MAC 2312 Quiz 7
March 13, 2012
SOLUTIONS
1.
Determine whether the series
∑
∞
n
=1
(
−
1)
n
(ln
n
)
/n
converges absolutely, converges
conditionally or diverges.
Solution:
Set
b
n
= (ln
n
)
/n
and
a
n
= (
−
1)
n
b
n
.
Observe that
b
n
≥
0 for all
n
=
1
,
2
,
3
,
· · ·
. I claim that
b
n
≥
b
n
+1
for all
n
. To see this, consider the related function
f
(
x
) =
ln
x
x
,
which satisfies
f
′
(
x
) =
x
·
1
x
−
ln
x
x
2
=
1
x
2
(1
−
ln
x
)
.
In particular,
f
′
(
x
)
≤
0 for all
x
≥
3 so
f
is decreasing on [3
,
∞
). Therefore,
b
n
≥
b
n
+1
for all
n
= 3
,
4
,
5
,
· · ·
. Finally, a routine exercise in L’Hospitals rule (the details of
which are omitted here) shows that lim
x
→∞
f
(
x
) = 0. Thus, lim
n
→∞
b
n
= 0. By the
alternating series test,
∑
a
n
converges.
It remains to determine whether
∑
a
n
converges absolutely. For this, observe that
for
n
≥
3,

a
n

=
ln
n
n
≥
1
n
.
Since
∑
∞
n
=3
1
/n
diverges (by
p
series),
∑
∞
n
=3

a
n

also diverges by comparison. There
fore,
∑
a
n
does not converge absolutely.
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 Spring '08
 Bonner
 Calculus, Mathematical Series, bn bn+1

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