CHAPTER 9 Infinite Series - Meyers' Mathmeyersmath.com/.../2014/01/Solutions-Larson-Ch.-9.pdf · 236 Chapter 9 Infinite Series 47. does not exist (oscillates between 1 and 1), diverges.
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The graph seems to indicate that the sequence convergesto 1. Analytically,
limn→�
an � limn→�
n � 1
n� lim
x→� x � 1
x� lim
x→� 1 � 1.
−1
−1
12
3 44.
The graph seems to indicate that the sequence convergesto 0. Analytically,
limn→�
an � limn→�
1
n3�2 � limx→�
1
x3�2 � 0.
−1 12
−1
2
45.
The graph seems to indicate that the sequence diverges.Analytically, the sequence is
Hence, does not exist.limn→�
an
�an� � �0, �1, 0, 1, 0, �1, . . .�.
12−1
−2
2 46.
The graph seems to indicate that the sequence convergesto 3. Analytically,
limn→�
an � limn→�
�3 �12n� � 3 � 0 � 3.
−1 12
−1
4
236 Chapter 9 Infinite Series
47.
does not exist (oscillates betweenand 1), diverges.�1
limn→�
��1�n� nn � 1� 48.
does not exist, (alternates between0 and 2), diverges.
limn→�
1 � ��1�n 49. convergeslimn→�
3n2 � n � 4
2n2 � 1�
32
,
50. convergeslimn→�
3�n
3�n � 1� 1, 51.
Thus, converges.limn→�
an � 0,
�1
2n�
32n
�52n
. . . 2n � 1
2n<
12n
an �1 � 3 � 5 . . . �2n � 1�
�2n�n
52. The sequence diverges. To prove this analytically, we use mathematical induction to show that Clearly, Assume that Then,
Since
we have which shows that for all n.an ≥ �32�n�1
ak�1 ≥ �32�k
,
2k � 1k � 1
≥32
,
� ak
2k � 1k � 1
≥ �32�
k�1�2k � 1k � 1 �. ak�1 �
1 � 3 � 5 . . . �2k � 1��2k � 1��k � 1�!
ak ≥ �32�k�1
.a1 � 1 ≥ �32�0
� 1.an ≥ �32�n�1
.
53. convergeslimn→�
1 � ��1�n
n� 0, 54. convergeslim
n→� 1 � ��1�n
n2 � 0,
55.
converges
(L’Hôpital’s Rule)
� limn→�
32 �
1n� � 0,
limn→�
ln�n3�
2n� lim
n→� 32
ln�n�
n56.
converges
(L’Hôpital’s Rule)
� limn→�
1
2n� 0,
limn→�
ln�n
n� lim
n→� 1�2 ln n
n
57. convergeslimn→�
�34�
n
� 0, 58. convergeslimn→�
�0.5�n � 0, 59.
� �, diverges
limn→�
�n � 1�!
n!� lim
n→� �n � 1�
60. convergeslimn→�
�n � 2�!
n!� lim
n→�
1n�n � 1� � 0, 61.
converges � limn→�
1 � 2nn2 � n
� 0,
limn→�
�n � 1n
�n
n � 1� � limn→�
�n � 1�2 � n2
n�n � 1�
62.
converges
limn→�
� n2
2n � 1�
n2
2n � 1� � limn→�
�2n2
4n2 � 1� �
12
, 63. converges
�p > 0, n ≥ 2�
limn→�
n p
en � 0,
Section 9.1 Sequences 237
64.
Let
(L’Hôpital’s Rule)
or,
Therefore limn→�
n sin 1n
� 1.limx→�
sin�1�x�
1�x� lim
y→0� sin� y�
y� 1.
limx→�
x sin 1x
� limx→�
sin�1�x�
1�x� lim
x→� ��1�x2� cos�1�x�
�1�x2 � limx→�
cos 1x
� cos 0 � 1
f �x� � x sin 1x.
an � n sin 1n
65.
where convergesu � k�n,
limn→�
�1 �kn�
n
� limu→0
�1 � u�1�uk � ek
an � �1 �kn�
n
66. convergeslimn→�
21�n � 20 � 1,
67.
converges (because isbounded)
�sin n�
limn→�
sin n
n� lim
n→� �sin n�1
n� 0, 68. convergeslim
n→� cos �n
n2 � 0, 69. an � 3n � 2
70. an � 4n � 1 71. an � n2 � 2 72. an ���1�n�1
n2 73. an �n � 1n � 2
74. an ���1�n�1
2n�275. an � 1 �
1n
�n � 1
n76.
�2n�1 � 1
2n
an � 1 �2n � 1
2n 77. an �n
�n � 1��n � 2�
81. an � �2n�!, n � 1, 2, 3, . . . 82. an � �2n � 1�!, n � 1, 2, 3, . . . 83.
monotonic; bounded�an� < 4,
an � 4 �1n
< 4 �1
n � 1� an�1,
84. Let Then
Thus, f is increasing which implies is increasing.
bounded�an� < 3,
�an�
f � �x� �6
�x � 2�2.f �x� �3x
x � 2.
78. an �1n! 79.
���1�n�1 2nn!
�2n�!
an ���1�n�1
1 � 3 � 5 . . . �2n � 1�80. an �
xn�1
�n � 1�!
238 Chapter 9 Infinite Series
85.
Hence,
True; monotonic; bounded�an� ≤ 18,
an ≥ an�1.
n
2n�2 ≥ n � 1
2�n�1��2
2n�3n ≥ 2n�2�n � 1�
2n ≥ n � 1
n ≥ 1
n ≥ 1
2n ≥?
n � 1
2n�3n ≥?
2n�2�n � 1�
n
2n�2 ≥? n � 1
2�n�1��2 86.
Not monotonic; bounded�an� ≤ 0.7358,
a3 � 0.6694
a2 � 0.7358
a1 � 0.6065
an � ne�n�2
87.
Not monotonic; bounded�an� ≤ 1,
a3 � �13
a2 �12
a1 � �1
an � ��1�n �1n� 88.
Not monotonic; bounded�an� ≤ 23,
a3 � �827
a2 �49
a1 � �23
an � ��23�
n
89.
Monotonic; bounded�an� ≤ 23,
an � �23�
n
> �23�
n�1
� an�1
90.
Monotonic; notbounded
limn→�
an � �,
an � �32�
n
< �32�
n�1
� an�1 91.
Not monotonic; bounded�an� ≤ 1,
a4 � 0.8660
a3 � 1.000
a2 � 0.8660
a1 � 0.500
an � sin�n�
6 � 92.
Not monotonic; bounded�an� ≤ 1,
a3 � cos�3�
2 � � 0
a2 � cos � � �1
a1 � cos �
2� 0
an � cos�n�
2 �
93.
Not monotonic; bounded�an� ≤ 1,
a4 � �0.1634
a3 � �0.3230
a2 � �0.2081
a1 � 0.5403
an �cos n
n94.
Not monotonic, bounded�an� ≤ 1,
a6 �sin�6�
6� �0.0466a3 �
sin�3�3
� 0.0470
a5 �sin�5�
5� �0.1918a2 �
sin�2�2
� 0.4546
a4 �sin�4�
4� �0.1892a1 �
sin�1�1
� 0.8415
an �sin�n
n
Section 9.1 Sequences 239
95. (a)
Therefore, converges.
(b)
limn→�
�5 �1n� � 5
−1
−1
12
7
�an�
� an�1 ⇒ �an�, monotonic
an � 5 �1n
> 5 �1
n � 1
�5 �1n� ≤ 6 ⇒ �an�, bounded
an � 5 �1n
96. (a)
Therefore, converges.
(b)
limn→�
�4 �3n� � 4
−10
12
8
�an�
an � 4 �3n
< 3 �4
n � 1� an�1 ⇒ monotonic
�4 �3n� < 4 ⇒ bounded
an � 4 �3n
97. (a)
Therefore, converges.
(b)
limn→�
13�1 �
13n�� �
13
−1
−0.1
12
0.4
�an�
� an�1 ⇒ �an�, monotonic
an �13�1 �
13n� <
13�1 �
13n�1�
�13�1 �13n�� <
13
⇒ �an�,bounded
an �13�1 �
13n� 98. (a)
Therefore, converges.
(b)
limn→�
�4 �12n� � 4
−1 12
−1
6
�an�
� an�1 ⇒ �an�, monotonic
an � 4 �12n > 4 �
12n�1
�4 �12n� ≤ 4.5 ⇒ �an�, bounded
an � 4 �12n
99. has a limit because it is a bounded, monotonicsequence. The limit is less than or equal to 4, and greaterthan or equal to 2.
2 ≤ limn→�
an ≤ 4
�an� 100. The sequence could converge or diverge. If is increasing, then it converges to a limit less than orequal to 1. If is decreasing, then it could converge(example: ) or diverge (example: ).an � �nan � 1�n
�an�
�an��an�
101.
(a) No, the sequence diverges,
The amount will grow arbitrarily large over time.
limn→�
An � �.�An�
An � P�1 �r
12�n
(b)
A10 � 9421.11A5 � 9208.15
A9 � 9378.13A4 � 9166.14
A8 � 9335.34A3 � 9124.32
A7 � 9292.75A2 � 9082.69
A6 � 9250.35A1 � 9041.25
P � 9000, r � 0.055, An � 9000�1 �0.055
12 �n
240 Chapter 9 Infinite Series
102. (a)
(b) (c) A240 � 32,912.28A60 � 6480.83
A6 � 605.27
A5 � 503.76
A4 � 402.51
A3 � 301.50
A2 � 200.75
A1 � 100.25
A0 � 0
An � 100�401��1.0025n � 1� 103. (a) A sequence is a function whose domain is the set ofpositive integers.
