1.2. Convergence in C 13

there are elementary ways to show that this result holds and to find R. One simply

uses the standard convergence tests from calculus – particularly the ratio test.

Example 1.2.11. Find the radius of convergence of the series

(1.2.3)

∞

n=1

zn

n

.

Solution: If we apply the ratio test to the series

∑∞

n=1

|z|n/n,

we conclude

that this series converges for |z| 1 and diverges for |z| 1. Hence, by Theorem

1.2.8, the series (1.2.3) also converges for |z| 1. If |z| 1, then the sequence of

terms of this series fails to converge to zero (since

|z|n/n

→ +∞) and so the series

diverges by the term test. Thus our series converges on D1(0) and diverges outside

of D1(0). We conclude that it has radius of convergence 1.

Example 1.2.12. Show that the radius of convergence of the series

(1.2.4)

∞

n=0

zn

n!

is +∞ – that is, the series converges for all z.

Solution: We apply the ratio test to the series

∑∞

n=0

|z|n/n!.

We have

|z|n+1

(n + 1)!

|z|n

n!

−1

=

|z|

n + 1

which converges to 0 for all z. Hence, by the ratio test, the power series (1.2.4)

converges for all z ∈ C. The radius of convergence is, thus, +∞.

Example 1.2.13. Find the radius of convergence of the power series

∞

n=0

anzn

where an =

2n

if n is prime and an = 0 if n is not prime.

Solution: We cannot use the ratio test on this one because most of the terms

of the series are 0. However, if we compare the series

∑∞

n=0

an|z|n

with the series

∑

∞

n=0

2n|z|n

– a series to which we can apply the ratio test – we conclude that the

series converges for |z| 1/2. If |z| 1/2, then the terms of our series do not tend

to 0 and so the series diverges by the term test. We conclude that the radius of

convergence is 1/2.

Exercise Set 1.2

1. Show that the sequence (2 + ni)−1 converges to 0.

2. Prove the second form of the triangle inequality for complex numbers: | |z| −

|w| | ≤ |z − w|.

3. Show that

1

z + 5

≤

1

|z| − 5

.

4. Does the sequence

(

1/

√

2 + i/

√

2

)n

converge?