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SOLUTIONS FOR HOMEWORK 7
14.1. (a)
a
n
=
n
4
/
2
n
. We know that lim
n
1
/n
= 1 (Example 9.7). There
fore, lim

a
n

1
/n
= 1
/
2
<
1. By Root Test, the series converges.
(b)
a
n
= 2
n
/n
!. lim(
n
!)
1
/n
= +
∞
(proved in class), hence lim

a
n

1
/n
= 0.
By Root Test, the series converges.
Alternatively
, note that

a
n
+1

/

a
n

= 2
/
(
n
+ 1), hence lim

a
n
+1

/

a
n

=
0. By Ratio Test, our series converges.
(e)
0
≤
a
n
= cos
2
n/n
2
≤
b
n
= 1
/n
2
.
∑
b
n
converges (Integral Test), hence,
by Comparison Test,
∑
a
n
converges.
14.4. (a)
a
n
= 1
/
(
n
+ (
−
1)
n
)
2
. The series
∑
n
a
n
converges. For
n
≥
2,
0
≤
a
n
≤
b
n
= 1
/
(
n
−
1)
2
, hence

a
n
 ≤
b
n
. However,
∑
n
b
n
converges
(Example 2 from p. 69), hence, by Comparison Test,
∑
n
a
n
also converges.
(b)
a
n
=
√
n
+ 1
−
√
n
. The series
∑
n
a
n
diverges to +
∞
. To see this,
consider the sequence of partial sums:
s
n
=
∑
n
k
=1
a
k
=
∑
n
k
=1
(
√
k
+ 1
−
√
k
) =
√
n
+ 1
−
1 (this is a “telescopic sum”). Thus, lim
s
n
= +
∞
.
Alternatively
, one could observe that
a
n
= (
√
n
+ 1
−
√
n
)
·
√
n
+ 1 +
√
n
√
n
+ 1 +
√
n
=
1
√
n
+ 1 +
√
n
>
1
2
√
n
+ 1
=
b
n
.
However,
∑
n
b
n
diverges, by Integral Test.
(c)
[
This is a bonus problem – very little partial credit is given
]
a
n
=
n
!
/n
n
.
The series
∑
n
a
n
converges. For instance, one can apply Root Test. Note
±rst (it was probably shown in the discussion) that, for any
n
∈
N
,
(1)
n
!
≤
(
n/
2)
n/
2
−
1
·
n
n/
2+1
=
2
n
n
√
2
n
.
Indeed, consider ±rst the case of
n
even, that is,
n
= 2
m
. Then
n
! =
(
1
·
2
·
. . .
·
m
)
·
(
(
m
+ 1)
·
. . .
·
n
)
.
Then 1
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 Winter '08
 staff

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