(b) A sequence converges if it has a limit. See thedefinition.
(c) A sequence is monotonic if its terms are nondecreas-ing, or nonincreasing.
(d) A sequence is bounded if it is bounded belowfor some and bounded above
for some M�.�an ≤ MN��an ≥ N
104. The first sequence because everyother point is below the x-axis.
105. an � 10 �1n
106. Impossible. The sequenceconverges by Theorem 9.5.
107. an �3n
4n � 1108. Impossible. An unbounded sequence diverges.
109. (a)
(b)
(c) limn→�
�0.8�n�2.5� � 0
A4 � $1.024 billion
A3 � $1.28 billion
A2 � $1.6 billion
A1 � $2 billion
An � �0.8�n �2.5� billion 110.
P5 � $19,938.91
P4 � $19,080.30
P3 � $18,258.66
P2 � $17,472.40
P1 � $16,720.00
Pn � 16,000�1.045�n
111. (a)
(b) In 2008, and endangered species.a18 � 979n � 18
1500
13500
5
an � �6.60n2 � 151.7n � 387 112. (a)
(b) For 2008, and million.a18 � 7646n � 18
7000
130
2
an � 244.2n � 3250
113.
(a)
(b) Decreasing
(c) Factorials increase more rapidly than exponentials.
�1,562,500
567
�1,000,000,000
362,880
a9 � a10 �109
9!
an �10n
n!114.
limn→�
�1 �1n�
n
� e
a6 � 2.5216
a5 � 2.4883
a4 � 2.4414
a3 � 2.3704
a2 � 2.2500
a1 � 2.0000
an � �1 �1n�
n
Section 9.1 Sequences 241
115.
Let
Since we have Therefore,lim
n→� n�n � 1.
y � e0 � 1.ln y � 0,
ln y � limn→�
�1n
ln n� � limn→�
ln n
n� lim
n→� 1�nn
� 0
y � limn→�
n1�n.
a6 � 6�6 � 1.3480
a5 � 5�5 � 1.3797
a4 � 4�4 � 1.4142
a3 � 3�3 � 1.4422
a2 � �2 � 1.4142
a1 � 11�1 � 1
�an� � � n�n � � �n1�n� 116. Since
there exists for each
an integer N such that for every
Let and we have,
or
for each n > N.
0 < sn < 2L�L < sn � L < L,�sn � L� < L,
� � L > 0
n > N.�sn � L� < �
� > 0,
limn→�
sn � L > 0,
117. True 118. True 119. True 120. True
121.
(a)
(b)
b10 �8955
b5 �85
b9 �5534
b4 �53
b8 �3421
b3 �32
b7 �2113
b2 �21
� 2
b6 �138
b1 �11
� 1
bn �an�1
an
, n ≥ 1
a12 � 89 � 55 � 144 a6 � 5 � 3 � 8
a11 � 55 � 34 � 89 a5 � 3 � 2 � 5
a10 � 34 � 21 � 55 a4 � 2 � 1 � 3
a9 � 21 � 13 � 34 a3 � 1 � 1 � 2
a8 � 13 � 8 � 21 a2 � 1
a7 � 8 � 5 � 13 a1 � 1
an�2 � an � an�1
(c)
(d) If then
Since we have
Since and thus is positive, �1 ��5
2� 1.6180.bn,an,
�1 ± �1 � 4
2�
1 ± �52
0 � 2 � � 1
� 1 � 2
1 � �1�� � .
limn→�
bn � limn→�
bn�1,
limn→�
�1 �1
bn�1� � .lim
n→� bn � ,
�an � an�1
an
�an�1
an
� bn
� 1 �an�1
an
1 �1
bn�1� 1 �
1an�an�1
122.
The limit of the sequence appears to be In fact, this sequence is Newton’s Method applied to f �x� � x2 � 2.�2.
x10 � 1.414214 x5 � 1.414214
x9 � 1.414214 x4 � 1.414214
x8 � 1.414114 x3 � 1.414216
x7 � 1.414214 x2 � 1.41667
x6 � 1.414214 x1 � 1.5
x0 � 1, xn �12
xn�1 �1
xn�1, n � 1, 2, . . .
242 Chapter 9 Infinite Series
123. (a) (b)
(c) We first use mathematical induction to show that clearly So assume Then
Now we show that is an increasing sequence. Since and
Since is a bounding increasing sequence, it converges to some number by Theorem 9.5.
⇒ �L � 2��L � 1� � 0 ⇒ L � 2 �L �1�
limn→�
an � L ⇒ �2 � L � L ⇒ 2 � L � L2 ⇒ L2 � L � 2 � 0
L,�an�
an ≤ an�1.
an ≤ �an � 2
an2 ≤ an � 2
an2 � an � 2 ≤ 0
�an � 2��an � 1� ≤ 0
an ≤ 2,an ≥ 0�an�
ak�1 ≤ 2.
�ak � 2 ≤ 2
ak � 2 ≤ 4
ak ≤ 2.a1 ≤ 2.an ≤ 2;
a5 � �2 � �2 � �2 � �2 � �2 � 1.9976
a4 � �2 � �2 � �2 � �2 � 1.9904
a3 � �2 � �2 � �2 � 1.9616
a2 � �2 � �2 � 1.8478
an � �2 � an�1, n ≥ 2, a1 � �2a1 � �2 � 1.4142
124. (a) (b)
(c) We first use mathematical induction to show that clearly So assume Then
Now we show that is an increasing sequence. Since and
Since is a bounded increasing sequence, it converges to some number Thus,
⇒ �L � 3��L � 2� � 0 ⇒ L � 3 �L �2�
�6 � L � L ⇒ 6 � L � L2 ⇒ L2 � L � 6 � 0
L: limn→�
an � L.�an�
an ≤ an�1.
an ≤ �an � 6
an2 ≤ an � 6
an2 � an � 6 ≤ 0
�an � 3��an � 2� ≤ 0
an ≤ 3,an ≥ 0�an�
ak�1 ≤ 3.
�6 � ak ≤ 3
6 � ak ≤ 9
ak ≤ 3.a1 ≤ 3.an ≤ 3;
a5 � �6 � �6 � �6 � �6 � �6 � 2.9996
a4 � �6 � �6 � �6 � �6 � 2.9974
a3 � �6 � �6 � �6 � 2.9844
a2 � �6 � �6 � 2.9068
an � �6 � an�1, n ≥ 2, a1 � �6a1 � �6 � 2.4495
Section 9.1 Sequences 243
125. (a) We use mathematical induction to show that
[Note that if then and if then ] Clearly,
Before proceeding to the induction step, note that
So assume Then
is increasing because
an ≤ an�1.
an ≤ �an � k
an2 ≤ an � k
an2 � an � k ≤ 0
�an �1 � �1 � 4k
2 ��an �1 � �1 � 4k
2 � ≤ 0
�an�
an�1 ≤1 � �1 � 4k
2.
�an � k ≤ �1 � �1 � 4k2
� k
an � k ≤1 � �1 � 4k
2� k
an ≤1 � �1 � 4k
2.
�1 � �1 � 4k2
� k �1 � �1 � 4k
2.
1 � �1 � 4k
2� k � 1 � �1 � 4k
2 �2
1 � �1 � 4k
2� k �
1 � 2�1 � 4k � 1 � 4k4
2 � 2�1 � 4k � 4k � 2 � 2�1 � 4k � 4k
a1 � �k ≤�1 � 4k
2≤
1 � �1 � 4k2
.
an ≤ 3.k � 6,an ≤ 3,k � 2,
an ≤1 � �1 � 4k
2.
(b) Since is bounded and increasing, it has a limit
(c) implies that
Since L > 0, L �1 � �1 � 4k
2.
⇒ L �1 ± �1 � 4k
2.
⇒ L2 � L � k � 0
L � �k � L ⇒ L2 � k � L
limn→�
an � L
L.�an�
126. (a)
The terms of are decreasing, and those of are increasing. They both seem to approach the same limit.
—CONTINUED—
�bn��an�
b5 � �a4b4 � 5.9777a5 �a4 � b4
2� 5.9777
b4 � �a3b3 � 5.9777a4 �a3 � b3
2� 5.9777
b3 � �a2b2 � 5.9777a3 �a2 � b2
2� 5.9777
b2 � �a1b1 � 5.9667a2 �a1 � b1
2� 5.9886
b1 � �a0b0 � �10�3� � 5.4772a1 �a0 � b0
2�
10 � 32
� 6.5
a0 � 10, b0 � 3
244 Chapter 9 Infinite Series
126. —CONTINUED—
(b) For we need to show Since
Since
Since Thus, we have shown that
Now assume Since
Since
Since
Thus, we have shown that
(c) converges because it is decreasing and bounded below by 0. converges because it is increasing and bounded above by
132. The entire rectangle has area 2 because the height is 1and the base is The squares alllie inside the rectangle, and the sum of their areas is
Thus, ��
n�1 1n2 < 2.
1 �122 �
132 �
142 � . . . .
1 �12 �
14 � . . . � 2.
