Practice Prelim 3 Solutions, Math 191, Fall 2005
Problem 1)
Decide, giving reasons, whether the following series converges ab
solutely, converges conditionally, or diverges?
Note: all the series have positive terms, so either converge absolutely, or diverge.
There is no conditional convergence.
a
)
∞
n
=1
(ln
n
)
2
n
3
/
2
[10
points
]
(ln
n
)
2
n
3
/
2
=
1
n
5
4
.
ln
n
n
1
8
2
Since
ln
n
n
1
8
→
0 by L’Hopital’s rule as
n
→ ∞
we see that
for
n
large enough
(ln
n
)
2
n
3
/
2
=
1
n
5
4
.
ln
n
n
1
8
2
≤
1
n
5
4
.
Since
∑
∞
n
=1
1
n
5
4
converges as it is a
p
series with
p
=
5
4
>
1, it follows by the Comparison Test that
∑
∞
n
=1
(ln
n
)
2
n
3
/
2
also
converges.
b
)
∞
n
=2
1
n
+ sin
n
[5
points
]
n
+sin
n
n
= 1 +
sin
n
n
→
1 as
n
→ ∞
since

sin
n

is bounded. Comparison with the
harmonic series
∑
∞
n
=2
1
n
,
which diverges, therefore shows by the Limit Comparison
Test that
∑
∞
n
=2
1
n
+sin
n
also diverges.
c
)
∞
n
=2
n
√
n
+ 1
n
3
+ 3
n
+ 1
[5
points
]
Let
a
n
be the
n
th
term of the above series, and let
b
n
be
1
n
3
2
.
Then
a
n
b
n
=
n
5
2
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 Fall '07
 BERMAN
 Math, Mathematical Series, lim

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