Thomas Calculus 13th [Solutions]

724 Chapter 10 Infinite Sequences and Series
40. converges by the Integral Test:
1 tanh 1 0.76
2 b 2
b
x dx x dx x b
1 b 1
b
1
b
sech lim sech lim tanh lim (tanh tanh 1)
41.
1
( 2) a
1
3 ( 2)
a
a
b b b
dx a x x
3
x 2 x 4
b 1 b
b 4 5
b
b 4 5
lim ln 2 ln 4 lim ln ln lim ln ln ;
a
( b 2) a 1 , a 1
lim a lim ( b 2)
the series converges to ln
5
if a 1 and diverges to if
b
b 4
b
1, a 1
3
a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From
that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.
a
a
b
1 2a x 1 b 1 2 b 1 2
3 x 1 x 1
b ( x 1) ( b 1) 4 ( b 1) 4
3 b b
42. dx
2a 2a 2a 2a 2a
b
lim
b 1
lim
1
( b 1) b 2 a( b 1)
2a
2a
1
lim ln lim ln ln lim ln ln ;
1,
,
a
a
1
2
1
2
the series converges to ln
4
ln 2 if a 1
and diverges to
2
2
if a 1
2 . If a 1
2 , the terms of the series eventually become negative and the Integral Test does not apply.
From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it
diverges.
43. (a)
(b) There are (13)(365)(24)(60)(60)
9
10 seconds in 13 billion years; by part (a) s 1 ln n where
n
9 9
n (13)(365)(24)(60)(60) 10 s n 1 ln (13)(365)(24)(60)(60) 10
1 ln(13) ln(365) ln(24) 2ln(60) 9ln(10) 41.55
44. No, because
1 1 1
nx x n
n 1 n 1
and
1
n 1 n
diverges
45. Yes. If
n
1
a is a divergent series of positive numbers, then
1
n
a n
a also diverges and 2
.
2 n 2
n 1 n 1
a n
a n
There is no smallest divergent series of positive number: for any divergent series
n
1
a of positive
n
numbers
n
1
a n
2
has smaller terms and still diverges.
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