MATH 138 Calculus 2, Solutions to Assignment 9
Problem 1:
For each series below determine whether it converges absolutely, conditionally, or not at all.
(a)
∞
X
n
=1
(

1)
n
n
10
2
n
Solution:
Let
a
n
=
n
10
2
n
. Then
a
n
+1
a
n
=
(
n
+ 1)
10
2
n
+1
·
2
n
n
10
=
n
+ 1
n
10
·
1
2
=
1
2
(
1 +
1
n
)
10
→
1
2
as
n
→ ∞
.
Thus
∑

a
n

converges by the Ratio Test. In other words,
∑
a
n
converges absolutely.
(b)
∞
X
n
=1
(

1)
n
+1
√
n
n
+ 1
Solution:
Let
a
n
= (

1)
n
+1
√
n
n
+ 1
. Let
b
n
=
√
n
n
=
1
√
n
. Then

a
n

b
n
=
√
n
n
+ 1
÷
√
n
n
=
1
1 +
1
n
→
1 as
n
→ ∞
,
and
∑
b
n
diverges
(
its a
p
series with
p
=
1
2
)
, so
∑

a
n

diverges by the Limit Comparison Test.
Thus
∑
a
n
is not absolutely convergent, but we still need to determine whether it is conditionally convergent. Let
f
(
x
) =
√
x
x
+ 1
so that

a
n

=
f
(
n
). Then
f
0
(
x
) =
1
2
√
x
(
x
+ 1)

√
x
(
x
+ 1)
2
=
1

x
2
√
x
(
x
+ 1)
≤
0 for
x
≥
1
,
so
f
(
x
) is decreasing for
x
≥
1, and hence
{
a
n
}
is decreasing for
n
≥
1. Also,
lim
n
→∞

a
n

= lim
n
→∞
√
n
n
+ 1
= lim
n
→∞
1
√
n
1 +
1
n
→
0 as
n
→ ∞
.
Thus
∑
a
n
converges by the Alternating Series Test, so we have shown that
∑
a
n
converges conditionally.
1
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Problem 2:
Find the radius as well as the interval of convergence of the following power series, and give
reasons for your answers.
(a)
∞
X
n
=0
2
n
n
!
x
n
Solution:
Let
a
n
=
2
n
n
!
x
n
. Then for all
x
, we have
a
n
+1
a
n
=
2
n
+1
x
n
+1
(
n
+ 1)!
·
n
!
2
n
x
n
=
2

x

n
+ 1
→
0 as
n
→ ∞
,
so
∑

a
n

converges for all
x
by the Ratio Test. Thus the radius of convergence is
R
=
∞
, and the interval
of convergence is
I
= (
∞
,
∞
).
(b)
∞
X
n
=0
1
3
n
(
x
+ 1)
n
Solution:
This series is geometric with ratio
r
=
x
+1
3
, so it converges if and only if
x
+1
3
<
1, that is when

x
+ 1

<
3. Thus the radius of convergence is
R
= 3 and the interval of convergence is
I
= (

4
,
2).
(c)
∞
X
n
=2
1
(ln
n
)
2
x
n
Solution:
Let
a
n
=
1
(ln
n
)
2
x
n
. Then for all
x
, we have
a
n
+1
a
n
=
x
n
+1
(ln(
n
+ 1))
2
·
(ln
n
)
2
x
n
=
ln
n
ln(
n
+ 1)
2

x
 → 
x

as
n
→ ∞
since replacing
n
by the continuous variable
t
then using l’Hˆ
opital’s Rule gives
lim
n
→∞
ln
n
ln(
n
+ 1)
= lim
t
→∞
ln
t
ln(
t
+ 1)
= lim
t
→∞
1
/t
1
/
(
t
+ 1)
= lim
t
→∞
t
+ 1
t
= lim
t
→∞
(
1 +
1
t
)
= 1
.
By the Ratio Test,
∑

a
n

converges when

x

<
1 and
∑
a
n
diverges when

x

<
1, so the radius of
convergence is
R
= 1. To determine the interval of convergence, we must consider the case that

x

= 1.
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 Winter '07
 Anoymous
 Calculus, Mathematical Series, Xn

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