3Convergence of infinite sums
IA Analysis I
3.4 Complex versions
Most definitions in the course so far carry over unchanged to the complex
numbers. e.g. z
n
→ z iff (∀ε > 0)(∃N)(∀n ≥ N) |z
n
− z| < ε.
Two exceptions are least upper bound and monotone sequences, because the
complex numbers do not have an ordering! (It cannot be made into an ordered
field because the square of every number in an ordered field has to be positive)
Fortunately, Cauchy sequences still work.
We can prove the complex versions of most theorems so far by looking at the
real and imaginary parts.
Example. Let (
z
n
) be a Cauchy sequence in
C
. Let
z
n
=
x
n
+
iy
n
. Then (
x
n
)
and (
y
n
) are Cauchy. So they converge, from which it follows that
z
n
=
x
n
+
iy
n
converges.
Also, the Bolzano-Weierstrass theorem still holds: If (
z
n
) is bounded, let
z
n
=
x
n
+
y
n
, then (
x
n
) and (
y
n
) are bounded. Then find a subsequence (
x
n
k
)
that converges. Then find a subsequence of (y
n
k
) that converges.
Then nested-intervals property has a “nested-box” property as a complex
analogue.
Finally, the proof that absolutely convergent sequences converge still works.
It follows that the ratio test still works.
Example. If |z| < 1, then
P
nz
n
converges. Proof is the same as above.
However, we do have an extra test for complex sums.
Lemma (Abel’s test). Let
a
1
≥ a
2
≥ ··· ≥
0, and suppose that
a
n
→
0. Let
z ∈ C such that |z| = 1 and z 6= 1. Then
P
a
n
z
n
converges.
Proof. We prove that it is Cauchy. We have
N
X
n=M
a
n
z
n
=
N
X
n=M
a
n
z
n+1
− z
n
z −1
=
1
z −1
N
X
n=M
a
n
(z
n+1
− z
n
)
=
1
z −1
N
X
n=M
a
n
z
n+1
−
N
X
n=M
a
n
z
n
!
=
1
z −1
N
X
n=M
a
n
z
n+1
−
N−1
X
n=M−1
a
n+1
z
n+1
!
=
1
z −1
a
N
z
N+1
− a
M
z
M
+
N−1
X
n=M
(a
n
− a
n+1
)z
n+1
!
We now take the absolute value of everything to obtain
N
X
n=M
a
n
z
n
≤
1
|z −1|
a
N
+ a
M
+
N−1
X
n=M
(a
n
− a
n+1
)
!
=
1
|z −1|
(a
N
+ a
M
+ (a
M
− a
M+1
) + ···+ (a
N−1
− a
N
))
=
2a
M
|z −1|
→ 0.
So it is Cauchy. So it converges
Note that here we transformed the sum
P
a
n
(
z
n+1
− z
n
) into
a
N
z
N+1
−
a
M
z
M
+
P
(
a
n
−a
n+1
)
z
n+1
. What we have effectively done is a discrete analogue
of integrating by parts.
Example. The series
P
z
n
/n
converges if
|z| <
1 or if
|z|
= 1 and
z 6
= 1, and it
diverges if z = 1 or |z| > 1.
The cases
|z| <
1 and
|z| >
1 are trivial from the ratio test, and Abel’s test
is required for the |z| = 1 cases.