133. Let H represent the half-life of the drug. If a patient receives n equal doses of P units each of this drug,administered at equal time interval of length t, the total amount of the drug in the patient’s system at the time the last dose is administered is given by where One time interval after the last dose is administered is given by Two time intervals after the last dose is administered is given by and so on. Since as where s is an integer.s →�,k < 0, Tn�s→ 0
and are either both odd or both even. If both even, then
If both odd,
Proof by induction that the formula for is correct. It is true for Assume that the formula is valid for
If is even, then and
The argument is similar if is odd.k
�k2 � 2k
4�
�k � 1�2 � 14
.
f �k � 1� � f �k� �k2
�k2
4�
k2
f �k� � k2�4kk.n � 1.
f �n�
� xy.
f �x � y� � f �x � y� ��x � y�2 � 1
4�
�x � y�2 � 14
f �x � y� � f �x � y� ��x � y�2
4�
�x � y�2
4� xy.
x � yx � y
f �n� � �n2�4,
�n2 � 1��4, n even
n odd.
f �1� � 0, f �2� � 1, f �3� � 2, f �4� � 4, . . .
Section 9.3 The Integral Test and p-Series
260 Chapter 9 Infinite Series
1.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
1x � 1
dx � �ln�x � 1��1
�� �
x ≥ 1.
f �x� �1
x � 1.
��
n�1
1n � 1
2.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
23x � 5
dx � �23
ln�3x � 5���
1� �
x ≥ 1.
f �x� �2
3x � 5.
��
n�1
2
3n � 5
3.
Let
f is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1 e�x dx � ��e�x�
�
1�
1e
x ≥ 1.
f �x� � e�x.
��
n�1e�n 4.
Let
f is positive, continuous, and decreasing for since
for
Converges by Theorem 9.10
��
3xe�x�2 dx � ��2�x � 2�e�x�2�
�
3� 10e�3�2
x ≥ 3.f � �x� �2 � x2e x�2 < 0
x ≥ 3
f �x� � xe�x�2.
��
n�1ne�n�2
5.
Let
is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1
1x2 � 1
dx � �arctan x��
1�
�
4
x ≥ 1.f
f �x� �1
x2 � 1.
��
n�1
1n2 � 1
6.
Let
f is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
12x � 1
dx � �ln �2x � 1��
1� �
x ≥ 1.
f �x� �1
2x � 1.
��
n�1
12n � 1
7.
Let
f is positive, continuous, and decreasing for since
Diverges by Theorem 9.10
��
1 ln�x � 1�
x � 1 dx � �ln�x � 1� 2
2 ��
1� �
f��x� �1 � ln�x � 1�
�x � 1�2 < 0 for x ≥ 2.
x ≥ 2
f �x� �ln�x � 1�
x � 1.
��
n�1 ln�n � 1�
n � 18.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
2 ln x�x
dx � �2�x�ln x � 2���
2� �,
x > e2 � 7.4.f
f �x� �ln x�x
, f ��x� �2 � ln x
2x3�2 .
��
n�2 ln n�n
Section 9.3 The Integral Test and p-Series 261
9.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x��x � 1� dx � �2 ln��x � 1��
�
1� �,
x ≥ 1.f
f �x� �1
�x��x � 1�, f ��x� � �1 � 2�x
2x3�2��x � 1�2 < 0.
��
n�1
1�n��n � 1�
10.
Let
is positive, continuous, and decreasing for since
Diverges by Theorem 9.10
��
1
xx2 � 3
dx � �ln�x2 � 3��
1� �
f��x� �3 � x2
�x2 � 3� < 0 for x ≥ 2.
x ≥ 2f �x�
f �x� �x
x2 � 3.
��
n�1
nn2 � 3
11.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x � 1
dx � �2�x � 1��
1� �,
x ≥ 1.f
f �x� �1
�x � 1, f ��x� �
�12�x � 1�3�2 < 0.
��
n�1
1�n � 1
12.
Let
is positive, continuous, and decreasing for
Hence, the series converges by Theorem 9.10.
�2 ln 2 � 1
16, converges
��
2 ln xx2 dx � ��
�2 ln x � 1�4x 4 �
�
2
x > 2.f
f �x� �ln xx3 , f ��x� �
1 � 3 ln xx4 .
��
n�2
ln nn3
13.
Let
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1 ln xx2 dx � ���ln x � 1�
x ��
1� 1,
x > e1�2 � 1.6.f
f �x� �ln xx2 , f ��x� �
1 � 2 ln xx3 .
��
n�1 ln nn2 14.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
2
1
x�ln x dx � �2�ln x�
�
2� �,
x ≥ 2.f
f �x� �1
x�ln x, f ��x� � �
2 ln x � 12x2�ln x�3�2.
��
n�2
1
n�ln n
15.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1 arctan xx2 � 1
dx � ��arctan x�2
2 ��
1�
3� 2
32,
x ≥ 1.f
x ≥ 1.f �x� �arctan xx2 � 1
, f ��x� �1 � 2x arctan x
�x2 � 1�2 < 0
��
n�1 arctan nn2 � 1
16.
Let
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
3
1x ln x ln�ln x� dx � �ln�ln�ln x���
�
3� �,
x ≥ 3.f
f ��x� ���ln x � 1� ln�ln x� � 1
x2�ln x�2�ln�ln x��2 .
f �x� �1
x ln x ln�ln x�,
��
n�3
1n ln n ln�ln n�
262 Chapter 9 Infinite Series
17.
Let for
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
2xx2 � 1
dx � �ln�x2 � 1���
1� �,
x > 1.f
x > 1.f �x� �2x
x2 � 1, f ��x� � 2
1 � x2
�x2 � 1�2 < 0
��
n�1
2nn2 � 1
18.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1
xx4 � 1
dx � �12
arctan�x2���
1�
�
8,
x > 1.f
x > 1.f �x� �x
x 4 � 1, f ��x� �
1 � 3x 4
�x 4 � 1�2 < 0
��
n�1
nn4 � 1
19.
Let
is positive, continuous, and decreasing forsince
Diverges by Theorem 9.10
��
1
xk�1
xk � c dx � �1
k ln�xk � c��
�
1� �
for x > k�c�k � 1�.
f � �x� �xk�2c�k � 1� � xk
�xk � c�2 < 0
x > k�c�k � 1�f
f �x� �xk�1
xk � c.
��
n�1
nk�1
nk � c20.
Let
f is positive, continuous, and decreasing for since
for
We use integration by parts.
Converges by Theorem 9.10
�1e
�ke
�k�k � 1�
e� . . . �
k!e
��
1 x ke�x dx � ��x ke�x�
�
1� k ��
1 x k�1e�x dx
x > k.f � �x� �x k�1�k � x�
ex < 0
x > k
f �x� �xk
ex .
��
n�1 n ke�n
21. Let
The function is not positive for x ≥ 1.f
f �x� ���1�x
x, f �n� � an.
22. Let
The function is not positive for x ≥ 1.f
f �x� � e�x cos x, f �n� � an.
23. Let
The function is not decreasing for x ≥ 1.f
f �x� �2 � sin x
x, f �n� � an. 24. Let
The function is not decreasing for x ≥ 1.f
f �x� � �sin xx 2
, f �n� � an.
25.
Let
f is positive, continuous, and decreasing for
Converges by Theorem 9.10
��
1
1x3 dx � ��
12x2�
�
1�
12
x ≥ 1.
f �x� �1x3.
��
n�1 1n3 26.
Let
is positive, continuous, and decreasing for
Diverges by Theorem 9.10
��
1
1x1�3 dx � �3
2 x2�3�
�
1� �
x ≥ 1.f
f �x� �1
x1�3.
��
n�1
1n1�3
Section 9.3 The Integral Test and p-Series 263
27.
Let for
is positive, continuous, and decreasing for
diverges
Hence, the series diverges by Theorem 9.10.
��
1
1�x
dx � �2�x��
1� �,
x ≥ 1.f
x ≥ 1.f �x� �1�x
, f ��x� ��1
2x3�2 < 0
��
n�1
1�n
28.
Let for
is positive, continuous, and decreasing for
converges
Hence, the series converges by Theorem 9.10.
��
1
1x2 dx � ��
1x�
�
1� 1,
x ≥ 1.f
x ≥ 1.f �x� �1x2, f ��x� �
�2x3 < 0
��
n�1 1n2
29.
Divergent p-series with p �15 < 1
��
n�1
15�n
� ��
n�1
1n1�5 30.
Convergent p-series with p �53 > 1
��
n�1
3n5�3 31.
Divergent p-series with p �12 < 1
��
n�1
1n1�2
32.
Convergent p-series with p � 2 > 1
��
n�1 1n2 33.
Convergent p-series with p �32 > 1
��
n�1
1n3�2 34.
Divergent p-series with p �23 < 1
��
n�1
1n2�3
35.
Convergent p-series withp � 1.04 > 1
��
n�1
1n1.04 36.
Convergent p-series withp � � > 1
��
n�1
1n� 37.
Matches (c), diverges
S4 � 4.7740
S3 � 4.0666
S2 � 3.1892
S1 � 2
��
n�1
2n3�4
38.
Matches (f), diverges
S4 � 4.1667
S3 � 3.6667
S2 � 3
S1 � 2
��
n�1 2n
39.
Matches (b), converges
Note: The partial sums for 39 and41 are very similar because��2 � 3�2.
S4 � 3.2560
S3 � 3.0293
S2 � 2.6732
S1 � 2
��
n�1
2�n�
� ��
n�1
2n��2 40.
Matches (a), diverges
S3 � 4.8045
S2 � 3.5157
S1 � 2
��
n�1
2n2�5
41.
Matches (d), converges
Note: The partial sums for 39 and 41 are very similarbecause ��2 � 3�2.
S4 � 3.3420
S3 � 3.0920
S2 � 2.7071
S1 � 2
��
n�1
2
n�n� �
�
n�1
2n3�2 42.
