Lecture 2: Convergence of Series
Hart Smith
Department of Mathematics
University of Washington, Seattle
Math 427, Autumn 2016
Motivation: functions given by power series
We will define:
ez = 1 + z +
1
2!
z2 +
1
3!
z3 + · · ·
for all z
We will prove:
√
1
= 1+
1−z
1
2
z+
13 1
2 2 2!
z2 +
135 1
2 2 2 3!
z3 + · · ·
for |z| < 1
To start, need to study when infinite sums converge...
Limits of sequences
Definition
If {zn } = (z1 , z2 , . . .) is a sequence of complex numbers, we say
lim zn = z
n→∞
lim |zn − z| = 0 .
if
n→∞
This is equivalent to the conditions
lim Re(zn ) = Re(z)
and
n→∞
lim Im(zn ) = Im(z) .
n→∞
Example: Consider a complex number z, and let zn = z n . Then
lim z n = 0
n→∞
if
|z| < 1 .
Comparison test: If bn ≥ 0 and lim bn = 0, then
n→∞
|zn − z| ≤ bn
implies
lim zn = z .
n→∞
Theorem
Suppose that lim zn = z and lim wn = w. Then
n→∞
n→∞
lim (zn + wn ) = z + w ,
n→∞
lim zn wn = zw .
n→∞
First result follows from
(zn + wn ) − (z + w) ≤ |zn − z| + |wn − w|
For second, we use
zn wn − zw ≤ |zn − z| |wn | + |z| |wn − w|
To show first term → 0, use |wn | ≤ |w| + 1 for large n.
Convergence of series
If (z0 , z1 , z2 , . . .) are complex numbers, we say
∞
X
zk = z
if
k =0
If
∞
X
lim
n→∞
n
X
zk
= z
k =0
zk = z for some z ∈ C we say that
k =0
!
∞
X
zk converges.
k =0
Necessary (but not sufficient) condition for convergence:
lim zk = 0
k →∞
Important example: zk = z k
∞
X
z k = (1 − z)−1
if
|z| < 1 .
k =0
Note: |z k | = |z|k → 0 only when |z| < 1 .
Above is example of an absolutely convergent series:
Definition
A series is absolutely convergent if
∞
X
|zk | converges.
k =1
Extremely important fact
An absolutely convergent series is convergent.
Comparison test
Suppose Mk ≥ 0 are real numbers, and
If |zk | ≤ Mk for every k , then
∞
X
∞
X
Mk converges.
k =0
zk converges absolutely.
k =0
Ratio test
∞
X
|zk +1 |
If lim
< 1, then
zk converges absolutely.
k →∞ |zk |
k =0
Proof. Choose r such that
Then |zk | < C
rk
lim
k →∞
|zk +1 |
< r < 1.
|zk |
for some C, and
∞
X
k =0
C r k converges.
Examples: let z be a complex number
∞
X
zk
= 1+z +
k!
1
2!
z2 +
1
3!
z3 + · · ·
k =0
is convergent for all complex numbers z .
∞
X
(2k )! z k
= 1+
22k (k !)2
1
2
z+
13 1
2 2 2!
z2 +
135 1
2 2 2 3!
z3 + · · ·
k =0
is convergent if |z| < 1 , and not convergent if |z| > 1.