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Infinite Series of Complex Numbers
Infinite Series
DefinitionAn infinite series
∞∑𝑘=0
𝑎_{𝑘}=𝑎_{0}+𝑎_{1}+𝑎_{2}+⋯+𝑎_{𝑛}+𝑎_{𝑛+1}
(with 𝑎_{𝑘}∈ℂ) converges to 𝑆 if the sequence of partial sums {𝑆_{𝑛}}, given by
𝑆_{𝑛}=𝑛∑𝑘=0
=𝑎_{𝑘}=𝑎_{0}+𝑎_{1}+𝑎_{2}+⋯+𝑎_{𝑛}
converges to 𝑆.
Example
Consider ∞∑𝑘=0
𝑧^{𝑘}, for some 𝑧∈ℂ. having that
𝑆_{𝑛}=1+𝑧+𝑧^{2}+⋯+𝑧^{𝑛}
Finding a closed formula for 𝑆_{𝑛} in order to find the limit as 𝑛→∞.
Trick:
𝑆_{𝑛}=1+𝑧+𝑧^{2}+⋯+𝑧^{𝑛}, so
𝑧⋅𝑆_{𝑛}=𝑧+𝑧^{2}+⋯+𝑧^{𝑛}+𝑧^{𝑛+1}, thus
𝑆_{𝑛}−𝑧⋅𝑆_{𝑛}=1−𝑧^{𝑛+1}
Hence 𝑆_{𝑛}=1−𝑧^{𝑛+1}1−𝑧
for 𝑧≠1, and since 𝑧^{𝑛+1}→0 as 𝑛→∞ as long as 𝑧<1. Having that
∞∑𝑘=0
𝑧^{𝑘}=11−𝑧
for 𝑧<1
For 𝑧≥1
TheoremIf a series ∞∑𝑘=0
𝑎^{𝑘} converges then 𝑎^{𝑘}→0 as 𝑘→∞.
If 𝑧≥1, then 𝑧^{𝑘}→
0 as 𝑘→∞, thus ∞∑𝑘=0
𝑧^{𝑘} does not converge for 𝑧≥1. Thus the series diverges for 𝑧≥1.
Let's now analyze the real and imaginary parts of the equation ∞∑𝑘=0
𝑧^{𝑘}=11−𝑧
for 𝑧<1;
For 𝑧<1
Writing 𝑧=𝑟ℯ^{𝑖𝜃}, then 𝑧^{𝑘}=𝑟^{𝑘}ℯ^{𝑖𝑘𝜃}=𝑟^{𝑘}cos(𝑘𝜃)+𝑖𝑟^{𝑘}sin(𝑘𝜃). Thus
∞∑𝑘=0
𝑧^{𝑘}=∞∑𝑘=0
𝑟^{𝑘}cos(𝑘𝜃)+𝑖∞∑𝑘=0
𝑟^{𝑘}sin(𝑘𝜃)
Furthermore,
11−𝑧
=11−𝑟ℯ^{𝑖𝜃}
=1−𝑟ℯ^{−𝑖𝜃}(1−𝑟ℯ^{𝑖𝜃})(1−𝑟ℯ^{−𝑖𝜃})
=1−𝑟cos 𝜃+𝑖𝑟sin 𝜃1−𝑟ℯ^{𝑖𝜃}−𝑟ℯ^{−𝑖𝜃}+𝑟^{2}
=1−𝑟cos 𝜃+𝑖𝑟sin 𝜃1−2𝑟cos 𝜃+𝑟^{2}
Thus
∞∑𝑘=0
𝑟^{𝑘}cos(𝑘𝜃)=1−𝑟cos 𝜃1−2𝑟cos 𝜃+𝑟^{2}
and ∞∑𝑘=0
𝑟^{𝑘}sin(𝑘𝜃)=𝑟sin 𝜃1−2𝑟cos 𝜃+𝑟^{2}
Another Example
Consider ∞∑𝑘=1
𝑖^{𝑘}𝑘
. Does this series converge?
 Note that
∞∑𝑘=1
𝑖^{𝑘}𝑘
=∞∑𝑘=1
1𝑘
is the harmonic series, which is known to diverge. One way to see this:
∞∑𝑘=1
1𝑘
=1+12
+13+14
≥1/2} +15+16+17+18
≥1/2} +19+⋯+116
≥1/2} +⋯
 But does the series itself (without the absolute values) converge? let's split it up into real and imaginary parts.
