

1

10

1 31 2

n n

nn

xn

+∞

+=

−+∑

85. ( )

2 3

0

2 4 8 2 1 3 3 31 , , 3 9 27 3 1 2 3 3 2 2 2

n

n

xx x xx x

∞

=

⎛ ⎞ ⎛ ⎞+ + + + = = = −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠∑

86. ( ) ( ) ( ) ( )( )

( ) ( )

2 3

0

31 1 88 2 3 3 3 82 8 4 1 3 4

32 32 , 1, 74 3 1

n

n

xx x x

x

x x

∞

=

⎡− − ⎤− − + − − − + = =⎢ ⎥ − ⎡− − ⎤⎣ ⎦ ⎣ ⎦

= = −+ − +

∑

87. ( )( )( )( )

( )( ) ( ) ( )

( ) ( ) ( )0

1 22

0

sin

cos

sin

cos ,

3 4sin

!

1 3 42 2 3 2 3 22 2 4 2 2! 4 2 !

nn

n

nn n

n

f x x

f x x

f x x

f x x

f x xx

n

xx x

n

π

ππ π

∞

=

+∞

=

=

′ =

′′ = −

′′′ = −

⎡ − ⎤⎣ ⎦=

− ⎡ − ⎤⎛ ⎞ ⎛ ⎞ ⎣ ⎦= − − − − + =⎜ ⎟ ⎜ ⎟⋅⎝ ⎠ ⎝ ⎠

∑

∑

88. ( )( )( )( )

( ) ( ) ( )

( ) ( ) ( )( )

2 3 4

0

11 2

1

cos

sin

cos

sin

4 4 2 2 2 2 2cos! 2 2 4 2 2! 4 2 3! 4 2 4! 4

1 42 12 4 1 !

nn

n

nn n

n

f x x

f x x

f x x

f x x

f xx x x x x

n

xx

n

π π π π π π

ππ

∞

=

++⎡ ⎤⎣ ⎦∞

=

=

′ = −

′′ = −

′′′ =

− ⎡ + ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎣ ⎦= = + + − + − + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⋅ ⋅ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎡ ⎤− ⎡ + ⎤⎛ ⎞ ⎣ ⎦⎢ ⎥= + + +⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

∑

∑

89. ( )( ) ( )ln 3 ln 33x xx e e= = and because

0,

!

nx

n

xen

∞

=

= ∑ you have

( ) [ ] [ ] [ ]2 3 42 3 4

0

ln 3 ln 3 ln 3ln 33 1 ln 3 .

! 2! 3! 4!

nx

n

x x xxx

n

∞

=

= = + + + + +∑



90. ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( )

3 2

3 3

4 5 3 2 4

2 4

0

csc

csc cot

csc csc cot

5 csc cot csc cot

5 csc 15 csc cot csc cot

2 2 1 5csc 1! 2! 2 4! 2

nn

n

f x x

f x x x

f x x x x

f x x x x x

f x x x x x x

f xx x x

nπ π π π∞

=

=

′ = −

′′ = +

′′′ = − −

= + +

⎡ − ⎤ ⎛ ⎞ ⎛ ⎞⎣ ⎦= = + − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

91. ( )

( )

( )

( )( ) ( )( ) ( ) ( )

2

3

4

0 0 0

1

1

2

6 ,

1 1 ! 11 1 , 2 0! !

n nnn

n n n

f xx

f xx

f xx

f xx

f x n xx x

x n n

∞ ∞ ∞

= = =

=

′ = −

′′ =

′′′ = −

− + − += = = − + − < <∑ ∑ ∑

92. ( )

( )

( )

( )

( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

1 2

1 2

3 2

5 2

4 7 2

2 3 4

2 5 8 110

1

2

12

1 12 2

1 1 32 2 2

1 1 3 5 ,2 2 2 2

4 4 4 4 1 3 4 1 3 5 42

! 2 2 2! 2 3! 2 4!

4 1 1 3 52

2

nn

nn

f x x

f x x

f x x

f x x

f x x

f x x x x xx

n

x

−

−

−

−

∞

=

+

=

′ =

⎛ ⎞⎛ ⎞′′ = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞⎛ ⎞′′′ = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

− − − ⋅ − ⋅ ⋅ −= = + − + − +

− − ⋅ ⋅= + +

∑

( )( )3 1

2

2 3 42 !

n

nn

n xn

∞

−=

− −∑

93. ( ) ( ) ( )( )

( ) ( )( ) ( )( )

( ) ( )

2 3

2 31 5

12 32 3

2 32

1 1 21 1

2! 3!1 5 4 5 1 5 4 5 9 5

1 15 2! 3!

