Math 1910: Prelim III Solutions (12/02/2008)
Problem 1.
(a)
( 7pts)
Determine whether the series
∞
∑
n
=1
(
−
1)
n
3
n
+ln
n
converges conditionally, converges ab
solutely or diverges. Justify your answer.
Solution
According to the Alternating Series Test, it converges. (Let
a
n
=
1
3
n
+ln
n
. Note
that for
a
n
> a
n
+1
and lim
n
→∞
a
n
= 0
.
)
However,
a
n
=
1
3
n
+ ln
n
≥
1
3
n
+
n
=
1
4
n
,
which implies that the series is not absolutely convergent; so the convergence
is conditional.
(b)
( 6pts)
Determine whether the series
∞
∑
n
=1
(
−
1)
n
(1 +
1
n
)
n
converges conditionally, con
verges absolutely or diverges. Justify your answer.
Solution
(1 +
1
n
)
n
>
1 for each
n
so lim
n
→∞
(
−
1)
n
(1 +
1
n
)
n
negationslash
= 0. Therefore the series
diverges by the
n
th term test.
(c)
( 7pts)
Find all the values of
x
such that the series
∞
∑
n
=1
(
x
−
3)
n
n
converges.
Solution
If

x
−
3

<
1, the series
∑

x
−
3

n
is a convergent geometric series.
Since

x
−
3

n
n
<

x
−
3

n
for all
n
≥
1, the original series converges absolutely if
2
< x <
4. If

x
−
3

>
1, then lim
n
→∞

x
−
3

n
n
=
∞
, so it diverges. If
x
= 4, it
is the harmonic series, which diverges. If
x
= 2, it is the alternating harmonic
series and thus converges. Hence, the series converges if 2
≤
x <
4 and diverges
if
x <
2 and
x
≥
4.
1
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Problem 2.
Decide whether or not each of the following integrals converges.
If it converges, determine the value of the integral.
(a)
( 10pts)
integraldisplay
∞
1
(ln(
x
))
3
x
3
dx.
Solution
This integral converges (this can be shown by the limit comparison test with
1
/x
2
). To compute its value, we use the substitution:
u
= 2 ln
x
,
du
=
2
x
dx,
and
e
u
=
x
2
.
I
=
integraldisplay
∞
1
(ln(
x
))
3
x
3
dx
=
1
16
integraldisplay
∞
0
u
3
e
−
u
du
The latter has to be repeatedly integrated by parts:
integraldisplay
u
3
e
−
u
du
=
−
e
−
u
u
3
−
integraldisplay
3
u
2
(
−
e
−
u
)
du
=
−
e
−
u
u
3
−
3
bracketleftbigg
u
2
e
−
u
−
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 Fall '07
 BERMAN
 Math, Mathematical Series

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