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

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Recall

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.

Introduction

In complex analysis, a power series is not merely a formal algebraic object; it defines a geometric and analytic entity. A central question is whether the infinite sum of a power series behaves like a finite polynomial. Specifically: can we differentiate a power series term by term?

To answer this, we must prove two distinct things:

Algebraic/Geometric: The new, term-by-term differentiated series converges in the exact same region (disk) as the original series.
Analytic: The function defined by this new series is, in fact, the true complex derivative of the original function.

Analyticity of Power Series: A Comprehensive Guide

Before we attempt to take a derivative, we must ensure that the “derivative series” actually converges.

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

In other words, term-by-term differentiation (or integration) does not change the “circle of convergence.” The new series (the derivative) will still converge absolutely 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 the center is $a = 0$. Let R and $R_{\text{deriv}}$ be the radii of convergence for $\sum a_n z^n$ and $\sum n a_n z^{n-1}$ respectively.

Part 1: Show $R_{\text{deriv}} \ge R$. Pick any $z$ with $|z| < R$. We will show that $\sum n a_n z^{n-1}$ converges absolutely.

Because $|z|$ is strictly less than $R$, we can “sacrifice” a bit of room and choose a real number $\rho$ such that $|z| < \rho < R$. Let $r := \frac{|z|}{\rho}.$ Since $|z| < \rho$, we have $0 < r < 1$.

We are going to rewrite the absolute value of the terms in the derived series to compare (or estimate) them to the original series as follows: $|na_nz^{n-1}| = \frac{n}{\rho}(\frac{|z|}{\rho})^{n-1} \cdot |a_n\rho^n| = \frac{nr^{n-1}}{\rho}|a_n\rho^n|$ $(\star)$

Consider the sequence $n r^{n-1}$. Because $0 < r < 1$, the ratio test shows that the series $\sum_{n=1}^\infty n r^{n-1}$ converges: $\lim_{n \to \infty} \frac{(n + 1) r^{n}}{n r^{n-1}} = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)r = r < 1$.

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.

Because the sequence $n r^{n-1}$ represents the terms of a convergent series, the sequence itself must approach $0$, meaning it is bounded. Thus, there exists a constant $M > 0$ such that $n r^{n-1} \le M$ for all $n \in \mathbb{N}$.

Substituting this bound M back into our equation $(\star)$: $|na_nz^{n-1}| \le \frac{M}{\rho} |a_n\rho^n|$

Since $\rho < R$, the original series evaluated at $\rho$, $\sum_{n=1}^\infty |a_n\rho^n|$, converges. By the Comparison Test, since our derived terms are bounded by a constant multiple of a convergent series, $\sum n a_n z^{n-1}$ must converge absolutely. Therefore, the derivative’s convergence zone is at least as big as the original’s: $R_{\text{deriv}} \ge R$.

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

Pick (or fix) any z with $|z| < R_{\text{deriv}}$. We are given that $\sum n a_n z^{n-1}$ converges absolutely. 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.

This means the original’s convergence zone must be at least as big as the derivative’s, $R \ge R_{\text{deriv}}$. Since $R_{\text{deriv}} \ge R$ and $R \ge R_{\text{deriv}}$, we conclude that $R = R_{\text{deriv}}$. $\blacksquare$

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 coefficient of the $n$-th degree term is $b_n := (n+1)a_{n+1}$. Let’s find the radius of convergence for the new coefficients $b_n$: $\limsup_{n\to\infty} |b_n|^{1/n} = \limsup_{n\to\infty} \bigl( (n+1)^{1/n} |a_{n+1}|^{1/n} \bigr)$

Recall from real analysis that $\lim_{n \to \infty} (n+1)^{1/n} = 1$. Therefore, the $(n+1)^{1/n}$ factor does not affect the limit superior: $\limsup_{n\to\infty} |b_n|^{1/n} = \limsup_{n\to\infty} |a_{n+1}|^{1/n} = \limsup_{n\to\infty} |a_n|^{1/n} = \frac{1}{R}$. Hence the derivative has the same radius R.

Corollaries for Higher Derivatives and Integrals

Because the first derivative shares the same radius of convergence, we can apply the Lemma inductively to higher derivatives and antiderivatives.

Corollary. All higher derivatives and integrals of a power series share the same radius of convergence $R$.

$k$-th Derivative: $\frac{d^k}{dz^k}\Bigl(\sum a_n(z-a)^n\Bigr) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1)a_n(z-a)^{n-k}$
Antiderivative: $\int \sum a_n(z-a)^n dz = C + \sum_{n=0}^\infty \frac{a_n}{n+1}(z-a)^{n+1}$

Equality of radii says nothing about what happens on the boundary circle |z - a| = R: one series may converge at some boundary points while its derivative does not (and vice versa), e.g., $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$. The radius is $R=1$, and because $\sum \frac{1}{n^2}$ converges, this series converges absolutely everywhere on the boundary $|z|=1$. However, its derivative is $f'(z) = \sum_{n=1}^\infty \frac{z^{n-1}}{n}$. This also has $R=1$, but on the boundary, it becomes the harmonic series $\sum \frac{1}{n}$, which diverges

Theorem. Term-by-Term Differentiation of Power series. Let f(z) = $\sum_{n=0}^\infty a_n(z-a)^n$ be a power series with radius of convergence $R > 0$. Then, f is analytic on the open disk $B(a; R)$ and its complex derivative is given by $f'(z) = \sum_{n=1}^\infty na_n(z-a)^{n-1}$ for all $|z| < R$.

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, we can actually get: $(z+h)^n, (z+h)^n = \sum_{k=0}^n {n \choose k}z^{n-k}h^k \leadsto \frac{(z+h)^n-z^n}{h}-nz^{n-1} = \frac{z^n + {n \choose 1}z^{n-1}h + \cdots + {n \choose n}h^n -z^n}{h} -nz^{n-1}$

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

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 entirely 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 rigorously proves that $f'(z)$ exists, and that $f'(z) = g(z)$ for every $z$ inside the radius of convergence. $\blacksquare$

Profound Consequences

The theorem we just proved unlocks the entirety of complex analytic function theory. The implications are staggering.

Infinite Differentiability.
In real calculus, a function can be differentiable once, but not twice (e.g., $f(x) = x|x|$). In complex analysis, because the derivative of a power series is another power series with the exact same radius of convergence, we can apply the theorem repeatedly.
If a function can be represented by a power series, it is infinitely complex-differentiable on that disk.
Recovery of Taylor Coefficients.
If $f(z) = a_0 + a_1(z-a) + a_2(z-a)^2 + a_3(z-a)^3 + \dots$, evaluate $f(a)$: all terms with $(z-a)$ vanish, yielding $f(a) = a_0$.
Take the first derivative and evaluate at $a$: $f'(a) = a_1$. Take the second derivative: $f''(z) = 2a_2 + 3\cdot 2 a_3(z-a) + \dots \implies f''(a) = 2! a_2$.
By induction, the $n$-th derivative evaluated at the center gives $f^{(n)}(a) = n! a_n$.
Therefore, the coefficients of the series are uniquely determined by the function’s derivatives: $\boxed{a_n = \frac{f^{(n)}(a)}{n!}}$
Every power series is exactly the Taylor Series of the function it defines.
Analyticity implies Holomorphy.
A function is called analytic if it can be represented by a local power series. A function is called holomorphic if it is complex-differentiable. This article proves that Analyticity implies Holomorphy.
Cauchy’s Integral Formula will prove the reverse: Holomorphy implies Analyticity —meaning any complex-differentiable function automatically has a valid power series representation.