1 4 9 14 5 61 1 4 1 4 9 2 61 1 15 5 2! 5 3! 5 5 ! 5 25 125

k

n n

nn

k k x k k k xx kx

x xxx

n xx x x xx x xn

+∞

=

− − −+ = + + + +

− − −+ = + + + +

− ⋅ ⋅ −⋅ ⋅ ⋅= + − + − = + + = + − + −∑

94. ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( )( ) ( ) ( )

( )( ) ( ) ( ) ( )( )

3

4

5

6

74

85

2 3 4 5

30 0

1

3 1

12 1

60 1

360 1

2520 1

1 2 ! 1 2 11 12 60 360 25201 32! 3! 4! 5! 2 ! 21

n nn n

n n

h x x

h x x

h x x

h x x

h x x

h x x

n x n n xx x x xxnx

−

−

−

−

−

−

∞ ∞

= =

= +

′ = − +

′′ = +

′′′ = − +

= +

= − +

− + − + += − + − + − + = =

+∑ ∑



95. ( ) ( )

( ) ( ) ( )

1

1

1 1

1 1

1ln 1 , 0 2

5 4 15 1ln 1 1 0.22314 4

nn

nn

n nn

n n

xx x

n

n n

∞+

=

∞ ∞+ +

= =

−= − < ≤

⎛ − ⎞⎛ ⎞ = − = − ≈⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

∑ ∑

96. ( ) ( )

( ) ( )

( )

1

1

1

1

1

1

1ln 1 , 0 2

6 5 16ln 15

11 0.18235

nn

nn

n

n

nn

n

xx x

n

n

n

∞+

=

∞+

=

∞+

=

−= − < ≤

⎛ − ⎞⎛ ⎞ = − ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

= − ≈

∑

∑

∑

97.

( )0

1 2

0 0

,!

1 2 1 1.6487! 2 !

nx

nn

nn n

xe xn

en n

∞

=

∞ ∞

= =

= −∞ < < ∞

= = ≈

∑

∑ ∑

98.

( )0

2 3

0 0

,!

2 3 2 1.9477! 3 !

nx

n

n n

nn n

xe xn

en n

∞

=

∞ ∞

= =

= −∞ < < ∞

= = ≈

∑

∑ ∑

99. ( ) ( )

( ) ( )

2

0

2

20

cos 1 ,2 !

2 2cos 1 0.78593 3 2 !

nn

n

nn

nn

xx xn

n

∞

=

∞

=

= − −∞ < < ∞

⎛ ⎞ = − ≈⎜ ⎟⎝ ⎠

∑

∑

100. ( ) ( )

( ) ( )

2 1

0

2 10

sin 1 ,2 1 !

1 1sin 1 0.32723 3 2 1 !

nn

n

nn

n

xx xn

n

+∞

=

∞

+=

= − −∞ < < ∞+

⎛ ⎞ = − ≈⎜ ⎟ +⎝ ⎠

∑

∑

101. The series in Exercise 45 converges very slowly because the terms approach 0 at a slow rate.

102. ( )

( ) ( )( ) ( )( )( ) ( ) ( )

1 233

33 6 9 3

2

1 11 ,21

1 2 3 2 1 2 3 2 5 2 1 1 3 2 111 12 2! 3! 2 2 !

nn

nn

x kx

nxx x x xn

−

∞

=

= + = −+

− − − − − − ⋅ −= − + + + = − + ∑

103. (a) ( ) ( )( ) ( )( ) ( )( ) ( )

2

2

2

2

0 12 0 24 0 48 0 8

x

x

x

x


= =

′ ′= =

′′ ′′= =

′′′ ′′′= =

( )2 3

2 34 8 41 2 1 2 22! 3! 3x xP x x x x x= + + + = + + +

(b) ( )

( )

2

0 0

2 3

2,

! !41 2 23

nnx x

n n

xxe en n

P x x x x

∞ ∞

= =

= =

= + + +

∑ ∑

(c)

( )

2 2

2 3

1 12! 2!