 Note: When 𝑘 is even (i.e. 𝑘 is of the form 𝑘=2𝑛), then 𝑖^{𝑘}=𝑖^{2𝑛}=(−1)^{𝑛} is real. When 𝑘 is odd (i.e. 𝑘 is of the form 𝑘=2𝑛+1), then 𝑖^{𝑘}=𝑖^{2𝑛+1}=𝑖(−1)^{𝑛} is purely imaginary. Thus
∞∑𝑘=1
𝑖^{𝑘}𝑘
=∞∑𝑛=1
𝑖^{2𝑛}2𝑛
+∞∑𝑛=0
𝑖^{2𝑛+1}2𝑛+1
=12
∞∑𝑛=1
(−1)^{𝑛}𝑛
+𝑖∞∑𝑛=0
(−1)^{𝑛}2𝑛+1
But
∞∑𝑛=1
(−1)^{𝑛}𝑛
=−1+12
−13
+14
−+⋯
is the alternating harmonic series, which converges.
Absolute Convergence
DefinitionA series ∞∑𝑘=0
𝑎_{𝑘} converges absolutely if the series ∞∑𝑘=0
𝑎_{𝑘} converges.
Examples
∞∑𝑘=0
𝑧_{𝑘} converges and converges absolutely for 𝑧<1.
∞∑𝑘=0
𝑖^{𝑘}𝑘
converges, but not absolutely.
TheoremIf ∞∑𝑘=0
𝑎_{𝑘} converges absolutely, then it also converges, and ∞∑𝑘=0
𝑎_{𝑘}≤∞∑𝑘=0
𝑎_{𝑘}.
Example
If 𝑧<1, then the series ∞∑𝑘=0
𝑧^{𝑘} converge absolutely, so
∞∑𝑘=0
𝑧^{𝑘}≤∞∑𝑘=0
𝑧^{𝑘}
But the lefthand side equals 11−𝑧
, and the righthand side equals 11−𝑧
, so that
11−𝑧
≤11−𝑧
Power Series (Taylor Series)
DefinitionA power series (also called Taylor series), centered at 𝑧_{0}∈ℂ, is a series of the form
∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}
Example
∞∑𝑘=0
𝑧_{𝑘} is a power series with 𝑎_{𝑘}=1, 𝑧_{0}=0. It converges for 𝑧<1.
∞∑𝑘=0
(−1)^{𝑘}2^{𝑘}
𝑧^{2𝑘}=1−𝑧^{2}2
+𝑧^{4}4
−𝑧^{6}8
+−⋯=∞∑𝑘=0
−𝑧^{2}2
^{𝑘}=∞∑𝑘=0
𝑤^{𝑘}, where 𝑤=−𝑧^{2}2
. This series converges when 𝑤<1, and diverges when 𝑤≥1. Therefore, the original series converges when 𝑧<2 and diverges when 𝑧≥2
The Radius of Convergence
For what values of 𝑧 does a power series converge?
TheoremLet ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} be a power series. Then there exists a number 𝑅, with 0≤𝑅≤∞, such that the series converges absolutely in {𝑧−𝑧_{0}<𝑅} and diverges in {𝑧−𝑧_{0}>𝑅}. Furthermore, the convergence is uniform in {𝑧−𝑧_{0}≤𝑟} for each 𝑟<𝑅
𝑅 is called the radius of convergence of the power series.
Examples
∞∑𝑘=0
𝑧^{𝑘} has the radius of convergence 1.
∞∑𝑘=0
(−1)^{𝑘}2^{𝑘}
𝑧^{2𝑘} has radius of convergence 2
∞∑𝑘=0
𝑘^{𝑘}𝑧^{𝑘} Pick an arbitrary 𝑧∈ℂ\{0}. Observe that 𝑘^{𝑘}𝑧^{𝑘}=(𝑘𝑧)^{𝑘}≥2^{𝑘} as soon as 𝑘≥2𝑧
, thus the series does not converge for any 𝑧≠0. The radius of convergence of this power series is 0!
∞∑𝑘=0
𝑧^{𝑘}𝑘^{𝑘}
Pick an arbitrary 𝑧∈ℂ. Observe that 𝑧^{𝑘}𝑘^{𝑘}
=𝑧𝑘
^{𝑘}≤12
^{𝑘} as soon as 𝑘≥2𝑧. Thus the series converges absolutely for all 𝑧∈ℂ, and so 𝑅=∞!
