Math 1B. Sample Answers to Second Midterm
1.
(12 points) Determine whether the series
∞
X
n
=1
(

1)
n
n
+ 1
√
n
2
+ 1
is absolutely convergent, conditionally convergent, or divergent.
Following Example 2 on pages 711–712, we note that the series is alternating, but
lim
n
→∞
n
+ 1
√
n
2
+ 1
= lim
n
→∞
1 + 1
/n
p
1 + 1
/n
2
= 1
,
so the Alternating Series Test cannot be applied.
Instead, we look at the limit of the
n
th
term of the series:
lim
n
→∞
(

1)
n
n
+ 1
√
n
2
+ 1
.
This limit does not exist (the terms alternate between being close to 1 and

1 ), so
the series diverges, by the Test for Divergence.
2.
(14 points) Determine whether the series
∞
X
n
=2
cos
n
n
2

1
is absolutely convergent, conditionally convergent, or divergent.
Since

1
≤
cos
n
≤
1 , we have
cos
n
n
2

1
≤
1
n
2

1
.
Therefore, we can test for absolute convergence by the Comparison Test, comparing it
with the series
∑
1
/
(
n
2

1) (
provided
the series
∑
1
/
(
n
2

1) converges). This latter
series has positive terms, so we can use the Limit Comparison Test to compare it with
the
p
series
∑
1
/n
2
:
lim
n
→∞
1
n
2

1
1
n
2
= lim
n
→∞
1
1

1
/n
2
= 1
.
Since the
p
series converges (
p
= 2
>
1 ), the series
∑
1
/
(
n
2

1) also converges (by
the Limit Comparison Test), and so the original series converges absolutely (by the
ordinary Comparison Test).
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 Fall '08
 Reshetiken
 Math, Mathematical Series

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