Thomas Calculus 13th [Solutions]

726 Chapter 10 Infinite Sequences and Series
1
b
1 1 1
b
x
n x 4 b n x 4 b
2 2 n
50. We want S sn
0.1 dx 0.1 lim dx lim tan
2 2
1 1 b 1 1 n 1 1 n
b
2 2 2 2 4 2 2
n
2
10
S s
1
10
0.57
2
n 4
n 1
lim tan tan tan 0.1 2 tan 0.2 9.867 n 10
1 1
b
10 b
1
10 10
n x n x b n x b x n b b n
10 60
n
n
51. S sn
0.00001 dx 0.00001 dx lim dx lim lim
1.1 1.1 1.1 0.1 0.1 0.1
10 0.00001 1000000 10
n
0.1
1 1 1 1
b
b
n x(ln x) n x(ln x) b n x(ln x) b 2(ln x)
n
52. S s n 0.01 dx 0.01 dx lim dx lim
3 3 3 2
b
1 1 1
2(ln b) 2(ln n) 2(ln n)
2 2 2
50
n e n
lim 0.01 1177.405 1178
53. Let
n
n
k
An
a k and n
k 1
k 1
A and { n }
n
Bn
a2 a4 a8 a
2
to 0. Note that { } n
2 4 8 2 n
B 2 a , where { a } is a nonincreasing sequence of positive terms converging
2
k
k
B are nondecreasing sequences of positive terms. Now,
2a2 2a4 2a4 2a8 2a8 2a8 2a8 2a 2a 2a
2 2 2
n n n
n 1
2 terms
2a 2a 2a 2a 2a 2a 2a 2a 2a 2a 2a 2A
1 2 3 4 5 6 7 8 n 1 n 1
n n
2 2 1 2 2
2 a k . Therefore if a k converges, then { B n } is bounded above
k 1
k
2 a k converges. Conversely,
2
1 2 3 4 5 6 7 1 2 2 4 4 2 n
2 1 1 2 k
An a a a a a a a an a a a a n a Bn
a a k
2
.
k 1
Therefore, if
k
1
k
a converges, then { A }
2
n is bounded above and hence converges.
2 k
54. (a) 1 1 n
n
a 1 1 1
n
2 a n 2 , which diverges
2 n n n
2 ln 2 2 (ln 2) 2
n
n
2 n(ln 2) ln 2 n
n 2 n 2 n 2
n
1
ln
2 n n
diverges.
n
(b) 1 n
n
a 1 1 1
n
2 a n 2 , a geometries series that converges if
2 np 2
np p 1 p 1
2 2 n
2
2
n 1 n 1 n 1 n 1
1 1 or p 1, but diverges if p 1.
p 1
2
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