Analyticity of Power Series
TheoremSuppose that ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} is a power series of radius of convergence 𝑅>0. Then
𝑓(𝑧)=∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} is analytic in {𝑧−𝑧_{0}<𝑅}
Furthermore, the series can be differentiated term by term, i.e.
𝑓′(𝑧)=∞∑𝑘=1
𝑎_{𝑘}⋅𝑘(𝑧−𝑧_{0})^{𝑘−1}, 𝑓″(𝑧)=∞∑𝑘=2
𝑎_{𝑘}⋅𝑘(𝑘−1)(𝑧−𝑧_{0})^{𝑘−2}, ⋯
In particular, 𝑓^{(𝑘)}(𝑧_{0})=𝑎_{𝑘}⋅𝑘!, i.e. 𝑎_{𝑘}=𝑓^{(𝑘)}(𝑧_{0})𝑘!
for 𝑘≥0.
Example
∞∑𝑘=0
𝑧^{𝑘} has the radius of convergence 1, and so by the theorem,
𝑓(𝑧)=∞∑𝑘=0
𝑧^{𝑘} is analytic in {𝑧<1}
Taking the derivative and differentiating term by term (as in the theorem), we find
𝑓′(𝑧)=∞∑𝑘=1
𝑘𝑧^{𝑘−1} (=∞∑𝑘=0
(𝑘+1)𝑧^{𝑘})
But also know that 𝑓(𝑧)=11−𝑧
, and so 𝑓′(𝑧)=1(1−𝑧)^{2}
. Thus
∞∑𝑘=0
(𝑘+1)𝑧^{𝑘}=1(1−𝑧)^{2}
Integration of Power Series
Note: Power series can similarly be integrated term by term:
FactIf ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} has radius of convergence 𝑅, then for any 𝑤 with 𝑤−𝑧_{0}<𝑅. we have that
𝑤∫𝑧_{0}
∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}𝑑𝑧=∞∑𝑘=0
𝑎_{𝑘}𝑤∫𝑧_{0}
(𝑧−𝑧_{0})^{𝑘}𝑑𝑧=∞∑𝑘=0
𝑎_{𝑘}1𝑘+1
(𝑤−𝑧_{0})^{𝑘+1}
Here, the integral is taken over any curve in the disk {𝑧−𝑧_{0}<𝑅} from 𝑧_{0} to 𝑤.
Example
Let's again look at the power series ∞∑𝑘=0
𝑧^{𝑘}, which has 𝑅=1. Then for any 𝑤 with 𝑤<1, thus have
𝑤∫𝑧_{0}
∞∑𝑘=0
𝑧^{𝑘}𝑑𝑧=∞∑𝑘=0
𝑤∫𝑧_{0}
𝑧^{𝑘}𝑑𝑧=∞∑𝑘=0
1𝑘+1
𝑤^{𝑘+1}=∞∑𝑘=1
𝑤^{𝑘}𝑘
Also know that ∞∑𝑘=0
𝑧^{𝑘}=11−𝑧
, hence
𝑤∫𝑧_{0}
∞∑𝑘=0
𝑧^{𝑘}𝑑𝑧=𝑤∫𝑧_{0}
11−𝑧
𝑑𝑧=−Log(1−𝑧)𝑤｜0
=−Log(1−𝑤)
Here, we used that Log z is analytic in ℂ\(−∞,0], hence −Log(1−z) is analytic in ℂ\[1,∞), in particular in {z<1}, where it is a primitive of 11−𝑧
We have shown:
𝑤∫𝑧_{0}
∞∑𝑘=0
𝑧^{𝑘}𝑑𝑧=∞∑𝑘=1
𝑤^{𝑘}𝑘
and 𝑤∫𝑧_{0}
∞∑𝑘=0
𝑧^{𝑘}𝑑𝑧=−Log(1−𝑤)
hence
∞∑𝑘=1
𝑤^{𝑘}𝑘
=−Log(1−𝑤) for 𝑤<1
Letting z=1−𝑤 this becomes
Log z=−∞∑𝑘=1
(1−z)^{𝑘}𝑘
=∞∑𝑘=1
(−1)^{𝑘+1}𝑘
(z−1)^{𝑘} for z−1<1
Ratio Test
Given a power series ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}, there exists a number 𝑅 with 0≤𝑅≤∞ such that the series converges (absolutely) in {𝑧−𝑧_{0}<𝑅} and diverges in {𝑧−𝑧_{0}>𝑅}.