41 2 23

x x x xe e x x

P x x x x

⎛ ⎞⎛ ⎞⋅ = + + + + + +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

= + + +

104. (a) ( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

4 4

5 5

6 6

7 7

2 3 4 5 6 73

sin 2 0 02 cos 2 0 2

4 sin 2 0 08 cos2 0 8

16 sin 2 0 0

32 cos 2 0 32

64 sin 2 0 0

128 cos 2 0 1280 0 32 0 128 4 4sin 2 0 2 22! 3! 4! 5! 6! 7! 3 15


f x x f

f x x f

f x x f

f x x fx x x x x xx x x x

= =

′ ′= =

′′ ′′= − =

′′′ ′′′= − = −

= =

= =

= − =

= − = −

8= + + − + + + − + = − + 5 78

315x x− +



(b) ( )( )

( ) ( )( )

( ) ( ) ( )

2 1

0

2 1 3 5 7

0

3 5 73 5 7

1sin

2 1 !

1 2 2 2 2sin 2 2

2 1 ! 3! 5! 7!

8 32 128 4 4 82 26 120 5040 3 15 315

n n

n

n n

n

xx

n

x x x xx x

n

x x xx x x x x

+∞

=

+∞

=

−=

+

−= = − + − +

+

= − + − + = − + − +

∑

∑

(c) 3 5 7 2 4 6

3 3 5 5 5 7 7 7 7

3 5 73 5 7

sin 2 2 sin cos

2 16 120 5040 2 24 720

22 6 24 12 120 720 144 240 5040

2 2 4 4 4 82 23 15 315 3 15 315

x x x

x x x x x xx

x x x x x x x x xx

x x xx x x x x

=

⎛ ⎞⎛ ⎞= − + − + − + − +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − − + + + + − − − − +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞= − + − + = − + − +⎜ ⎟

⎝ ⎠

105. ( )( )

( )( )

( )( )( )

( )( )( )

2 1

0

2

0

2 1

00 0

2 1

0

1sin

2 1 !

1sin2 1 !

1sin2 1 2 1 !

12 1 2 1 !

n n

n

n n

n

xn nx

n

n n

n

tt

n

ttt n

tt dtt n n

xn n

+∞

=

∞

=

+∞

=

+∞

=

−=

+

−=

+

⎡ ⎤−⎢ ⎥=

+ +⎢ ⎥⎣ ⎦

−=

+ +

∑

∑

∑∫

∑

106. ( )( )

( )( )

( )( ) ( )

( )( ) ( )

2

0

20

1

200 0

1

20

1cos

2 !

1cos

2 2 2 !

1cos

2 2 2 ! 1

12 2 ! 1

n n

n

n n

nn

xn nx

nn

n n

nn

tt

n

ttn

tt dtn n

xn n

∞

=

∞

=

+∞

=

+∞

=

−=

−=

⎡ ⎤−⎢ ⎥=

+⎢ ⎥⎣ ⎦

−=

+

∑

∑

∑∫

∑

107. ( )

( ) ( )

( ) ( )

( ) ( )( )

( )( )

0

1

0

0

1 1

2 200 00

1 11

11ln 11 1

ln 1 11

ln 1 1 1

1 1

n n

n

n n

n

n n

n

xn nn nx

n n

tt

tt dt

t n

t tt n

t t xdt

t n n

∞

=

+∞

=

∞

=

+ +∞ ∞

= =

= −+

−+ = =

+ +

+ −=

+

⎡ ⎤+ − −⎢ ⎥= =⎢ ⎥+ +⎣ ⎦

∑

∑∫

∑

∑ ∑∫

108. 0

1

1

1

01 10

!

1!

1!

1! !

nt

n

nt

n

t n

n

xt n nx

n n

ten

ten

e tt n

e t xdtt n n n n

∞

=

∞

=

−∞

=

∞ ∞

= =

=

− =

−=

⎡ ⎤−= =⎢ ⎥⋅ ⋅⎣ ⎦

∑

∑

∑

∑ ∑∫

109. 3 5 7 9

5 2 9 2 13 2 17 2

0

arctan3 5 7 9

arctan3 5 7 9

arctanlim 0x

x x x xx x

x x x x xxx

xx+→

= − + − + −

= − + − + −

=

By L'Hôpital's Rule,

2

20 0 0

1arctan 21lim lim lim 0.

1 12

x x x

x xxxx

x+ + +→ → →

⎛ ⎞⎜ ⎟+⎝ ⎠= = =

+⎛ ⎞⎜ ⎟⎝ ⎠

110. 3 5 7

2 4 6

0

1 3 1 3 5arcsin2 3 2 4 5 2 4 6 7

arcsin 1 3 1 3 512 3 2 4 5 2 4 6 7

arcsinlim 1x

x x xx x

x x x xx

xx→

⋅ ⋅ ⋅= + + + +

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅

= + + + +⋅ ⋅ ⋅ ⋅ ⋅ ⋅

=

By L'Hôpital's Rule,

2

0 0

1arcsin 1lim lim 1.

