eview R Exercises for Chapter 8
167
Review Exercises for Chapter 8
1. an 1 n! 3. an 2 : 6, 5, 4.67, . . . n Matches (a) 4 5. an 10 0.3
n 1:
10, 3, . . .
Matches (d) n3 n
2
7. an
8
5n n
2
9. lim
n
n n
2
1
0
11. lim
n
1
Converges
0 0
12
The sequence seems to converge to 5.
n
lim an
n
lim
5n n
2 2 n
n
lim 5
5
13. lim
n
n
1
n
n
lim
n
1
n
n n
1 1
n n
n
lim
n
1 1
n 0.05 4
0
Converges
15. lim
n
sin n n Converges
n
0
17. An n
5000 1 1, 2, 3
5000 1.0125
n
(a) A1 A2 A3 A4 (b) A40 19. (a)
5062.50 5125.78 5189.85 5254.73 8218.10 (b)
A5 A6 A7
5320.41 5386.92 5454.25
A8 5522.43
k Sk
5 13.2
10 113.3
15 873.8
20 6448.5
3 2
25 50,500.3
120
(c) The series diverges geometric r
>1
0 10
12
21. (a)
k Sk
5 0.4597
10 0.4597
15 0.4597
20 0.4597
25 0.4597
(b)
1
(c) The series converges by the Alternating Series Test.
0 0.1
12
23. Converges. Geometric series, r
0.82, r < 1.
25. Diverges. nth Term Test. lim an
n
0.
168
Chapter 8 2 3
n
Infinite Series 1 2n 1 3n 1n 2 1 12 1 3
n
27.
n 0
29.
n 0
n
0
n
0
Geometric series with a S a 1 r 1
1 and r 1 13
2 3.
1
1
1 13
2
3 2
1 2
1 23
3
31. 0.09
0.09
0.0009
0.000009
...
0.09 1
0.01
0.0001
...
n 0
0.09 0.01
n
1
0.09 0.01
1 11
33. D1 D2 D
8 0.7 8 0.7 8 16 0.7
2
35. See Exercise 86 in Section 8.2. A ... 8 16 0.7 16 0.7
b n
P e rt er 12
1 1
8 8
16 0.7
16 0.7
n
...
200 e 0.06 2 1 e0.06 12 1 $5087.14
16 0.7
n 0
1
451 meters 3
37.
1
x
4
ln x dx
b
lim 1 9
ln x 3x3 1 9
1 9x3
39.
1 n 1
1 n2
1 n
n
1 n2 1
n
1 n 1
0
By the Integral Test, the series converges. 1
n 1
Since the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges.
41. lim 1
43. 2n 2n
2 n n 1
1 2 1
3
n3
5 . . . 2n 1 4 6 . . . 2n 3 5 . . . 2n 1 4 6 . . . 2n
n
n3 1 n3
lim
n3 n3
2
2n
1
an
2 3 2
By a limit comparison test with the convergent p-series 1 , the series converges. n3 2 1
n
5 . . . 2n 1 1 1 > 4 2n 2 2n 2n 1 1 1 Since diverges (harmonic series), 2n 2 n 1 n n1 so does the original series.
45. Converges by the Alternating Series Test (Conditional convergence) n n2 1e an 1 an
n
47. Diverges by the nth Term Test
49.
n
51.
n
2n 3 1n an 1 an
n
n
lim
lim lim lim 01
n1 2 en 1
2 en 2 en
2 en
n 1
1n
n
lim
lim lim
2n 1 n1 2n3 n1
3
n3 2n 2
n
2n
n
n
3
n
1 e2n
n
1
1 n
Therefore, by the Ratio Test, the series diverges.
0<1
By the Ratio Test, the series converges.
Review Exercises for Chapter 8 an 1 an n n n 1 35 n3 5n 1 3 5
n 1
169
53. (a) Ratio Test: lim
n
n
lim lim
n
3 <1 5
Converges (b) x Sn (c)
4
5 2.8752
10 3.6366
15 3.7377
20 3.7488
25 3.7499
(d) The sum is approximately 3.75.
0 1
12
55. (a)
N
1 dx x2
1 x
N
1 N
N
N n
5 1 n2 1 1 dx x2 1.4636
10 1.5498
20 1.5962
30 1.6122
40 1.6202
0.2000
0.1000
0.0500
0.0333
0.0250
N
(b)
N
1 dx x5
1 4x 4
N
1 4N4
N
N n 1
5 1 n5 1 dx x5 1.0367
10 1.0369
20 1.0369
30 1.0369
40 1.0369
0.0004
0.0000
0.0000
0.0000
0.0000
N
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums. 57. f x fx fx fx P3 x e
x2
f0
x2
1 1 2 1 4 1 8 x2 2! f 0 x3 3!