Matches (e), converges
S3 � 2.7222
S2 � 2.5
S1 � 2
��
n�1 2n2
49. The area under the rectangles is greater than the area under the graph of
Since diverges, the series diverges.��
n�1
1�n
��
1
1�x
dx � �2�x��
1� �
��
n�1
1�n
> ��
1
1�x
dx
y � 1��x, x ≥ 1.
x
y
1 2 3 4 5 n
0.5
1.0
264 Chapter 9 Infinite Series
43. (a)
The partial sums approach the sum 3.75 very rapidly.
(b)
The partial sums approach the sum slower than the series in part (a).
� 2�6 � 1.6449 00
11
5
00
11
8
n 5 10 20 50 100
3.7488 3.75 3.75 3.75 3.75Sn
n 5 10 20 50 100
1.4636 1.5498 1.5962 1.6251 1.635Sn
44.
(a)
�N
n�1 1n
� 1 �12
�13
�14
� . . . �1N
> M
(b) No. Since the terms are decreasing (approaching zero),more and more terms are required to increase the partialsum by 2.
M 2 4 6 8
N 4 31 227 1674
45. Let f be positive, continuous, and decreasing for and Then,
and
either both converge or both diverge (Theorem 9.10). See Example 1, page 618.
��
1f �x� dx�
�
n�1an
an � f �n�.x ≥ 1 46. A series of the form is a p-series,
The p-series converges if and diverges if0 < p ≤ 1.
p > 1
p > 0.��
n�1
1np
47. Your friend is not correct. The series
is the harmonic series, starting with the term,and hence diverges.
10,000th
��
n�10,000 1n
�1
10,000�
110,001
� . . .
48. No. Theorem 9.9 says that if the series converges, then the terms tend to zero. Some of the series in Exercises37–42 converge because the terms tend to 0 very rapidly.
an
51.
If then the series diverges by the Integral Test. If
Converges for �p � 1 < 0 or p > 1
��
2
1x�ln x� p dx � ��
2�ln x��p
1x dx � ��ln x��p�1
�p � 1 ��
2.
p � 1,p � 1,
��
n�2
1n�ln n� p50.
1 2 3 4 5 6 7
1
x
y
�6
n�1an ≥ �7
1f �x� dx ≥ �
7
n�2an
Section 9.3 The Integral Test and p-Series 265
52.
If then the series diverges by the Integral Test. If
(Use integration by parts.)
Converges for �p � 1 < 0 or p > 1
��
2 ln xx p dx � ��
2x�p ln x dx � � x�p�1
��p � 1�2 �1 � ��p � 1� ln x��
2.
p � 1,p � 1,
��
n�2 ln nn p
53.
If diverges (see Example 1). Let
For a fixed is eventually negative. is positive, continuous, and eventually decreasing.
For this integral converges. For it diverges.0 < p < 1,p > 1,
��
1
x�1 � x2� p dx � � 1
�x2 � 1� p�1�2 � 2p���
1
fp > 0, p � 1, f ��x�
f ��x� �1 � �2p � 1�x2
�1 � x2� p�1 .
f �x� �x
�1 � x2� p , p � 1
��
n�1
n1 � n2p � 1,
��
n�1
n�1 � n2� p
54.
Since the series diverges for all values of p.p > 0,
��
n�1n�1 � n2� p 55. converges for
Hence, diverges.��
n�2
1n ln n
,
p > 1.��
n�2
1n�ln n� p
56. converges for
Hence, diverges.��
n�2
1
n 3��ln n�2� �
�
n�2
1n�ln n�2�3,
p > 1.��
n�2
1n�ln n� p 57. converges for
Hence, converges.��
n�2
1n�ln n�2,
p > 1.��
n�2
1n�ln n� p
58. converges for
�12
��
n�2
1n ln�n�, diverges
Hence, ��
n�2
1n ln�n2� � �
�
n�2
12n ln n
p > 1.��
n�2
1n�ln n� p 59. Since f is positive, continuous, and decreasing for
and we have,
Also,
Thus, 0 ≤ RN ≤ ��
N
f �x� dx.
≤ ��
N
f �x� dx.
RN � S � SN � ��
n�N�1
an ≤ aN�1 � ��
N�1 f �x� dx
RN � S � SN � ��
n�1 an � �
N
n�1 an � �
�
n�N�1 an > 0.
an � f �n�,x ≥ 1
266 Chapter 9 Infinite Series
60. From Exercise 59, we have:
�N
n�1 an ≤ S ≤ �
N
n�1 an � ��
N
f �x� dx
SN ≤ S ≤ SN � ��
N
f �x� dx
0 ≤ S � SN ≤ ��
N
f �x� dx
61.
1.0811 ≤ ��
n�1 1n4 ≤ 1.0811 � 0.0015 � 1.0826
R6 ≤ ��
6 1x4 dx � ��
13x3�
�
6� 0.0015
S6 � 1 �124 �
134 �
144 �
154 �
164 � 1.0811
62.
1.0363 ≤ ��
n�1 1n5 ≤ 1.0363 � 0.0010 � 1.0373
R4 ≤ ��
4 1x5 dx � ��
14x 4�
�
4� 0.0010
S4 � 1 �125 �
135 �
145 � 1.0363
63.
0.9818 ≤ ��
n�1
1n2 � 1
≤ 0.9818 � 0.0997 � 1.0815
R10 ≤ ��
10
1x2 � 1
dx � �arctan x��
10�
�
2� arctan 10 � 0.0997
� 1
50�
165
�1
82�
1101
� 0.9818 S10 �12
�15
�1
10�
117
�1
26�
137
64.
1.9821 ≤ ��
n�1
1�n � 1�ln�n � 1�3 ≤ 1.9821 � 0.0870 � 2.0691
R10 ≤ ��
10
1�x � 1�ln�x � 1�3 dx � ��
12ln�x � 1�2�
�
10�
12�ln 11�3 � 0.0870
S10 �1
2�ln 2�3 �1
3�ln 3�3 �1
4�ln 4�3 � . . . �1
11�ln 11�3 � 1.9821
65.
0.4049 ≤ ��
n�1 ne�n2
≤ 0.4049 � 5.6 � 10�8
R4 ≤ ��
4 xe�x2
dx � ��12
e�x2��
4�
e�16
2� 5.6 � 10�8
S4 �1e
�2e4 �
3e9 �
4e16 � 0.4049 66.
0.5713 ≤ ��
n�0 e�n ≤ 0.5713 � 0.0183 � 0.5896
R4 ≤ ��
4 e�x dx � ��e�x�
�
4� 0.0183
S4 �1e
�1e2 �
1e3 �
1e4 � 0.5713
67.
N ≥ 7
N > 6.93
N 3 > 333.33
1
N 3 < 0.003
0 ≤ RN ≤ ��
N
1x4 dx � ��
13x3�
�
N�
13N 3 < 0.001 68.
N ≥ 4,000,000
�N > 2000
N�1�2 < 0.0005
0 ≤ RN ≤ ��
N
1
x3�2 dx � ��2
x1�2��
N�
2�N
< 0.001
Section 9.3 The Integral Test and p-Series 267
69.
N ≥ 2
N > 1.0597
N > ln 200
5
5N > ln 200
e5N > 200
1
e5N < 0.005
RN ≤ ��
N
e�5x dx � ��15
e�5x��
N �
e�5N
5 < 0.001 70.
N ≥ 16
N > 2 ln 2000 � 15.2
N2
> ln 2000
eN�2 > 2000
2
eN�2 < 0.001
RN ≤ ��
N
e�x�2 dx � ��2e�x�2��
N�
2eN�2 < 0.001
71.
N ≥ 1004
N > tan 1.5698
arctan N > 1.5698
�arctan N < �1.5698
��
2� arctan N < 0.001
RN ≤ ��
N
1
x2 � 1 dx � �arctan x�
�
N72.
N ≥ 2004
N�5
> tan 1.56968
1.56968 < arctan� N�5
�
2� arctan� N
�5 < 0.001118
�2�5��
2� arctan� N
�5 < 0.001
Rn ≤ ��
N
2x2 � 5
dx � 2� 1�5
arctan� x�5 �
�
N
73. (a) This is a convergent p-series with a divergent series. Use the Integral Test.
is positive, continuous, and decreasing for
(b)
For the terms of the convergent series seem to be larger than those of the divergent series!
(c)
This inequality holds when Or, Then ln e40 � 40 < �e40�0.1 � e4 � 55.n > e40. n ≥ 3.5 � 1015.
f is positive, continuous, and decreasing for since for
(Use integration by parts.)
Converges by Theorem 9.10. See Exercise 30.
��
2 ln xx3 dx � ��
ln x2x2��
2�
12
��
2 1x3 dx �
ln 28
� ��1
4x2��
2�
ln 28
�1
16
x ≥ 2.f��x� �1 � 3 ln x
x4 < 0x ≥ 2
f �x� �ln xx3 .
��
n�2 ln nn3
81.
p-series with
Converges by Theorem 9.11
p �54
��
n�1
1
n 4�n� �
�
n�1
1n5�4 82.
p-series with
Diverges by Theorem 9.11
p � 0.95
3 ��
n�1
1n0.95
85.
Diverges by Theorem 9.9
limn→�
n
�n2 � 1� lim
n→�
1�1 � �1�n2�
� 1 � 0
��
n�1
n�n2 � 1
86.
Since these are both convergent p-series, the difference is convergent.
��
n�1 � 1
n2 �1n3 � �
�
n�1 1n2 � �
�
n�1 1n3
87.
Fails nth-Term Test
Diverges by Theorem 9.9
limn→�
�1 �1n
n
� e � 0
��
n�1 �1 �
1n
n
88.
