1.2. Convergence in C 11

divergence is called the term test. Its proof for complex series is the same as the

proof for real series. We leave it as an exercise (Exercise 1.2.9).

Example 1.2.7. For what values of z does the complex geometric series

∞

k=0

zk

converge and what does it converge to?

Solution: We first note that if |z| ≥ 1, then

|z|k

≥ 1 for all k and the series

diverges by the term test. For |z| 1 we use the same trick that is used to study

the real geometric series. The nth partial sum of the series is

sn =

n

k=0

zk.

If we multiply this by (1 − z), a vast cancellation of terms occurs and we obtain

(1 − z)sn = 1 −

zn+1,

so that

sn =

1 − zn+1

1 − z

.

This sequence converges to

1

1 − z

if |z| 1, since

sn −

1

1 − z

=

|z|n

|1 − z|

→ 0

in this case.

series

∑∞

k=0

zk is said to converge absolutely if the series of positive terms ∑∞A

k=0

|zk| converges. From calculus, we know a great deal about convergence of pos-

itive termed series (comparison test, ratio test, root test, etc.), and so the following

theorem will obviously play an important role.

Theorem 1.2.8. If a series

∑∞

k=0

zk of complex numbers converges absolutely,

then it converges. Furthermore,

∞

k=0

zk ≤

∞

k=0

|zk|.

Proof. By hypothesis, the series

∑∞

k=0

|zk| converges. If zk = xk + iyk, then

|xk| |zk| and |yk| |zk|. By the comparison test, the two series

∞

k=0

|xk| and

∞

k=0

|yk|

both converge. This, in turn, implies that

∞

k=0

xk and

∞

k=0

yk

converge, since an absolutely convergent series of real numbers converges. This

tells us that the real and imaginary parts of the sequence {sn} of partial sums of