Theorem (Ratio Test)If the sequence 𝑎_{𝑘}𝑎_{𝑘+1}
has a limit as 𝑘→∞ then this limit is the radius of convergence, 𝑅, of the poser series ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}.
Note: "∞" is an allowable limit.
Examples
∞∑𝑘=0
𝑧^{𝑘}: Here,
𝑎_{𝑘}=1, so 𝑎_{𝑘}𝑎_{𝑘+1}
=1→1 as 𝑘→∞. Thus 𝑅=1.
∞∑𝑘=0
𝑘𝑧^{𝑘}: Here,
𝑎_{𝑘}=𝑘, so 𝑎_{𝑘}𝑎_{𝑘+1}
→1 as 𝑘→∞. Thus 𝑅=1.
∞∑𝑘=0
𝑧^{𝑘}𝑘!
: Here, 𝑎_{𝑘}=1𝑘!
, so 𝑎_{𝑘}𝑎_{𝑘+1}
=(𝑘+1)!𝑘!
=𝑘+1→∞ as 𝑘→∞. Thus 𝑅=∞.
∞∑𝑘=0
𝑧^{𝑘}𝑘^{𝑘}
: Here, 𝑎_{𝑘}=1𝑘^{𝑘}
, so 𝑎_{𝑘}𝑎_{𝑘+1}
=(𝑘+1)^{𝑘+1}(𝑘)^{𝑘}
=(𝑘+1)1+1𝑘
→∞?? as 𝑘→∞
The Root Test
Theorem (Root Test)If the sequence {^{𝑘}𝑎_{𝑘}} has a limit as 𝑘→∞ then 𝑅=1
Lim𝑘→∞{^{𝑘}𝑎_{𝑘}}
.
Note:
 If
Lim𝑘→∞{^{𝑘}𝑎_{𝑘}}=0 then 𝑅=∞.
 If
Lim𝑘→∞{^{𝑘}𝑎_{𝑘}}=∞ then 𝑅=0.
Examples
∞∑𝑘=0
𝑧^{𝑘}𝑘^{𝑘}
: Here, 𝑎_{𝑘}=1𝑘^{𝑘}
, so ^{𝑘}𝑎_{𝑘}=1𝑘
→0 as 𝑘→∞. Thus 𝑅=∞.
∞∑𝑘=0
𝑘𝑧^{𝑘}: Here, 𝑎_{𝑘}=𝑘, so ^{𝑘}𝑎_{𝑘}=^{𝑘}𝑘→1 as 𝑘→∞. Thus 𝑅=1.
∞∑𝑘=0
2^{𝑘}𝑧^{𝑘}: Here, 𝑎_{𝑘}=2^{𝑘}, so ^{𝑘}𝑎_{𝑘}=2→2 as 𝑘→∞. Thus 𝑅=12
.
∞∑𝑘=0
(−1)^{𝑘}2^{𝑘}
𝑧^{2𝑘}: Here, 𝑎_{2𝑘}=(−1)^{𝑘}2^{𝑘}
, and 𝑎_{2𝑘+1}=0, so ^{2𝑘}𝑎_{2𝑘}=12^{1/2}
for 𝑘≥1 and ^{2𝑘+1}𝑎_{2𝑘+1}
=0, and so the sequence ^{𝑘}𝑎_{𝑘}
(𝑘≥1) is
0, 12
, 0, 12
, 0, 12
, 0,⋯
and this sequence does not have a limit. Note: 𝑎_{𝑘}𝑎_{𝑘+1} has no limit either.
The Cauchy Hadamard Criterion
For the series ∞∑𝑘=0
(−1)^{𝑘}2^{𝑘}
𝑧^{2𝑘}, neither the root test nor the ratio test "work".
Yet, letting 𝑤=𝑧^{2}, the series becomes ∞∑𝑘=0
(−1)^{𝑘}2^{𝑘}
𝑤^{𝑘}, and ^{𝑘}(−1)^{𝑘}2^{𝑘}
=12
→12
as 𝑘→∞, so the new series converges for 𝑤<2. Thus the original series converges for 𝑧<2. Is there a formula that finds this?