Diverges by Theorem 9.9
limn→�
ln�n� � �
��
n�2 ln�n�
Section 9.4 Comparisons of Series
270 Chapter 9 Infinite Series
1. (a)
(b) The first series is a p-series. It converges
(c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. Hence, the other two series converge.
(d) The smaller the magnitude of the terms, the smaller the magnitude of the termsof the sequence of partial sums.
n
2
2
4
4
6
6 8
8
10
10
12
6k3/2Σ
k = 1
n
Σ 6k3/2 + 3k = 1
n
Σ 6
k2+ 0.5kk = 1
n
Sn
� p �32 > 1�.
��
n�1
6
n�n2 � 0.5�
6
1�1.5�
62�4.5
� . . . ; S1 �6
�1.5� 4.9
��
n�1
6n3�2 � 3
�64
�6
23�2 � 3� . . . ; S1 �
32
n
2
1
2
4
3
4
6
5
6 8 10
6n3/2
an =
6n3/2 + 3
an =
6
n2+ 0.5an =
n
an��
n�1
6n3�2 �
61
�6
23�2 � . . . ; S1 � 6
2. (a)
(b) The first series is a p-series. It diverges
(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. Hence, the other two series diverge.
(d) The larger the magnitude of the terms, the larger the magnitude of the terms of thesequence of partial sums.
n4
4
8
8
10
12
16
20
62
n + 0.54
2n
n − 0.52Σ
Σ
Σ
Sn
� p �12 < 1�.
��
n�1
4�n � 0.5
�4
�1.5�
4�2.5
� . . . S1 � 3.3
��
n�1
2�n � 0.5
�2
0.5�
2�2 � 0.5
� . . . S1 � 4
n
2
2
4
4 6 8 10
1
3an =
n + 0.54
an =n − 0.5
2
an = 2n
an��
n�1
2�n
� 2 �2�2
� . . . S1 � 2
3.
Therefore,
converges by comparison with theconvergent p-series
��
n�1 1n2.
��
n�1
1n2 � 1
0 <1
n2 � 1 <
1n2 4.
Therefore,
converges by comparison with theconvergent p-series
13
��
n�1 1n2.
��
n�1
13n2 � 2
13n2 � 2
< 1
3n2 5. for
Therefore,
diverges by comparison with thedivergent p-series
��
n�2 1n
.
��
n�2
1n � 1
n ≥ 21
n � 1 >
1n
> 0
Section 9.4 Comparisons of Series 271
6. for
Therefore,
diverges by comparison with thedivergent p-series
��
n�2
1�n
.
��
n�2
1�n � 1
n ≥ 21
�n � 1 >
1�n
7.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0 �1
3�n
.
��
n�0
13n � 1
0 <1
3n � 1 <
13n 8.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0�3
4�n
.
��
n�0
3n
4n � 5
3n
4n � 5< �3
4�n
9. For
Therefore,
diverges by comparison with thedivergent series
Note: diverges by the
Integral Test.
��
n�1
1n � 1
��
n�1
1n � 1
.
��
n�1
ln nn � 1
ln nn � 1
> 1
n � 1> 0.n ≥ 3, 10.
Therefore,
converges by comparison with theconvergent p-series
��
n�1
1n3�2.
��
n�1
1�n3 � 1
1�n3 � 1
< 1
n3�2 11. For
Therefore,
converges by comparison with theconvergent p-series
��
n�1 1n2.
��
n�0 1n!
1n2 >
1n!
> 0.n > 3,
12.
Therefore,
diverges by comparison with thedivergent p-series
14
��
n�1
14�n
.
��
n�1
1
4 3�n � 1
1
4 3�n � 1>
1
4 4�n13.
Therefore,
converges by comparison with theconvergent geometric series
��
n�0 �1
e�n
.
��
n�0
1en2
0 <1
en2 ≤ 1en 14.
Therefore,
diverges by comparison with thedivergent geometric series
��
n�1 �4
3�n
.
��
n�1
4n
3n � 1
4n
3n � 1 >
4n
3n
15.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
nn2 � 1
limn→�
n��n2 � 1�
1�n� lim
n→�
n2
n2 � 1� 1 16.
Therefore,
converges by a limit comparison with the convergentgeometric series
��
n�1�1
3�n
.
��
n�1
23n � 5
limn→�
2��3n � 5�
1�3n � limn→�
2 � 3n
3n � 5� 2
272 Chapter 9 Infinite Series
17.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�0
1�n2 � 1
limn→�
1��n2 � 1
1�n� lim
n→�
n�n2 � 1
� 1 18.
Therefore,
diverges by a limit comparison with the divergentharmonic series
��
n�3 1n
.
��
n�3
3�n2 � 4
limn→�
3��n2 � 4
1�n� lim
n→�
3n�n2 � 4
� 3
19.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n3.
��
n�1
2n2 � 13n5 � 2n � 1
limn→�
2n2 � 13n5 � 2n � 1
1�n3 � limn→�
2n5 � n3
3n5 � 2n � 1�
23
20.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
5n � 3n2 � 2n � 5
limn→�
5n � 3n2 � 2n � 5
1�n� lim
n→�
5n2 � 3nn2 � 2n � 5
� 5
21.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
n � 3n�n � 2�
limn→�
n � 3n�n � 2�
1�n� lim
n→� n2 � 3nn2 � 2n
� 1 22.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n3.
��
n�1
1n�n2 � 1�
limn→�
1n�n2 � 1�
1�n3 � limn→�
n3
n3 � n� 1
23.
Therefore,
converges by a limit comparison with the convergent p-series
��
n�1 1n2.
��
n�1
1
n�n2 � 1
limn→�
1��n�n2 � 1�
1�n2 � limn→�
n2
n�n2 � 1� 1 24.
Therefore,
converges by a limit comparison with the convergentgeometric series
��
n�1�1
2�n�1
.
��
n�1
n�n � 1�2n�1
limn→�
n��n � 1�2n�1
1��2n�1� � limn→�
n
n � 1� 1
Section 9.4 Comparisons of Series 273
25.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1
nk�1
nk � 1
limn→�
�nk�1���nk � 1�
1�n� lim
n→�
nk
nk � 1� 1 26.
Therefore,
diverges by a limit comparison with the divergentharmonic series
��
n�1 1n
.
��
n�1
5
n � �n2 � 4
limn→�
5��n � �n2 � 4�
1�n� lim
n→�
5n
n � �n2 � 4�
52
27.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1 sin�1
n�
� limn→�
cos�1n� � 1
limn→�
sin�1�n�
1�n� lim
n→� ��1�n2� cos�1�n�
�1�n2 28.
Therefore,
diverges by a limit comparison with the divergent p-series
��
n�1 1n
.
��
n�1 tan�1
n�
� limn→�
sec2�1n� � 1
limn→�
tan�1�n�
1�n� lim
n→� ��1�n2� sec2�1�n�
�1�n2
29.
Diverges
p-series with p �12
��
n�1 �nn
� ��
n�1
1�n
30.
Converges
Geometric series with r � �15
��
n�0 5��
15�
n
31.
Converges
Direct comparison with ��
n�1�1
3�n
��
n�1
13n � 2
32.
Converges
Limit comparison with ��
n�1 1n2
��
n�4
13n2 � 2n � 15
33.
Diverges; nth-Term Test
limn→�
n
2n � 3�
12
� 0
��
n�1
n2n � 3
34.
Converges; telescoping series
��
n�1� 1
n � 1�
1n � 2� � �1
2�
13� � �1
3�
14� � �1
4�
15� � . . . �
12
35.
Converges; Integral Test
��
n�1
n�n2 � 1�2 36.
Converges; telescoping series
��
n�1 �1
n�
1n � 3�
��
n�1
3n�n � 3�
274 Chapter 9 Infinite Series
37. By given conditions
is finite and nonzero. Therefore,
diverges by a limit comparison with the p-series
��
n�1 1n
.
��
n�1an
limn→�
nanlimn→�
an
1�n� lim
n→� nan. 38. If then The p-series with
converges and since
the series
converges by the Limit Comparison Test. Similarly, ifthen which implies that
diverges by the Limit Comparison Test.
��
n�1 P�n�Q�n�
k � j ≤ 1j ≥ k � 1,
��
n�1 P�n�Q�n�lim
n→� P�n��Q�n�
1�nk� j � L > 0,
p � k � jk � j > 1.j < k � 1,
39.
which diverges since the degree of the numerator is onlyone less than the degree of the denominator.
12
�25
�3
10�
417
�5
26� . . . � �
�
n�1
nn2 � 1
, 40.
which converges since the degree of the numerator is twoless than the degree of the denominator.
13
�18
�1
15�
124
�1
35� . . . � �
�
n�2
1n2 � 1
,
41.
converges since the degree of the numerator is three lessthan the degree of the denominator.
��
n�1
1n3 � 1
42.
diverges since the degree of the numerator is only one lessthan the degree of the denominator.
��
n�1
n2
n3 � 1
43.
Therefore, diverges.��
n�1
n3
5n4 � 3
limn→�
n� n3
5n4 � 3� � limn→�
n4
5n4 � 3�
15
� 0 44.
Therefore, diverges.��
n�2
1ln n
limn→�
n
ln n� lim
n→�
11�n
� limn→�
n � � � 0
45.
diverges, (harmonic)
1200
�1
400�
1600
� . . . � ��
n�1
1200n
46.
diverges
1200
�1
210�
1220
� . . . � ��
n�0
1200 � 10n
47.
converges
1201
�1
204�
1209
�1
216� �
�
n�1
1200 � n2
48.
converges
1201
�1
208�
1227
�1
264� . . . � �
�
n�1
1200 � n3
49. Some series diverge or converge very slowly. You cannot decide convergence or divergence of a series by comparing the first few terms.