1 e 2 1x e 4 1 e 8 f0 1 1
f0 f0 f0 f0
2
x2
f 0x 1 x 2 1 x 2 95 180 1 x2 4 2! 12 x 8
1 x3 8 3! 13 x 48 95 180 95 3 18033! 0.75 3
3
59. sin 95
sin
95 5 18055! 0.75 4
4
95 7 18077! 0.75 5
5
95 9 18099! 0.75 6
6
0.996
61. ln 1.75
0.75
0.75 2
2
...
0.75 15
15
0.560
170 63. f x
Chapter 8 cos x, c f
n 1
Infinite Series 0
1
Rn x f
n 1
n z
zn x 1!
1 Rn x
xn 1 n 1! (b) Rn x 4. 1 n
n 1
(a) Rn x
0.5 n 1 < 0.001 n 1!
1!
< 0.001 6.
This inequality is true for n (c) Rn x 0.5 n 1 < 0.0001 n 1!
This inequality is true for n (d) Rn x 2n 1 < 0.0001 n 1!
This inequality is true for n x 10
n
5.
This inequality is true for n
10.
65.
n 0
Geometric series which converges only if x 10 < 1 or 1nx 2 n 12 un 1 un
n
10 < x < 10.
67.
n 0
69.
n 0
n! x lim
2
n
n
lim
n
lim x 2
1
n
1
x 2
2
2
n
1
n
n 12 1nx 2
n
n
un 1 un
n
lim
n
1!x 2 n! x 2 n
n
1
R1 Center: 2 Since the series converges when x 1 and when x the interval of convergence is 1 x 3. x2n n! 3,
which implies that the series converges only at the center x 2.
71.
y
n 0
1 1
n 1
n
4n
n
2 1 n 0
y y
n 0
4n 1 1
n
2n x2n n! 2
1
1n 4n 1 x2n
1 1
2n n
2 x2n 1!2
1
2n
1
4n
n 1
2 2n n 1!
2 2 n 0 n 1 1
x2y
xy
x2y
n 0
2n 2 2n 1 x2n 4n 1 n 1 ! 2
1
1n 4n 2n 1!
n
1 1
2n n
2 x2n 1!2 1 4n n!
n 2
2
1
n 0
n
x2n 1 4n n! 2
1
n 0
n
2n 4n
1
2 2n 1 n 1!2 1
2
1 4n 1 1 4n x n! 2n
2
2
2
n 1 x 2n 0
x 2n
2
1
n 0
n
2n 2 2n 4n 1 n 1 !
14
1
1 4n n!
2
2
x 2n
2
1
n 0
n 1
n 1!
1
2
2
4n
n
1
n
n
2
n
0
1 n 11 4n n! 2 a
1
1 4n n!
2
73.
2 3 x 2x 33
23 1 x3
n n 0
1 2x n
1
r
75. Derivative:
n
2nx n 1 n1 13
n
0
3n
Review Exercises for Chapter 8 2 x 3 fx fx fx fx sin x
n 0
171
77. 1
42 x 9 sin x cos x sin x
83 x 27
...
n 0
2x 3
n
1
1 2x 3
3 3 2x
,
3 3 <x< 2 2
79.
cos x , . . . f 2 2
n
xx n! 2 x 2
3 3 4
4
n
2
2 x 2!
3 4
2
...
2 2n
1
0
nn
12
x n!
3
4
n
81. 3x 3x
eln 3
x
ex ln 3 and since ex
n
xn , we have 0 n!
83.
fx fx
1 x 1 x2 2 x3 6 ... , x4 f
n 0 n
n
0
x ln 3 n! x ln 3
n
1
x2 ln2 3 2!
x3 ln3 3 3!
x 4 ln4 3 4!
. . ..
fx f x 1 x
1x n! n! x 1 n!
n
1n
x
n 0
1
n
n
0
85. 1 1
x x
k
1 1 1 1 1
kx x 5 1 x 5 x 5 x 5
kk 2! 15 1
1 x2
kk
1k 3! 15
2 x3 45 3! ...
... 9 5 x3 ...
15
4 5 x2 2! 1
n 14
4x2 52 2! 1
4 9x3 533! 9
n
2
14 . . . 5n 5nn! ...
6 xn
22 x 25
63 x 125
87. ln
ln x
n 1
1 1
n 1
n
1
x n
1
n
, 1
n
0<x2
89.
ex
n
xn , n! 0 1 2nn!
<x< 1.6487
5 4
n
1
54 n 1 4nn
e1
2 n 0
1
n 1
n
1
0.2231
172
Chapter 8
Infinite Series x2n , 2n ! 22n 32n 2n !
91.
cos x
n 0
1 1
n 0
n
<x< 0.7859
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
cos
2 3
n
95. (a) f x fx fx fx e2x 1 1 (c) e2x ex 1
e2x 2e2x 4e2x 8e2x 2x 2x ex x
f0 f0 f0 f0 4x2 2! 2x2 1 x x x2
1 2 4 8 8x3 3! 43 x 3 x2 2 x2 2 ... ... x3 6 x2 2 ... x3 6 x2 2 x3 2 x3 6 x3 2
(b) ex
n
xn 0 n! 2x n n! 2x 1 2x2 2x 43 x 3 4x2 2! ... 8x3 3! ...
e2x
n 0
1
1
x x3 6
... ... 43 x 3 ...