1x x

x xx→ →

⎛ ⎞⎜ ⎟

−⎝ ⎠= =



Problem Solving for Chapter 9

1. (a) ( )0

1 1 1 1 2 1 31 2 4 13 9 27 3 3 1 2 3

n

n

∞

=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑

(b) 1 20, , , 1, etc.3 3

(c) 0

1 2lim 1 1 1 03 3

n

nn n

C∞

→∞ =

⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

∑

2. (a) Let 0ε > be given. 2lim nn

a L→∞

= means there exists 1M such that 2na L ε− < for 1.n M> 2 1lim nn

a L+→∞

= means

there exists 2M such that 2 1na L ε+ − < for 2.n M> Let { }1 2max 2 , 2 1 .M M M= + Then for ,n M> and 2n m=

even, you have 1 1 22 2 .mm M M m M a L ε> > ⇒ > ⇒ − < And for 1, 2mn M n +> = odd, you have

1 2 2 2 12 2 1 .m nM M m M a L ε+ +> > + ⇒ > ⇒ − < So, lim .nn

a L→∞

=

(b)

( )

1 1

21

32

11, 1 .1

1 1 31 1 1.51 1 1 2

1 1 71 1 1.41 1 3 2 5

nn

a aa

aa

aa

+= = ++

= + = + = =+ +

= + = + = =+ +

4

5

6

7

8

17 1.4161241 1.41402999 1.4142970239 1.414201169577 1.414216408

a

a

a

a

a

= =

= ≈

= ≈

= ≈

= ≈

Using mathematical induction, you can show that the odd terms are increasing and the even terms are decreasing. Both sequences are bounded in [ ]1, 2 . So, both sequences converge.

Let 2lim .nn

a L→∞

= Then 2 2lim ,nn

a L+→∞

= and

21

1 1 1 1 1 4 31 1 1 1 11 3 2 3 21 2 3 21 1 1

1 1 1

n nn

n n nn n

n n n

a aaa a aa a

a a a

++

+ += + = + = + = + = + =

+ + +⎡ ⎤ ⎡ ⎤ ⎛ ⎞+ ++ + + ⎜ ⎟⎢ ⎥ ⎢ ⎥+ + +⎣ ⎦ ⎣ ⎦ ⎝ ⎠

22 2

2

4 33 2

nn

n

aaa+

+⇒ =

+

So, 24 3 2 4 2.3 2

LL L LL

+= ⇒ = ⇒ =

+Similarly, 2 1lim 2.n

na +

→∞= So by part (a), lim 2n

na L

→∞= =

3. Let ( )2 2 2 2

1

1 1 1 1 .1 3 52 1n

Sn

∞

=

= = + + +−

∑

Then 2

2 2 2 2

2 2

2

2 2 2 2

1 1 1 16 1 2 3 4

1 12 41 1 1 11 .2 2 3 2 6

S

S S

π

π

= + + + +

= + + +

⎛ ⎞⎡ ⎤= + + + + = + ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

So, 2 2 2 21 3 .

6 4 6 6 4 8S π π π π⎛ ⎞= − = =⎜ ⎟

⎝ ⎠

Problem Solving for Chapter 9 391


4. If there are n rows, then ( )1 .2n

n na

+=

For one circle, 1 11 3 3 11 and .3 2 6 2 3

a r⎛ ⎞

= = = =⎜ ⎟⎜ ⎟⎝ ⎠

For three circles, 2 2 23 and 1 2 3 2a r r= = +

21 .

2 2 3r =

+

For six circles, 3 3 36 and 1 2 3 4a r r= = +

31 .

2 3 4r =

+

Continuing this pattern, ( )

1 .2 3 2 1nr n

=+ −

( ) ( )( )

( )( )

2

2

2

11Total Area22 3 2 1

12 2 3 2 1

1lim2 4 8

n n

n

nn

n nr a

n

n nA

n

A

π π

π

π π→∞

⎛ ⎞ += = ⎜ ⎟⎜ ⎟+ −⎝ ⎠

+=

⎡ ⎤+ −⎣ ⎦

= ⋅ =

5. (a) Position the three blocks as indicated in the figure. The bottom block extends 1 6 over the edge of the

table, the middle block extends 1 4 over the edge of the bottom block, and the top block extends 1 2 over the edge of the middle block.