50. See Theorem 9.12, page 624. One example is
converges because and
converges ( p-series).��
n�1 1n2
1n2 � 1
<1n2�
�
n�1
1n2 � 1
51. See Theorem 9.13, page 626. One example is
diverges because and
diverges ( p-series).��
n�2
1�n
limn→�
1��n � 1
1��n� 1�
�
n�2
1�n � 1
52. This is not correct. The beginning terms do not affect theconvergence or divergence of a series. In fact,
diverges (harmonic)
and converges ( p-series).1 �14
�19
� . . . � ��
n�1 1n2
11000
�1
1001� . . . � �
�
n�1000 1n
Section 9.4 Comparisons of Series 275
53.
For Hence, the lower terms are those of � an2.0 < an < 1, 0 < an
2 < an < 1.
0.2
0.4
0.6
0.8
1.0
n4 8 12 16 20
Σan∞
n = 1
Σan∞
n = 1
2
Terms of
Terms of
54. (a)
converges since the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.)
(c) ��
n�3
1�2n � 1�2 �
� 2
8� S2 � 0.1226
��
n�1
1�2n � 1�2 � �
�
n�1
14n2 � 4n � 1
(b)
(d) ��
n�10
1�2n � 1�2 �
� 2
8� S9 � 0.0277
n 5 10 20 50 100
1.1839 1.02087 1.2212 1.2287 1.2312Sn
55. False. Let and and both and converge.��
n�1 1n2�
�
n�1 1n30 < an ≤ bnbn �
1n2.an �
1n3
58. False. Let Then,
but converges.��
n�1cnan ≤ bn � cn,
an � 1�n, bn � 1�n, cn � 1�n2. 59. True
56. True 57. True
60. False. could converge or diverge.
For example, let which diverges.
and diverges, but
and converges.��
n�1 1n20 <
1n2 <
1�n
��
n�1 1n
0 <1n
<1�n
��
n�1bn � �
�
n�1
1�n
,
��
n�1an 61. Since converges, There exists N such
that for Thus, for and
converges by comparison to the convergent
series ��
i�1 an .
��
n�1 an bn
n > Nanbn < ann > N.bn < 1
limn→�
bn � 0.��
n�1 bn
62. Since converges, then
converges by Exercise 61.
��
n�1an an � �
�
n�1 an
2��
n�1 an 63. and both converge, and hence so does
�� 1n2�� 1
n3� � � 1n5.
� 1n3�
1n2
64. converge, and hence so does �� 1n2�2
� � 1n4.�
1n2
65. Suppose and converges.
From the definition of limit of a sequence, there existssuch that
whenever Hence, for From theComparison Test, converges.� an
n > M.an < bnn > M.
�an
bn
� 0� < 1
M > 0
� bnlimn→�
an
bn
� 0
276 Chapter 9 Infinite Series
66. Suppose and diverges. From the definition of limit of a sequence, there exists such that
for Thus, for By the Comparison Test, diverges.� ann > M.an > bnn > M.
an
bn
> 1
M > 0� bnlimn→�
an
bn
� �
67. (a) Let and converges.
By Exercise 65, converges.��
n�1
1�n � 1�3
limn→�
an
bn
� limn→�
1���n � 1�3�
1��n2� � limn→�
n2
�n � 1�3 � 0
� bn � � 1n2,� an � �
1�n � 1�3, (b) Let and converges.
By Exercise 65, converges.��
n�1
1�n� n
limn→�
an
bn
� limn→�
1���n� n�
1��� n� � limn→�
1�n
� 0
� bn � � 1
� n,� an � � 1
�n� n,
68. (a) Let and diverges.
By Exercise 66, diverges.��
n�1 ln n
n
limn→�
an
bn
� limn→�
�ln n��n
1�n� lim
n→� ln n � �
� bn � � 1n
,� an � � ln n
n, (b) Let and diverges.
By Exercise 66, diverges.� 1
ln n
limn→�
an
bn
� limn→�
n
ln n� �
� bn � � 1n
,� an � � 1
ln n,
69. Since the terms of are positive for
sufficiently large Since
and
converges, so does � sin�an�.
� anlimn→�
sin�an�
an
� 1
n.
� sin�an�limn→�
an � 0, 70.
Since converges, and
converges. � 1
1 � 2 � . . . � n
limn→�
2��n�n � 1��
1��n2� � limn→�
2n2
n�n � 1� � 2,
� 1�n2
� ��
n�1
2n�n � 1�
��
n�1
11 � 2 � . . . � n
� ��
n�1
1�n�n � 1���2
71. The series diverges. For
Since diverges, so does � 1
n�n�1��n.� 12n
1
n�n�1��n >12n
1
n1�n >12
n1�n < 2
n < 2n
n > 1, 72. Consider two cases:
If then and
If then and
combining,
Since converges, so does by
the Comparison Test.
��
n�1an
n��n�1���
n�1�2an �
12n
ann��n�1� ≤ 2an �
12n.
ann��n�1� ≤ � 1
2n�1n��n�1��
12n,an ≤
12n�1,
ann��n�1� �
an
an1��n�1� ≤ 2an.
an1��n�1� ≥ � 1
2n�11��n�1�
�12
,an ≥1
2n�1,
Section 9.5 Alternating Series
Section 9.5 Alternating Series 277
1.
Matches (d).
S3 � 8.1667
S2 � 7.5
S1 � 6
��
n�1 6n2 2.
Matches (f ).
S3 � 5.1667
S2 � 4.4
S1 � 6
��
n�1
��1�n�16n2 3.
Matches (a).
S3 � 5.0
S2 � 4.5
S1 � 3
��
n�1 3n!
4.
Matches (b).
S3 � 2.0
S2 � 1.5
S1 � 3
��
n�1 ��1�n�13
n!5.
Matches (e).
S2 � 6.25
S1 � 5
��
n�1 10n2n 6.
Matches (c).
S2 � 3.75
S1 � 5
��
n�1 ��1�n10
n2n
7.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term of the series.
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term in the series.
0 110.6
1.1
��
n�1 ��1�n�1
n2 ��2
12� 0.8225
10.
(a)
(b)
(c) The points alternate sides of the horizontal line that represents the sum of the series. The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
61. The graph of the approximating polynomial P and theelementary function f both pass through the point and the slopes of P and f agree at Depending onthe degree of P, the nth derivatives of P and f agree at�c, f �c��.
�c, f �c��.�c, f �c��
62. f �c� � P2�c�, f��c� � P2��c�, and f��c� � P2��c�
63. See definition on page 650. 64. See Theorem 9.19, page 654.
65. The accuracy increases as the degree increases (for valueswithin the interval of convergence).
66.
x20−20 10
2
−2
−4
4
6
8
10 f
y
P1
P3
P2
67. (a)
(b)
(c) g�x� �sin x
x�
1x P5�x� � 1 �
x2
3!�
x 4
5!
Q6�x� � x P5�x� � x2 �x 4
3!�
x6
5!
g�x� � x sin x
P5�x� � x �x3
3!�
x5
5!
f �x� � sin x
Q5�x� � x P4�x�
Q5�x� � x � x2 �12
x3 �16
x 4 �1
24 x5
g�x� � xex
P4�x� � 1 � x �12
x2 �16
x3 �1
24 x 4
f �x� � ex 68. (a) for
This is the Maclaurin polynomial of degree 4 for
(b) for
(c)
The first four terms are the same!
R��x� � 1 � x �x2
2!�
x3
3!
R�x� � 1 � x �x2
2!�
x3
3!�
x 4
4!
Q6��x� � �x �x3
3!�
x5
5!� �P5�x�
cos xQ6�x� � 1 �x2
2�
x 4
4!�
x6
6!
g�x� � cos x.
P5��x� � 1 �x2
2!�
x 4
4!
f �x� � sin xP5�x� � x �x3
3!�
x5
5!
308 Chapter 9 Infinite Series308 Chapter 9 Infinite Series
69. (a) Q2�x� � �1 ��2�x � 2�2
32(b) R2�x� � �1 �
�2�x � 6�2
32(c) No. The polynomial will be linear.
Translations are possible atx � �2 � 8n.
70. Let f be an odd function and be the nth Maclaurin polynomial for f. Since f is odd, is even:
Similarly, is odd, is even, etc. Therefore, f, etc. are all odd functions, which implies that Hence, in the formula
all the coefficients of the even power of x are zero.Pn�x� � f �0� � f��0�x �f ��0�x2
2!� . . .
f �0� � f ��0� � . . . � 0.f �4�,f�,f��f�
f���x� � limh→0
f ��x � h� � f ��x�
h� lim
h→0
�f �x � h� � f �x�h
� limh→0
f �x � ��h�� � f �x�
�h� f��x�.
f�Pn
71. Let f be an even function and be the nth Maclaurin polynomial for f. Since f is even, is odd, is even, is odd, etc. All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. will only haveterms with even powers of x.
Pn
f���f�f�Pn
72. Let
For Pn�k��c� � an k! � �f �k��c�
k! �k! � f �k��c�.1 ≤ k ≤ n,
Pn�c� � a0 � f �c�
Pn �x� � a0 � a1�x � c� � a2�x � c�2 � . . . � an �x � c�n where ai �f �i��c�
i!.
73. As you move away from the Taylor Polynomial becomes less and less accurate.x � c,
Section 9.8 Power Series
1. Centered at 0 2. Centered at 0 3. Centered at 2 4. Centered at �
5.
