All those who seem stupid, they are, and also so are half of those who do not, Quevedo

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Introduction

Definition. Complex sequence A sequence of complex numbers is a function $a: \mathbb{N} \to \mathbb{C}$. We usually denote it by $(a_n)_{n \in \mathbb{N}}$ or simply $(a_n)$, where $a_n := a(n)$. The value $a_1$ is called the first term of the sequence, $a_2$ the second term, and in general $a_n$ the n-th term of the sequence.

Definition. Convergent complex sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is said to converge to a complex number $L \in \mathbb{C}$ if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n \geq N$ one has $|a_n - L| < \varepsilon$. In this case we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$, and L is called the limit of the sequence $(a_n)_{n \in \mathbb{N}}$.

Definition. Cauchy sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is called a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n, m \geq N$ one has $|a_n - a_m| < \varepsilon$.

Definition. Series and partial sums.Let $(a_n)_{n \in \mathbb{N}}$ be a complex sequence. For each n $\in \mathbb{N}$, the finite sum $s_n := a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ is called the n-th partial sum of the (infinite) series $\sum_{k=1}^\infin a_k$ which we also denote simply by $\sum a_n$ when the index is clear from the context.

Definition. Convergent series. The series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $s \in \mathbb{C}$ if the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ defined by $s_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ converges to s, that is, $\lim_{n \to \infty} s_n = s$. In this case we write $s := \sum_{n=1}^\infin a_n$. If the sequence $(s_n)_{n \in \mathbb{N}}$ does not converge, we say that the series $\sum_{n=1}^{\infty} a_n$ diverges (or does not converge).

Definition. A complex power series centered at 0 in the variable z is a series of the form $a_0 + a_1z + a_2z^2 + \cdots = \sum_{n=0}^\infty a_n z^n$ with coefficients $a_i \in \mathbb{C}$

Definition. A complex power series centered at a complex number $a \in \mathbb{C} $ is an infinite series of the form: $\sum_{n=0}^\infty a_n (z - a)^n,$ where each $a_n \in \mathbb{C}$ is a coefficient, z is a complex variable, and $(z - a)^n$ is the nth power about the center.

Theorem. Given a power series $\sum_{n=0}^\infty a_n z^n$, there exists a unique value R, $0 \le R \le \infin$ (called the radius of convergence) such that:

For any z with |z| < R (inside the circle), the series $\sum_{n=0}^\infty a_n z^n$ converges absolutely (this is a “green light” zone).
For any z with |z| > R, the series diverges (this is a “red light” zone).
On the Circle (|z| = R), this theorem gives no information. This is the yellow light zone —the series could converge or diverge.

Analyticity of Power Series: A Comprehensive Guide

Lemma. The power series $\sum_{n=0}^\infty a_n(z-a)^n$ and $\sum_{n=1}^\infty na_n(z-a)^{n-1}$ have the same radius of convergence.

In other words, term-by-term differentiation (or integration) does not change the “circle of convergence.” The new series (the derivative) will still converge inside the same radius R and diverge outside it. The only thing that might change is the behavior on the boundary circle itself, but the radius itself R remains the same.

Proof.

Assume without loss of generality that a = 0.

Let R and $R_{\text{deriv}}$ be the radii for $\sum a_n z^n$ and $\sum n a_n z^{n-1}$ respectively.

Part 1: Show $R_{\text{deriv}} \ge R$. Fix z with |z| < R. We will show that $\sum n a_n z^{n-1}$ must also converge.

Since $|z| < R$, we can “sacrifice” a point $\rho$ such that $|z| < \rho < R$, and let $r := \frac{|z|}{\rho}, 0 \lt r \lt 1$

To estimate the terms in the differentiated power series by the terms in the original series, we rewrite their absolute values as follows: $|na_nz^{n-1}| = \frac{n}{\rho}(\frac{|z|}{\rho})^{n-1} \cdot |a_n\rho^n| = \frac{nr^{n-1}}{\rho}|a_n\rho^n|$.

The ratio test shows that the series $\sum_{n=1}^\infty n r^{n-1}$ converges (indeed $\sum n r^{n-1}=\frac{1}{(1-r)^2}$), since $\lim_{n \to \infin} \frac{(n + 1) r^{n}}{n r^{n-1}} = \lim_{n \to \infin} (1+\frac{1}{n})r = r \lt 1$, so the sequence $n r^{n-1}$ is bounded for some M, i.e., ∃M > 0 s.t. $n r^{n-1} \le M, \forall n \in \mathbb{N}$.

