Complex Analysis - Maths KU

6.1 Sequences and Series 305
EXAMPLE 3 Convergent Geometric Series
The infinite series
∞� (1 + 2i) k
k=1
5 k
= 1+2i
5
+ (1+2i)2
5 2
+ (1+2i)3
5 3
+ ···
is a geometric series.It has the form given in (2) with a = 1
5 (1 + 2i) and
z = 1
5 (1 + 2i).Since |z| = √ 5 � 5 < 1, the series is convergent and its sum is
given by (5):
∞� (1+2i)
k=1
k
5k 1+2i
= 5
1 − 1+2i
5
= 1+2i 1
=
4 − 2i 2 i.
We turn now to some important theorems about convergence and divergence
of an infinite series.You should have seen the analogues of these
theorems in a course in elementary calculus.
Theorem 6.2 ANecessary Condition for Convergence
If � ∞
k=1 zk converges, then lim
n→∞ zn =0.
Proof Let L denote the sum of the series.Then Sn → L and Sn−1 → L as
n →∞.By taking the limit of both sides of Sn − Sn−1 = zn as n →∞we
obtain the desired conclusion. ✎
A Test for Divergence The contrapositive ∗ of the proposition in
Theorem 6.2 is the familiar nth term test for divergence of an infinite series.
Theorem 6.3 The nth Term Test for Divergence
If lim
n→∞ zn �= 0, then � ∞
k=1 zk diverges.
For example, the series � ∞
k=1 (ik +5)/k diverges since zn =
(in +5)/n → i �= 0asn →∞.The geometric series (2) diverges if |z| ≥1
because even in the case when limn→∞ |z n | exists, the limit is not zero.
∗ See the footnote on page 154.