�x� < 1 ⇒ R � 1
� limn→�
�n � 1n � 2��x� � �x�
L � limn→�
�un�1
un � � limn→�
���1�n�1xn�1
n � 2�
n � 1��1�n xn�
��
n�0��1�n
xn
n � 1
7.
2�x� < 1 ⇒ R �12
� limn→�
� 2n2x�n � 1�2� � 2�x�
L � limn→�
�un�1
un � � limn→�
� �2x�n�1
�n � 1�2 �n2
�2x�n���
n�1 �2x�n
n2
6.
2�x� < 1 ⇒ R �12
L � limn→�
�un � 1un � � lim
n→� ��2x�n�1
�2x�n � � 2�x�
��
n�0�2x�n
8.
12
�x� < 1 ⇒ R � 2
�12�x�
L � limn→�
�un�1
un � � limn→�
���1�n�1xn�1
2n�1 �2n
��1�n xn���
n�0 ��1�n xn
2n
Section 9.8 Power Series 309
9.
Thus, the series converges for all x. R � �.
� limn→�
� �2x�2
�2n � 2��2n � 1�� � 0
L � limn→�
�un�1
un � � limn→�
��2x�2n�2��2n � 2�!�2x�2n��2n�! �
��
n�0
�2x�2n
�2n�!
11.
Since the series is geometric, it converges only ifor �2 < x < 2.�x�2� < 1
��
n�0�x
2�n
10.
The series only converges at R � 0.x � 0.
� limn→�
��2n � 2��2n � 1�x2
�n � 1� � � �
L � limn→�
�un � 1un � � lim
n→� ��2n � 2�!x2n�2��n � 1�!
�2n�!x2n�n! ���
n�0
�2n�!x2n
n!
12.
is geometric, so it converges if
�x�5� < 1 ⇒ �x� < 5 ⇒ �5 < x < 5.
��
n�0�x
5�n
13.
Interval:
When the alternating series converges.
When the p-series diverges.
Therefore, the interval of convergence is �1 < x ≤ 1.
��
n�1 1n
x � �1,
��
n�1 ��1�n
n x � 1,
�1 < x < 1
� limn→�
� nxn � 1� � �x�
limn→�
�un�1
un � � limn→�
���1�n�1xn�1
n � 1�
n��1�n xn�
��
n�1 ��1�n xn
n
15.
The series converges for all x. Therefore, the interval of convergence is �� < x < �.
� limn→�
� xn � 1� � 0
limn→�
�un�1
un � � limn→�
� xn�1
�n � 1�! �n!xn�
��
n�0 xn
n!
14.
Interval:
When the series diverges.
When the series diverges.
Therefore, the interval of convergence is �1 < x < 1.
��
n�0��n � 1�x � �1,
��
n�0��1�n�1�n � 1�x � 1,
�1 < x < 1
� limn→�
��n � 2�xn � 1 � � �x�
limn→�
�un � 1un � � lim
n→� ���1�n�2�n � 2�xn�1
��1�n�n � 1�xn ���
n�0��1�n�1�n � 1�xn
16.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� 3x�2n � 2��2n � 1�� � 0
limn→�
�un�1
un � � limn→�
� �3x�n�1
�2n � 1�! ��2n�!�3x�n�
��
n�0 �3x�n
�2n�!
17.
Therefore, the series converges only for x � 0.
� limn→�
��2n � 2��2n � 1� x2 � � � lim
n→� �un�1
un � � limn→�
��2n � 2�!xn�1
2n�1 �2n
�2n�!xn� ��
n�0�2n�!�x
2�n
310 Chapter 9 Infinite Series
19.
Since the series is geometric, it converges only if or �4 < x < 4.�x�4� < 1
��
n�1 ��1�n�1xn
4n
18.
Interval:
When the alternating series converges.
When the series converges by limit comparison to
Therefore, the interval of convergence is �1 ≤ x ≤ 1.
��
n�1 1n2.�
�
n�0
1�n � 1��n � 2�x � �1,
��
n�0
��1�n
�n � 1��n � 2�x � 1,
�1 < x < 1
� limn→�
��n � 1�xn � 3 � � �x� lim
n→� �un�1
un � � limn→�
� ��1�n�1xn�1
�n � 2��n � 3� ��n � 1��n � 2�
��1�n xn ���
n�0
��1�n xn
�n � 1��n � 2�
20.
Center:
Therefore, the series converges only for x � 4.
x � 4
R � 0
� limn→�
��n � 1��x � 4�3 � � � lim
n→� �un�1
un � � limn→�
���1�n�1�n � 1�!�x � 4�n�1
3n�1 �3n
��1�n n!�x � 4�n���
n�0 ��1�n n!�x � 4�n
3n
21.
Center:
Interval: or
When the p-series diverges. When the alternating series converges.
Therefore, the interval of convergence is 0 < x ≤ 10.
��
n�1
��1�n�1
nx � 10,�
�
n�1 �1n
x � 0,
0 < x < 10�5 < x � 5 < 5
x � 5
R � 5
� limn→�
�n�x � 5�5�n � 1�� �
15�x � 5� lim
n→� �un�1
un � � limn→�
���1�n�2�x � 5�n�1
�n � 1�5n�1 �n5n
��1�n�1�x � 5�n���
n�1 ��1�n�1�x � 5�n
n5n
22.
Center:
Interval: or
When the alternating series converges.
When the series diverges.
Therefore, the interval of convergence is �2 ≤ x < 6.
��
n�0
1n � 1
x � 6,
��
n�0
��1�n�1
�n � 1�x � �2,
�2 < x < 6�4 < x � 2 < 4
x � 2
R � 4
� limn→�
��x � 2��n � 1�4�n � 2� � �
14�x � 2� lim
n→� �un�1
un � � limn→�
� �x � 2�n�2
�n � 2�4n�2 ��n � 1�4n�1
�x � 2�n�1 ���
n�0
�x � 2�n�1
�n � 1�4n�1
Section 9.8 Power Series 311
23.
Center:
Interval: or
When the series diverges by the integral test.
When the alternating series converges.
Therefore, the interval of convergence is 0 < x ≤ 2.
��
n�0 ��1�n�1
n � 1x � 2,
��
n�0
1n � 1
x � 0,
0 < x < 2�1 < x � 1 < 1
x � 1
R � 1
� limn→�
��n � 1��x � 1�n � 2 � � �x � 1�lim
n→� �un�1
un � � limn→�
���1�n�2�x � 1�n�2
n � 2�
n � 1��1�n�1�x � 1�n�1�
��
n�0 ��1�n�1�x � 1�n�1
n � 1
24.
At
diverges.
At
converges.
Interval of convergence: 0 < x ≤ 4
��
n�1
��1�n�1 2n
n2n � ��
n�1
��1�n�1
n,
x � 4,
��
n�1
��1�n�1��2�n
n2n � ��
n�1
��1��2n�n2n � �
�
n�1
��1�n
,
x � 0,
�x � 22 � < 1 ⇒ �2 < x � 2 < 2 ⇒ 0 < x < 4
� limn→��x � 2
2
nn � 1� � �x � 2
2 � limn→��an�1
an � � limn→����1�n�2�x � 2�n�1
�n � 1�2n�1 ���1�n�1�x � 2�n
n2n ���
n�1
��1�n�1�x � 2�n
n2n25. is geometric. It converges if
Interval convergence: 0 < x < 6
�x � 33 � < 1 ⇒ �x � 3� < 3 ⇒ 0 < x < 6.
��
n�1�x � 3
3 �n�1
26.
Interval:
When converges.
When converges.
Therefore, the interval of convergence is �1 ≤ x ≤ 1.
x � �1, ��
n�0
��1�n�1
2n � 1
x � 1, ��
n�0
��1�n
2n � 1
�1 < x < 1
R � 1
� limn→�
��2n � 1��2n � 3�x
2� � �x2�limn→�
�un�1
un � � limn→�
���1�n�1x2n�3
�2n � 3� ��2n � 1�
��1�nx2n�1���
n�0
��1�nx2n�1
2n � 1
312 Chapter 9 Infinite Series
28.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� x2
n � 1� � 0
limn→�
�un�1
un � � limn→�
���1�n�1x2n�2
�n � 1�! �n!
��1�n x2n� ��
n�0 ��1�n x2n
n!
29.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� x2
�2n � 2��2n � 3�� � 0
limn→�
�un�1
un � � limn→�
� x2n�3
�2n � 3�! ��2n � 1�!
x2n�1 ���
n�0
x2n�1
�2n � 1�!30.
Therefore, the interval of convergence is �� < x < �.
� limn→�
� �n � 1�x�2n � 2��2n � 1�� � 0
limn→�
�un�1
un � � limn→�
��n � 1�!xn�1
�2n � 2�! ��2n�!n!xn �
��
n�1 n!xn
�2n�!
32.
When the series diverges by comparing it to
which diverges. Therefore, the interval of convergence is �1 < x < 1.
58. The set of all values of x for which the power series converges is the interval of convergence. If the powerseries converges for all x, then the radius of convergenceis If the power series converges at only c, then
Otherwise, according to Theorem 8.20, thereexists a real number (radius of convergence) suchthat the series converges absolutely for anddiverges for �x � c� > R.
�x � c� < RR > 0
R � 0.R � �.
59. A single point, an interval, or the entire real line. 60. You differentiate and integrate the power series term byterm. The radius of convergence remains the same.However, the interval of convergence might change.
61. Answers will vary.
converges for . At the convergence is conditional because diverges.
converges for . At the convergence is absolute.x � ±1,�1 ≤ x ≤ 1��
(c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum.
