Thomas Calculus 13th [Solutions]

Section 10.4 Comparison Tests 733
35. converges because 1 n 1 1
n n n
n2 n2 2
n 1 n 1 n 1
which is the sum of two convergent series: 1
n2
n 1
n
converges by
the Direct Comparison Test since 1 1 , n n
n2 2
and 1
n
2
n 1
is a convergent geometric series
36. converges by the Direct Comparison Test: n 2 1 1
2 n
n 2
n 2 n2
n
n 1 n 1
terms of a convergent geometric series and a convergent p-series
n
and 1 1 1 1 ,
n 2 n 2
n2 n 2 n
the sum of the nth
37. converges by the Direct Comparison Test: 1 1 , which is the nth term of a convergent geometric series
n 1 n 1
3 1 3
38. diverges;
n 1
lim 3 1 lim 1 1 1
3 3 3 3
0
n
n
n
n
39. converges by Limit Comparison Test: compare with 1
5
n 1
n 1 1
n
2 3 n 5
n
n
n
2
| r | 1 1
1
5 1, lim lim lim (1/5) 3 2n
3
0.
n n n n n
n
,
which is a convergent geometric series with
40. converges by Limit Comparison Test: compare with 3
4
n 1
2
n
3
n
8
n
3
n
4
n
n n
12
n n n
9
12
n
| 1
r | 3 8 12 1
4 1, lim lim lim (3/4) 9 12 1
1 0.
n n n 1
n
,
which is a convergent geometric series with
41. diverges by Limit Comparison Test: compare with 1 , which is a divergent p-series
n
n 1
2
n n
n
n 2
n
n
n
2 n 2 ln 2 1
n n n
2 (ln 2)
lim lim lim lim 1 0.
1/ 2
n
n
n 2 n 2 ln 2 n 2 (ln 2)
2
42. Since n grows faster than ln n and 2 ln 2,
ln n
n
n e ln n
lim lim 0.
n
n
e n n
Since e 1,
is a convergent geometric series, so 1
n
n
1 e n 1
ln n
n
n e
converges.
43. converges by Comparison Test with 1
( 1)
n 2 n n which converges since 1 1 1
n( n 1) n 1 n
n 2 n 2
s 1 1 1 1 1 1 1 1
k 1 1 lim 1;
2 2 3 k 2 k 1 k 1 k k
s k for n 2, ( n 2)! 1
k
n( n 1)( n 2)! n( n 1) n! n( n 1) 1 1
n! n( n 1)
,
and
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