The centers of gravity are located at

bottom block: 1 1 16 2 3− = −

middle block: 1 1 1 16 4 2 12+ − = −

top block: 1 1 1 1 5 .6 4 2 2 12+ + − =

The center of gravity of the top 2 blocks is

11 5 2 ,12 12 6

⎛ ⎞− + =⎜ ⎟⎝ ⎠

which lies over the bottom

block. The center of gravity of the 3 blocks is 1 1 5 3 03 12 12

⎛ ⎞− − + =⎜ ⎟⎝ ⎠

which lies over the table.

So, the far edge of the top block lies 1 1 1 116 4 2 12+ + = beyond the edge of the table.

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

lies 1

12

n

i i=∑ from the edge of the table. Using 4 blocks,

4

1

1 1 1 1 1 252 2 4 6 8 24i i=

= + + + =∑

which shows that the top block extends beyond the table.

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

1 1

1 1 1 .2 2i ii i

∞ ∞

= =

= = ∞∑ ∑

6. (a)

( ) ( ) ( )( )( ) ( )

2 3 4 5

3 6 4 7 2 5 8

3 6 2 23

1 2 3 2 3

1 2 3

11 1 2 3 1 2 31

nna x x x x x x

x x x x x x x x

x x x x x xx

= + + + + + +

= + + + + + + + + + + +

= + + + + + = + +−

∑

1R = because each series in the second line has 1.R =

(b) ( ) ( )( ) ( ) ( )

( )( ) ( )

1 10 1 1 0 1

10 1 1

1 10 1 1 0 1 1

1 1 1

111

1

n p p pn p

p p p pp

p p pp p p

a x a a x a x a x a x

a x a x x a x x

a a x a x x a a x a xx

R

− +−

−−

− −− −

= + + + + + + +

= + + + + + + + + +

= + + + + + = + + +−

=

∑

(Assume all 0.)na >

1 32

12

r1

1 3 r2

r2r22

1 3 r3

r3r32

16

512

1112

14

16

12

0



7. ( ) ( ) ( )1

1

12 3 4 2

n

n

a b a bb a ban

+∞

=

− + + −− + − + = ∑

If ( ) ( ) ( )1 1

1 1

1 2 1,

2

n n

n n

aa b a

n n

+ +∞ ∞

= =

− −= =∑ ∑ converges conditionally.

If ( ) ( )1

1 1

1,

2 2

n

n n

a b a ba bn n

+∞ ∞

= =

− + −≠ +∑ ∑ diverges.

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

8. (a)

( )

2

01

02

0

12! !

!

2 !

nx

nn

x

nn

x x x

n

x xe xn

xxen

xxe dx xe e Cn n

∞

=∞ +

=∞ +

=

= + + + =

=

= − + =+

∑

∑

∑∫

Letting 0,x = you have 1.C = Letting 1,x =

( ) ( )0 1

1 1 11 .2 ! 2 2 !n n

e en n n n

∞ ∞

= =

− + = = ++ +∑ ∑

So, ( )1

1 1.2 ! 2n n n

∞

=

=+∑

(b) Differentiating, ( )0

1.

!

nx x

n

n xxe e

n

∞

=

++ = ∑

Letting 1,x = 0

12 5.4366.!n

nen

∞

=

+= ≈∑

9.

( ) ( ) ( ) ( )

2

4 122 2

1212

12!

12! 6!

0 1 12!0 665,28012! 6! 6!

x

x

xe x

x xe x

ff

= + + +

= + + + + +

= ⇒ = =

10. Let 2

1 20

sin sin, ,x xa dx a dxx x

π π

π= = −∫ ∫

3

3 2

sin , etc.xa dxx

π

π= ∫ Then,

1 2 3 40

sin .x dx a a a ax

∞= − + − +∫

Because lim 0nn

a→∞

= and 1 ,n na a+ < this series converges.

11. (a) If [ ] 22

11, ln lnln

p dx xx x

∞ ∞= =∫ diverges.

If

( )

( ) ( )1 1

2

ln ln 211, lim1 1ln

p p

p b

bp dx

p px x

− −∞

→∞

⎡ ⎤> = ⎢ − ⎥

− −⎢ ⎥⎣ ⎦∫

converges. If 1,p < diverges.

(b) ( )2

4 4

1 1 12 lnlnn n n nn n

∞ ∞

= =

=∑ ∑ diverges by part (a).