The ratio test determines the convergence of a series by calculating the limit L of the absolute value of the ratio of consecutive terms $|\frac{a_{n+1}}{a_{n}}|$. If L < 1, the series is absolutely convergent. If L > 1, the series is divergent. If L = 1, the test is inconclusive, and other methods are needed.

It follows that $|na_nz^{n-1}| = \frac{nr^{n-1}}{\rho}|a_n\rho^n| \le \frac{M}{\rho}\cdot |a_n\rho^n|, \forall n \in \mathbb{N}$

Since $\rho < R$ (radius of converge), the series $\sum_{n=1}^\infty |a_n\rho^n|$ converges.

By the Comparison Test, since $|n a_n z^{n-1}|$ is less than or equal to the terms of a convergent series (i.e., $\frac{M}{\rho} \sum |a_n\rho^n|$), the series $\sum n a_n z^{n-1}$ must converge absolutely (multiplying by M/ρ preserves convergence). This means the derivative’s convergence zone must be at least as big as the original’s, $R_{\text{deriv}} \ge R$.

Conversely, show $R \ge R_{\text{deriv}}$.

Pick any z with $|z| < R_{\text{deriv}}$. We will show that $\sum a_n z^n$ must also converge.

We are given that $\sum n a_n z^{n-1}$ converges absolutely for $|z| < R_{\text{deriv}}$. We want to test $\sum |a_n z^n|$. Let’s compare its terms to the terms of the series we know converges. $|a_n z^{n}|=\frac{|z|}{n}|n a_n z^{n-1}|\le |z||n a_n z^{n-1}|, \forall n \ge 1$. So by Comparison test with a constant multiple of a convergent series, $\sum_{n=1}^\infty a_nz^n$ is absolutely convergent.

We proved that convergence for $|z| < R_{\text{deriv}}$ implies convergence for $\sum a_n z^n$. This means the original’s convergence zone must be at least as big as the derivative’s, $R \ge R_{\text{deriv}}$.

An alternative proof using Cauchy–Hadamard is as follows. Let $R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}.$ For the derivative series the coefficients are (after re-indexing) $b_n:=n a_n$. Since $n^{1/n}\to 1, \limsup_{n\to\infty}|b_n|^{1/n} =\limsup_{n\to\infty}\bigl(n^{1/n}|a_n|^{1/n}\bigr) =\limsup_{n\to\infty}|a_n|^{1/n}.$

Hence the derivative has the same radius R.

Corollary. All higher derivatives/integrals share the same radius. For $k \ge 1, \frac{d^k}{dz^k}\Bigl(\sum a_n(z-a)^n\Bigr) = \sum_{n\ge k} n(n-1)\cdots(n-k+1)a_n(z-a)^{n-k}$ and $\int \sum a_n(z-a)^n dz = C + \sum \frac{a_n}{n+1}(z-a)^{n+1}.$

Equality of radii says nothing about what happens on |z - a| = R: one series may converge at some boundary points while its derivative does not (and vice versa).

Theorem. Term-by-Term Differentiation of Power series. Let f(z) = $\sum_{n=0}^\infty a_n(z-a)^n$ and assume that this power series has radius of convergence R > 0. Then, f is analytic on B(0; R) and f'(z) = $\sum_{n=1}^\infty na_n(z-a)^{n-1}$ for |z| < R.

This is a cornerstone theorem of complex analysis. It states that a power series is not just a function; it’s an infinitely differentiable analytic function inside its circle of convergence.

Proof.

Assume for simplicity, without loss of generality, that a = 0.

By the previous demonstrated lemma (the power series $\sum_{n=0}^\infty a_n(z-a)^n$ and $\sum_{n=1}^\infty na_n(z-a)^{n-1}$ have the same radius of convergence), we can define g(z) = $\sum_{n=1}^\infty na_nz^{n-1}$ (it converges, and it converges to the same radius of convergence).

We aim to prove that f’(z) exists and equals g(z) for any z inside the radius of convergence R, |z| < R.

We will use the formal limit definition of the derivative and show that the error goes to zero. $f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}$, We must prove that $\lim_{h \to 0} \left[ \frac{f(z+h) - f(z)}{h} - g(z) \right] = 0$.

$\forall z, z + h \in B(0;R)$, let’s analyze the “error” term we want to show is zero. Error = $\frac{f(z+h)-f(z)}{h} -g(z) =[\text{Substitute the series definitions:}] \frac{\sum_{n=0}^\infty a_n(z+h)^n-\sum_{n=0}^\infty a_nz^n}{h}-\sum_{n=1}^\infty na_nz^{n-1}$

Simplify the Sum. n = o, the first term vanishes $a_0(z+h)^0 - a_0z^0 = a_0 -a_0 = 0$.

