Thomas Calculus 13th [Solutions]

Section 10.3 The Integral Test 727
55. (a)
dx
2 x(ln x)
1
p 1
p
;
p 1
ln , 1
1 1 1
ln 2 lim p b
dx
p u
1 lim p
ln 2
1
(ln 2) p
u x du u du b
x
b
p
b
p
(ln 2) p 1
, p 1
the improper integral converges if p 1 and diverges if p 1. For p 1:
2
dx
x ln x
b
lim ln(ln x) lim ln(ln b ) ln(ln 2) , so the improper integral diverges if p 1.
b
2
b
(b) Since the series and the integral converge or diverge together, 1
n(ln n)
n 2
p
converges if and only if p 1.
56. (a) p 1 the series diverges
(b) p 1.01 the series converges
(c) 1 1 1 ; p 1 the series diverges
3
n ln n 3 n(ln n)
n 2 n 2
(d) p 3 the series converges
57. (a) From Fig. 10.11 (a) in the text with ( ) 1
x
a k
1 k
,
n 1
we have 1 dx 1 1 1 1
1 x 2 3 n
1 n
1 1 1 1 1 1
1
2 3 n
1 ln 0 ln ( 1) ln 1 2 3 n
ln n 1.
Therefore the sequence 1 1 1 1
2 3 n
ln n is bounded above by 1 and below by 0.
(b) From the graph in Fig. 10.11 (b) with 1 1 1
1
x n 1 n x
ln ( 1) ln n
0 1 ln ( n 1) ln n 1 1 1 1 ln ( n 1) 1 1 1 1
n 1 2 3 n 1 2 3 n
ln n .
If we define a 1 1 1
n 1 2 3 n
ln n , then 0 an 1 an an 1 an { a n}
is a decreasing
sequence of nonnegative terms.
58.
2
x x
e e for x 1, and
2
x x
b
b 1 1 x
e dx lim e lim e e e e dx converges by the
1 b
1 b
1
2 2
n
n
Comparison Test for improper integrals e 1 e converges by the Integral Test.
n 0 n 1
59. (a) s10
2
1 1.97531986; 1
b
b
3
dx lim x dx lim x lim 1 1 1
3
3 2
n
11 x
11 2 2 242 242
n 1
b b 11 b b
and
2
1
b
b
3
dx lim x dx lim x lim 1 1 1
10
3 2
x b 10 b
2
10 b 2b
200 200
1.97531986 1 s 1.97531986 1
242 200
1.20166 s 1.20253
(b) s 1 1.20166 1.20253 1.202095; error 1.20253 1.20166
3
n
2
2
n 1
0.000435
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