Thomas Calculus 13th [Solutions]

Section 10.4 Comparison Tests 729
4. Compare with
1,
n
n 2
which is a divergent p-series since p 1 1. Both series have nonnegative terms for
2 2 1 1 n n 1 n 2 n 1 .
n n n n n n n n n n n n
n
n
n 2
n 2. For n 2, we have n n n Thus 2 2 2 2 2 2
2
diverges.
2
n
1
n
n 1
5. Compare with ,
3/2
which is a convergent p-series since p
3
1. Both series have nonnegative terms for
2
n 1. For n 1, we have
2 2
cos n 1
3/2
n
3/2
n
0 cos n 1 . Then by Comparison Test,
n
2
cos n
3/2
n
1
converges.
6. Compare with
1
,
n
3
n 1
which is a convergent geometric series, since | r |
1
1. Both series have nonnegative
3
n
n
terms for n 1. For n 1, we have n 3 3
1 1
. Then by Comparison Test,
1
n n
n 3 3
n
n 3
n 1
converges.
5
7. Compare with
3/2
n
n 1
.
The series
1
3/2
n 1 n
is a convergent p-series since p
3
1, and the series
3/2
2
5
n 1 n
5
we have
1
3/2
n
n 1
converges by Theorem 8 part 3. Both series have nonnegative terms for n 1. For n 1,
3 4 3 4 4 3 4 4 4 4 3 4 4
n n 4n 4n n 4n n 4n 5n n 4n 5n 20 5 n 4
4
n
3
4n 3
n ( n 4) n 4 5 n 4 5 5
4
n 4
4
n 4
4
n 4
3
n
4
n 4
3
n
3/2
n
converges.
5 5 . Then by Comparison Test,
4
n 1
n
n
4
4
8. Compare with
1
, which is a divergent p-series since p
1
1. Both series have nonnegative terms for
n 1
n
n 1. For n 1, we have n 1 2 n 2 2 n 1 3 n 2 n 1 3n 3 2n n n 3
2 2
2 2 n n 2 n 1
n 2 n 1
1 1
1 n 1 n 1 n 1 1
2 2 2 2
n 3 n 3 n n 3 n n 3 n 2
n 3 n
n 2n n n n 3 1 .
2
Then by Comparison Test,
n
1
n
2
n
1
3
diverges.
1
n
n 1
9. Compare with ,
2
n
1.
which is a convergent p-series since p 2 1. Both series have positive terms for
n 2
a 3 2 3 2 2
n n n 3 n 2n 3n 4n 6n
4 6
2 3 2 2
n
bn
n 1/ n n n n 3 n 3n 2n
n
6n
2
n
6
lim lim lim lim lim lim 1 0. Then by Limit
n 2
n n
n 1
Comparison Test,
3 2
3
converges.
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