12. (a)

[ ]

1

2

3

4

5

6

3.01.732052.175332.274932.296722.30146

1 13lim See part (b) for proof.2n

n

aaaaaa

a→∞

=

≈

≈

≈

≈

≈

+=

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

2 1 1.a a a a a a a= + = > = Now assume 1.n na a −> Then

1

1

1 .

n n

n n

n n

a a a a

a a a a

a a

−

−

+

+ > +

+ > +

>

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

1 .a a a= < Now assume .na a< Then

na a> and 1 1a − > implies

( ) ( )2

2

1

1 1

.

n

n

n

n n

a a a

a a a

a a a

a a a a +

− >

− >

> +

> + =

So, the sequence converges to some number L. To find L, assume 1 :n na a L+ ≈ ≈

2 2 0

1 1 4 .2

L a L L a L L L a

aL

= + ⇒ = + ⇒ − − =

± +=

So, 1 1 4 .2

aL + +=

13. Let

( ) ( )11 1

.

as .

1 1.

nn n

nn n nn n n

b a r

b a r a r Lr n

Lr rr

=

= = ⋅ → → ∞

< =

By the Root Test, nb∑ converges nna r⇒ ∑ converges.

Problem Solving for Chapter 9 393


14. (a) ( ) 1 1 2 1 3 1 4 1 5 111

1 0

1

3

4

5

1 1 1 1 1 12 2 2 2 221 12

1 918 8

9 1 118 4 811 1 458 32 3245 1 4732 16 32

nnn

S

S

S

S

S

∞

− + − + −+ −=

= + + + + +

= =

= + =

= + =

= + =

= + =

∑

(b) ( )

( ) ( )

( )( )

11

1 11 1 1 1

2 12

2 2

n nnn

n nnn

aa

+ −+

+ ++ + − + −

−= =

This sequence is 1 1, 2, , 2,8 8

… which diverges.

(c) ( ) ( ) ( )

1

1 1 1

1 1 1 1 122 2 2 2 2

n

n n n nn n n+ − − −

⎛ ⎞= = → <⎜ ⎟⎜ ⎟

⋅⎝ ⎠ ⋅

converges because ( ){ }1 1 12 , 2, , 2,2 2

n…− =

and 1 2 1n → and 2 1.n →

15. (a) ( )

( )0

2

1 1 0.010.99 1 0.01

1 0.01 0.01

1.010101

n

n

∞

=

= =−

= + + +

=

∑

(b) ( )

( )0

2

1 1 0.020.98 1 0.02

1 0.02 0.02

1 0.02 0.00041.0204081632

n

n

∞

=

= =−

= + + +

= + + +

=

∑

16. 6

7

8

9

10

130 70 40 240240 130 70 440440 240 130 810810 440 240 14901490 810 440 2740

SSSS

S

= + + =

= + + =

= + + =

= + + =

= + + =

17. (a)

1 21

1 1Height 2 12 3

1 12 -series, 12n

p pn

∞

=

⎡ ⎤= + + +⎢ ⎥⎣ ⎦

⎛ ⎞= = ∞ = <⎜ ⎟⎝ ⎠

∑

(b) 1

1 1 14 1 42 3 n

Sn

π π∞

=

⎡ ⎤= + + + = = ∞⎢ ⎥⎣ ⎦∑

(c) 3 2 3 2

3 21

4 1 113 2 3

4 1 converges.3 n

W

n

π

π∞

=

⎡ ⎤= + + +⎢ ⎥⎣ ⎦

= ∑

18. (a) 1 1

1 1 12 2n nn n

∞ ∞

= =

=∑ ∑ diverges (harmonic)

(b) Let ( ) sin .f x x= By the Mean Value Theorem,

( ) ( ) ( )( )cos ,

f x f y f c x y

c x y x y

′− = −

= − ≤ −

where c is between x and y. So,

( )

1 10 sin sin2 2 1

1 1 12 2 1 2 2 1

n n

n n n n

⎛ ⎞ ⎛ ⎞≤ −⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

≤ − =+ +

Because ( )1

12 2 1n n n

∞

= +∑ converges, the Comparison

Theorem tells us that

1

1 1sin sin2 2 1n n n

∞

=

⎡ ⎤⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎢ ⎥+⎝ ⎠ ⎝ ⎠⎣ ⎦∑ converges.

Documents

CHAPTER 9 Infinite Series · Infinite Series Section 9.1 Sequences.....258 Section 9.2 Series and Convergence .....272 Section 9.3 ... Section 9.5 Alternating Series.....306 Section