1
2x
2x2
97.
sin t
n 0
1 nt 2n 1 2n 1 ! 1 n t 2n 2n 1 ! 2n 2n 1 n t 2n 1 2n 1 nx 2n 1 2n
1 1 x
99.
1 1 ln 1 ln t t
x
t t 1
1 ntn
n 0
sin t t
x 0
1 1 n t
dt
n nt n 0
n
0
1 ntn n1
1
sin t dt t
1
n 0
n
0
1! 1!
0
1 1 nt n 1 n 12
x 0 n 0
ln t t
1
n
0
dt
n 0
0
1 nxn 1 n 12
101.
arctan x arctan x x
x0
x x 0
x3 3
x5 5 x5 2 3
x7 7 x9 2 5
x9 9 x13 7
2
... x17 9
2
...
lim
arctan x x
1 arctan x By LHpitals Rule, lim x0 x
x0
lim
1 1
x2
x0
lim
2x 1 x2
0.
2x
Problem Solving for Chapter 8
1. (a) 1 1 3 2 1 9 4 1 27 ...
n
12 03 3
n
1
13 23
1
12 (b) 0, , , 1, etc. 33 (c) lim Cn
n
1
n 0
12 33
n
1
1
0
Problem Solving for Chapter 8 nn 2 1
173
3. If there are n rows, then an For one circle,
.
1 3 2 r1
1 2
a1
1 and r1
1 3
3 2
3 6
1 23
For three circles, a2 r2 3 and 1 2 1 23
1
2 3r2
2r2
2r2 3 r2
r2
For six circles, a3 r3 6 and 1 1 23 4
1
2 3 r3
4r3
2r3 3 r3
r3
Continuing this pattern, rn Total Area rn2 an An lim An 1 4 1 1 1 1 R (b)
23 23
1 2n 1 2n
1 1 1
.
2
nn 2
1
nn 1 2 2 3 2n
2
n
2
8 2x x3 x3 2x 3x2 x6 x6 3x2 x3 ... ... 1 1 1 x3 1. a1x p ... ...
1
5. (a)
an x n
2x4 2x 2x
3x5 x4 3x2
... x7 ... 3 x2 x5 x8 ...
1 because each series in the second line has R an x n a0 a0 1 a0 a0 R 1 a1x xp a1 x a1 x ... ... ... ... ap
1x p 1
a0 x p xp ... xp 1 1 xp .
... ap
1x p
...
1
a1 x 1 ap ap
1x 1x p 1
1
xp
...
1
p
1
174 7.
Chapter 8 ex xex xex dx Letting x 1
n 0
Infinite Series x x2 ex x2 2! x3 2! C
n 0
1 x xex
... ...
n
xn 1 0 n!
2
xn n 1,
2 n!
0, C 1 n 1
1. Letting x 2 n! 1 2 1 . 2
1
n 1
n
2 n!
.
Thus,
n 1
n
2 n!
9. Let a1
0
sin x dx, a2 x
2
sin x dx, a3 x
3 2
sin x dx, etc. x
Then, sin x dx x a1 0 and an a2
1
a3
a4
. . ..
0
Since lim an
n
< an, this series converges.
11. (a) a1 a2 a3 a4 a5 a6
n
3.0 1.73205 2.17533 2.27493 2.29672 2.30146 1 2 13 [See part (b) for proof.] a a1 a a> a a1.
lim an
(b) Use mathematical induction to show the sequence is increasing. Clearly, a2 Now assume an > an an an an a > an a>
1 1 1.
Then a
an
1
a
> an.
Use mathematical induction to show that the sequence is bounded above by a. Clearly, a1 Now assume an < a. Then a > an and a aa a2 1 > an 1 a > an a2 > an a> an L 1 2 1 a a an
1. 1
a < a.
1 > 1 implies
Hence, the sequence converges to some number L. To find L, assume an L L a 1 L2 . 4a . a L L2 L a 0
an
L:
4a 1 2
Hence, L
Problem Solving for Chapter 8 1
n n 12 1
n
175
13. (a)
1 21 1 20 1 9 8 11 8 45 32
1
n
1
1 22 1 1 8 1 4 1 32 1 16 9 8
1
1 23
1
1 24
1
1 25
1
...
S1 S1 S3 S4 S5 (b) an 1 an 2n 2n
1
11 8 45 32 47 32 2
1 1
n n 1
1
n
1
21
This sequence is 1, 2, 1, 2, . . . which diverges. 8 8 (c)
n
1 2n
1
n
2n 2
n
1 2 1 2
1n 1
n
1
n
1 < 1 converges because 2 2
1
n
1 2,
2, 1, 2, . . . and 2
n
1 2 1 and
n
2 1.
15. S6 S7 S8 S9 S10
130 240 440 810 1490
70 130 240 440 810
40 70 130 240
240 440 810 1490 2740
440