Math 220A Complex Analysis
Solutions to Homework #4
Prof: Lei Ni
TA: Kevin McGown
Conway, Page 33, Problem 7.
Show that the radius of convergence of the power series
∞
n
=1
(

1)
n
n
z
n
(
n
+1)
is 1, and discuss convergence for
z
= 1,

1,and
i
.
Proof.
The sequence
b
n
=
1
n
1
n
(
n
+1)
is the subsequence of

a
n

1
/n
which consists of exactly the nonzero terms. It is
easy to see that lim
n
→∞
b
n
= 1. (For example, take logarithms and use results
from calculus.) Since the lim sup of a sequence is its largest subsequential limit,
we have lim sup
n
→∞

a
n

1
/n
= 1 and hence
R
= 1

1
= 1.
At
z
= 1 and
z
=

1, the series reduces to
∞
n
=1
(

1)
n
n
,
which converges by the alternating series test. Of course the convergence here
is conditional since
∑
1
/n
diverges. At
z
=
i
, we obtain the sum
1

1
2

1
3
+
1
4
+
1
5

1
6

1
7
+
. . . .
We consider two groupings of the above sum, both of which converge by the
alternating series test:
1

1
2
+
1
3
+
1
4
+
1
5

1
6
+
1
7
+
. . . ,
1

1
2

1
3

1
4
+
1
5

1
6

. . .
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 Three '09
 Cong
 Math, Calculus, Power Series, Taylor Series, principal branch, Kevin McGown Conway

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