−1 60
1.80
�1
1 � �3�8� �85
� 1.6 ��
n�0�3�4
2 �n
� ��
n�0�3
8�n
��
n�0�x
2�n
(b)
(d) �N
n�0�3
2�n
> M
0.60−1 6
1.10
�1
1 � ��3�8� �8
11 0.7272
��
n�0��3�4
2 �n
� ��
n�0��
38�
n
M 10 100 1000 10,000
N 4 9 15 21
78. converges on
(a)
(c) The alternating series converges more rapidly. The partial sums in (a) approach the sum 2 from below. The partial sums in (b) alternate sides of the horizontal line y �
23.
3
60
−1
��
n�0�3�1
6��n
� ��
n�0�1
2�n
�1
1 � �1�2� � 2x �16
:
��13
, 13�.�
�
n�0�3x�n
(b)
(d) �N
n�0�3 �
23�
n
� �N
n�02n > M
1.5
6
−0.5
−1
�1
1 � �1�2� �23
��
n�0�3��
16��
n
� ��
n�0��
12�
n
x � �16
:
M 10 100 1000 10,000
N 3 6 9 13
79. False;
converges for but diverges for x � �2.x � 2
��
n�0 ��1�nxn
n2n
81. True; the radius of convergence is for both series.R � 1
80. True; if
converges for then we know that it must convergeon ��2, 2.
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
95. (a)
� 1 � 2x � 2x2 �43
x3P�x� � 1 � 2x �4x2
2!�
8x3
3!
f ���x� � 8e2x f ���0� � 8
f��x� � 4e2x f ��0� � 4
f��x� � 2e2x f��0� � 2
f �x� � e2x f �0� � 1 (b)
(c)
P � 1 � 2x � 2x 2 �43
x 3
e x e x � �1 � x �x2
2!� . . .��1 � x �
x2
2!� . . .�
P�x� � 1 � 2x � 2x2 �43
x3
e x � ��
n�0 xn
n!, e2x � �
�
n�0 �2x�n
n!
92.
sin�13� � �
�
n�0��1�n
132n�1�2n � 1�! 0.3272
�� < x < � sin x � ��
n�0��1�n
x2n�1
�2n � 1�!,
94.
� 1 �x 3
2� �
�
n�2
��1�n1 3 . . . �2n � 1�2nn!
x 3n
� 1 �12
�x 3� ���1�2���3�2�
2!x6 �
��1�2���3�2���5�2�3!
x9 � . . .
1
�1 � x 3� �1 � x 3��1�2, k � �
12
96. (a)
—CONTINUED—
� 2x �43
x3 �4
15x5 �
8315
x7 � . . .sin 2x � 0 � 2x �0x2
2!�
8x3
3!�
0x4
4!�
32x5
5!�
0x6
6!�
128x7
7!� . . .
f �7��x� � �128 cos 2x f �7��0� � �128
f �6��x� � �64 sin 2x f �6��0� � 0
f �5��x� � 32 cos 2x f �5��0� � 32
f �4��x� � 16 sin 2x f �4��0� � 0
f���x� � �8 cos 2x f���0� � �8
f��x� � �4 sin 2x f��0� � 0
f��x� � 2 cos 2x f��0� � 2
f �x� � sin 2x f �0� � 0
Review Exercises for Chapter 9 353
97.
� ��
n�0
��1�nx2n�1
�2n � 1��2n � 1�!
�x
0 sin t
t dt � � �
�
n�0
��1�nt2n�1
�2n � 1��2n � 1�!�x
0
sin t
t� �
�
n�0 ��1�nt2n
�2n � 1�!
sin t � ��
n�0 ��1�nt2n�1
�2n � 1�!
99.
� ��
n�0 ��1�nxn�1
�n � 1�2�x
0 ln�t � 1�
t dt � � �
�
n�0 ��1�nt n�1
�n � 1�2 �x
0
ln�t � 1�
t� �
�
n�0 ��1�nt n
n � 1
ln�1 � t� � � 11 � t
dt � ��
n�0 ��1�ntn�1
n � 1
1
1 � t� �
�
n�0��1�ntn
96. —CONTINUED—
(b)
(c)
� 2�x �2x3
3�
2x5
15�
4x7
315� . . .� � 2x �
43
x3 �4
15x5 �
8315
x7 � . . .
� 2�x � ��x3
2�
x3
6 � � � x5
24�
x5
12�
x5
120� � ��x7
720�
x7
144�
x7
240�
x7
5040� � . . .�
� 2�x �x3
6�
x5
120�
x7
5040� . . .��1 �
x2
2�
x4
24�
x6
720� . . .�
sin 2x � 2 sin x cos x
� 2x �8x3
6�
32x5
120�
128x7
5040� . . . � 2x �
43
x3 �4
15x5 �
8315
x7 � . . .
sin 2x � ��
n�0 ��1�n�2x�2n�1
�2n � 1�! � 2x ��2x�3
3!�
�2x�5
5!�
�2x�7
7!� . . .
sin x � ��
n�0 ��1�nx2n�1
�2n � 1�!
98.
� ��
n�0
��1�nx n�1
22n�2n�!�n � 1�
�x
0cos
�t2
dt � � ��
n�0
��1�n t n�1
22n�2n�!�n � 1��x
0
cos �t2
� ��
n�0 ��1�n t n
22n�2n�!
cos t � ��
n�0 ��1�nt 2n
�2n�!
100.
�x
0 et � 1
t dt � � �
�
n�1
tn
n n!�x
0� �
�
n�1
xn
n n!
et � 1
t� �
�
n�1 tn�1
n!
et � 1 � ��
n�1 tn
n!
et � ��
n�0 tn
n!
101.
By L’Hôpital’s Rule, limx→0�
arctan x
�x� lim
x→0� � 1
1 � x2�� 1
2�x�� lim
x→0�
2�x1 � x2 � 0.
limx→0�
arctan x
�x� 0
arctan x
�x� �x �
x5�2
3�
x 9�2
5�
x13�2
7�
x17�2
9� . . .
arctan x � x �x3
3�
x5
5�
x7
7�
x9
9� . . .
354 Chapter 9 Infinite Series
102.
By L’Hôpital’s Rule, limx→0
arcsin x
x� lim
x→0 � 1�1 � x2�
1� 1.
limx→0
arcsin x
x� 1
arcsin x
x� 1 �
x2
2 � 3�
1 � 3x4
2 � 4 � 5�
1 � 3 � 5x6
2 � 4 � 6 � 7� . . .
arcsin x � x �x3
2 � 3�
1 � 3x5
2 � 4 � 5�
1 � 3 � 5x7
2 � 4 � 6 � 7� . . .
Problem Solving for Chapter 9
1. (a)
(b)
(c) limn→�
Cn � 1 � ��
n�0 13 �
23�
n
� 1 � 1 � 0
0, 13
, 23
, 1, etc.
�1�3
1 � �2�3�� 1
1�1
3� � 2�1
9� � 4� 1
27� � . . . � ��
n�0
1
3 �2
3�n
2. Let
Thus, S ��2
6�
14
�2
6�
�2
6 �34� �
�2
8.
� S �122 ��2
6 �. � S �122�1 �
122 �
132 � . . .
� S �122 �
142 � . . .
Then �2
6�
112 �
122 �
132 �
142 � . . .
S � ��
n�1
1
�2n � 1�2�
1
12�
1
32�
1
52� . . ..
3. If there are n rows, then
For one circle,
and
For three circles,
and
For six circles,
and
Continuing this pattern,
limn→�
An ��
2�
14
��
8
An ��
2
n�n � 1�2�3 � 2�n � 1��2
Total Area � ��rn2�an � �� 1
2�3 � 2�n � 1��2
n�n � 1�
2
rn �1
2�3 � 2�n � 1�.
r3 �1
2�3 � 4.
1 � 2�3r3 � 4r3a3 � 6
1 3 r3
r3r32
r2 �1
2 � 2�3.
1 � 2�3r2 � 2r2a2 � 3
1 3 r2
r2r22
r1 �1
3��3
2 � ��3
6�
1
2�3.a1 � 1
1 32
12
r1
an �n�n � 1�
2.
Problem Solving for Chapter 9 355
4. (a) Position the three blocks as indicated in the figure. The bottom block extends overthe edge of the table, the middle block extends over the edge of the bottom block,and the top block extends over the edge of the middle block.
The centers of gravity are located at
bottom block:
middle block:
top block:16
�14
�12
�12
�5
12.
16
�14
�12
� �1
12
16
�12
� �13
1�21�4
1�6
16
512
1112
14
16
12
0
The center of gravity of the top 2 blocks is
which lies over the bottom block. The center of gravity of the 3 blocks is
which lies over the table. Hence, the far edge of the top block lies
beyond the edge of the table.
(b) Yes. If there are n blocks, then the edge of the top block lies from the
edge of the table. Using 4 blocks,
which shows that the top block extends beyond the table.
(c) The blocks can extend any distance beyond the table because the series diverges:
��
i�1 12i
�12�
�
i�1 1i
� �.
�4
i�1 12i
�12
�14
�16
�18
�2524
�n
i�1 12i
16
�14
�12
�1112
��13
�1
12�
512��3 � 0
��1
12�
512��2 �
16
,
5. (a)
because each series in the second line has
(b)
�Assume all an > 0.� R � 1
� �a0 � a1x � . . . � ap�1x p�1� 11 � x p
� �a0 � a1x � . . . � ap�1xp�1��1 � xp � . . .�
� a0�1 � xp � . . .� � a1x�1 � xp � . . .� � . . . � ap�1xp�1�1 � xp � . . .