12.8. Power Series
Find the radius of convergence and interval of convergence of the series.
10.
∞
X
n
=1
x
n
n
3
n
solution:
If
a
n
=
x
n
n
3
n
, then
lim
n
→∞

a
n
+1
a
n

= lim
n
→∞

x
n
+1
(
n
+ 1)3
n
+1
n
3
n
x
n

= lim
n
→∞

xn
(
n
+ 1)3

=

x

3
lim
n
→∞
n
n
+ 1
=

x

3
.
By the Ratio test, the series converges when

x

3
<
1. That is,

x

<
3. So,
R
= 3.
When
x
=

3, the series is the alternating harmonic series, which converges by the Alter
nating Series test. When
x
= 3, it is the harmonic series, which diverges. Thus, the interval
of convergence is [

3
,
3).
14.
∞
X
n
=0
(

1)
n
x
2
n
(2
n
)!
.
solution:
a
n
= (

1)
n
x
2
n
(2
n
)!
, so
lim
n
→∞

a
n
+1
a
n

= lim
n
→∞

x

2
n
+2
(2
n
+ 2)!
·
(2
n
)!

x

2
n
= lim
n
→∞

x

2
(2
n
+ 1)(2
n
+ 2)
= 0
Thus, by the Ratio test, the series converges for all real
x
and we have
R
=
∞
and the
interval of convergence (
∞
,
∞
).
12.9. Representation of Functions as Power Series
8. Find a power series representation for the function
f
(
x
) =
x
4
x
+ 1
and determine the
interval of convergence.
solution:
f
(
x
) =
x
4
x
+ 1
=
x
·
1
1

(

4
x
)
=
x
∞
X
n
=0
(

4
x
)
n
=
∞
X
n
=0
(

1)
n
4
n
x
n
+1
.
1
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2
The series converges when
 
4
x

<
1 ( that is,

x

<
1
4
) since
∑
∞
n
=0
(

4
x
)
n
is a geometric
series. It is easy to see that the series diverges at the end points. So, the interval of conver
gence is (

1
4
,
1
4
).
14. (a) Find a power series representation for
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 Fall '09
 odell
 Power Series, Mathematical Series, lim

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