$= \frac{\sum_{n=1}^\infty a_n((z+h)^n-z^n)}{h}-\sum_{n=1}^\infty na_nz^{n-1} =[\text{Combine them all into a single sum}] \sum_{n=1}^\infty a_n(\frac{(z+h)^n-z^n}{h}-nz^{n-1})$

$n = 1, \frac{z+h-z}{h} - 1 = 1 -1 = 0$

So, our sum actually starts at n = 2: $\text{Error } = \sum_{n=2}^\infty a_n(\frac{(z+h)^n-z^n}{h}-nz^{n-1})$

By using the Binomial expansion for complex numbers, $(z+h)^n, (z+h)^n = \sum_{k=0}^n {n \choose k}z^{n-k}h^k \leadsto \frac{(z+h)^n-z^n}{h}-nz^{n-1} = \frac{z^n + {n \choose 1}z^{n-1}h + \cdots + {n \choose n}h^n -z^n}{h} -nz^{n-1}$

$= \frac{{n \choose 1}z^{n-1}h + \cdots + {n \choose n}h^n}{h} -nz^{n-1} = nz^{n-1} + {n \choose 2}z^{n-2}h + \cdots h^{n-1} -nz^{n-1} = {n \choose 2}z^{n-2}h + {n \choose 3}z^{n-3}h + \cdots h^{n-1}$

The key takeaway is that the entire expression has a factor of $h$ that we can pull out:

$= h\sum_{r=0}^{n-2} \frac{n!}{(n-(r+2)!(r+2)!)}h^rz^{n-2-r}$

Therefore, $\text{|Error| } = |\frac{f(z+h)-f(z)}{h} -g(z)| = | \sum_{n=2}^\infty a_n(h\sum_{r=0}^{n-2} \frac{n!}{(n-(r+2)!(r+2)!)}h^rz^{n-2-r})| \le \sum_{n=2}^\infty |a_n| |h|\sum_{r=0}^{n-2} \frac{n!}{(n-(r+2))!(r+2)!)}|h|^r|z|^{n-2-r}$ where we are using an infinite version of the triangular inequality $|\sum_{n=2}^\infty a_b| \le \sum_{n=2}^\infty |a_n|$.

We lessen the denominator and replace (r+2)! by r! and obviously (n-(r+2))! = (n-r-2)!

$\text{|Error| } \le |h|\sum_{n=2}^\infty |a_n| n(n-1)\sum_{r=0}^{n-2} \frac{(n-2)!}{(n-r-2)!r!}|h|^r|z|^{n-2-r} = |h|\sum_{n=2}^\infty |a_n| n(n-1) (|z|+|h|)^{n-2}$

Take our z with |z| < R. We “sacrifice” a bit of room, choose a $\rho$ such that z is still well inside the circle of radius $\rho$, and $\rho$ is still inside R: $|z| \lt \rho \lt R$.

Choose h small. We are taking a limit as $h \to 0$, so we can assume h is small enough that z + h is also inside the $\rho$-circle, $|z|+|h| \lt \rho, \text{ whenever } |h| \lt \rho - |z|$

We know our original series $\sum a_n z^n$ has radius R. By the previous lemma, the first derivative, $g(z) = \sum n a_n z^{n-1}$, also has radius R. By the same lemma again, the second derivative, $g'(z) = \sum n(n-1) a_n z^{n-2}$, also has radius R. Since we chose $\rho < R$, the point $\rho$ is inside the circle of convergence for the second derivative. Therefore, the series $\sum n(n-1) a_n \rho^{n-2}$ converges absolutely. This means $\sum |a_n| n(n-1) \rho^{n-2}$ is a constant independent of h.

Therefore, $0 \le \text{ |Error| } \le |h|\sum_{n=2}^\infty |a_n| n(n-1) (|z|+|h|)^{n-2} \le |h|\sum_{n=2}^\infty |a_n| n(n-1) \rho^{n-2}$

$0 \le \lim_{h \to 0} |\text{Error}| \le \lim_{h \to 0} (|h| \cdot C)$

By the Squeeze Theorem, the limit of the error is 0.$\lim_{h \to 0} \left[ \frac{f(z+h) - f(z)}{h} - g(z) \right] = 0$. This proves that f’(z) exists and f’(z) = g(z) for every ∣z∣ < R.