Fact (CauchyHadamard)The radius of convergence of the poser series ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} equals
𝑅=1
Lim sup𝑘→∞ ^{𝑘}𝑎_{𝑘}
Analytic Functions and Power Series
Recall: ∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘} is analytic in {𝑧−𝑧_{0}<𝑅}, where 𝑅 is the radius of convergence of the poser series, Now:
TheoremLet 𝑓:𝑈→ℂ be analytic and let {𝑧−𝑧_{0}<𝑟}⊂𝑈. Then in this disk, 𝑓 has a power series representation
𝑓(𝑧)=∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}, 𝑧−𝑧_{0}<𝑟, where 𝑎_{𝑘}=𝑓^{(𝑘)}(𝑧_{0})𝑘!
, 𝑘≥0.
The radius of convergence of this power series is 𝑅≥𝑟
Examples
𝑓(𝑧)=∞∑𝑘=0
𝑎_{𝑘}(𝑧−𝑧_{0})^{𝑘}, 𝑧−𝑧_{0}<𝑟, where 𝑎_{𝑘}=𝑓^{(𝑘)}(𝑧_{0})𝑘!
, 𝑘≥0.
 𝑓(𝑧)=ℯ^{𝑧}, then 𝑓^{(𝑘)}(𝑧)=ℯ^{𝑧}. Letting 𝑧_{0}=0, we have 𝑓^{(𝑘)}(𝑧_{0})=ℯ^{0}=1 for all 𝑘. Thus 𝑎_{𝑘}=
1𝑘!
for all 𝑘, and so
ℯ^{𝑧}=∞∑𝑘=0
𝑧^{(𝑘)}𝑘!
for all 𝑧∈ℂ
 𝑓(𝑧)=ℯ^{𝑧} as above, but now let 𝑧_{0}=1. Then 𝑓^{(𝑘)}(𝑧_{0})=ℯ^{1}=ℯ for all 𝑘. Thus 𝑎_{𝑘}=
ℯ𝑘!
for all 𝑘, and so
ℯ^{𝑧}=∞∑𝑘=0
ℯ𝑘!
(𝑧−1)^{𝑘} for all 𝑧∈ℂ

𝑓(𝑧)=
sin 𝑧 is analytic in ℂ. Let 𝑧_{0}=0. Then
𝑓(𝑧)=sin 𝑧, 𝑓(0)=0
𝑓′ (𝑧)=cos 𝑧, 𝑓′(0)=1
𝑓″(𝑧)=−sin 𝑧, 𝑓″(0)=0
𝑓^{(3)}(𝑧)=−cos 𝑧, 𝑓^{(3)}(0)=−1
𝑓^{(4)}(𝑧)=sin 𝑧, 𝑓^{(4)}(0)=0
⋯
Thus
sin 𝑧=0+11!
𝑧+02!
𝑧^{2}+−13!
𝑧^{3}+04!
𝑧^{4}+15!
𝑧^{5}+⋯
=𝑧−𝑧^{3}3!
+𝑧^{5}5!
−𝑧^{7}7!
+𝑧^{9}9!
−+⋯
=∞∑𝑘=0
(−1)^{𝑘}(2𝑘+1)!
𝑧^{2𝑘+1}
 𝑓(𝑧)=
cos 𝑧 is analytic in ℂ. Let 𝑧_{0}=0. Then
cos 𝑧=𝑑𝑑𝑧
sin 𝑧=𝑑𝑑𝑧
∞∑𝑘=0
(−1)^{𝑘}(2𝑘+1)!
𝑧^{2𝑘+1}
=∞∑𝑘=0
(−1)^{𝑘}(2𝑘+1)!
(2𝑘+1)𝑧^{2𝑘}
=∞∑𝑘=0
(−1)^{𝑘}(2𝑘)!
𝑧^{2𝑘}
=1−𝑧^{2}2!
+𝑧^{4}4!
−𝑧^{6}6!
+𝑧^{8}8!
−+⋯
A Corollary
Note: The theorem implies that an analytic function is entirely determined in a disk by all of its derivatives 𝑓^{(𝑘)}(𝑧_{0}) at the center 𝑧_{0} of the disk.
CorollaryIf 𝑓 and 𝑔 are analytic in {𝑧−𝑧_{0}<𝑟} and if 𝑓^{(𝑘)}(𝑧_{0})=𝑔^{(𝑘)}(𝑧_{0}) for all 𝑘, then 𝑓(𝑧)=𝑔(𝑧) for all 𝑧 in {𝑧−𝑧_{0}<𝑟}.