If you find yourself in a hole, stop digging, Will Rogers

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Introduction

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.

The infinite series $\sum_{k=1}^{\infty} a_k$ is not defined as the result of “adding up infinitely many numbers”. Rather, it is defined through the sequence of partial sums: each $s_n$ is a perfectly well-defined finite sum, and we study the behavior of the sequence $\{ s_n \}_{n \in \mathbb{N}}$.

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).

To study the convergence of a complex series $\sum_{n=1}^{\infty} a_n$, it is often convenient to apply standard convergence tests (e.g., the comparison test, the ratio test, and Cauchy’s n-th root test) from real analysis to the associated real series $\sum_{n=1}^{\infty} |a_n|$. If $\sum_{n=1}^{\infty} |a_n|$ is found to converge by one of these tests, then $\sum_{n=1}^{\infty} a_n$ is absolutely convergent and hence convergent by the previous proposition.

Complex Power Series: A Comprehensive guide

Power series lie at the heart of complex analysis. They provide concrete representations of functions, reveal convergence behavior governed by elegant geometric structure, and connect the algebraic act of summing terms to the analytic properties of holomorphic functions. In real analysis, a power series converges on an interval; in complex analysis, it converges on a disk. This single shift in geometry —from intervals to disks— unlocks a richer and more unified theory.

Definition. A complex power series centered at $z_0 = 0$ is an infinite series of the form $\sum_{n=0}^\infty a_n z^n = a_0 + a_1z + a_2z^2 + \cdots,$ where each $a_n \in \mathbb{C}$ is a coefficient and $z$ is a complex variable.

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.

A power series centered at $z_0$ is simply one centered at $0$ with $z$ replaced by $z - z_0$. For this reason, most proofs are stated for $z_0 = 0$ without loss of generality —every result extends immediately by the substitution $z \mapsto z - z_0$.

Definition. The $N$th partial sum of a power series is the finite sum $S_N(z) = \sum_{n=0}^N a_n(z - z_0)^n$. The series converges at a point $z$ if the sequence of partial sums $\{ S_N(z) \}_{N=0}^\infty$ converges in $\mathbb{C}$; it converges absolutely at $z$ if $\sum_{n=0}^\infty |a_n(z-z_0)^n|$ converges.

Why absolute convergence is so important? In $\mathbb{C}$, absolute convergence implies convergence (since $\mathbb{C}$ is complete). Moreover, absolutely convergent series can be rearranged without changing the sum —a property that fails for conditionally convergent series. Power series, as we shall prove later on, always converge absolutely inside their disk of convergence.

The Geometric Series

Example. The geometric series $\sum_{n=0}^\infty z^n = 1 + z + z^2 + \cdots $ is a power series centered at $z_0 = 0$ with coefficients $a_n = 1 $ for all n.

Multiply the partial sum by $(1 - z)$:

$(1 - z)(1 + z + z^2 + \cdots + z^n) = 1 - z^{n + 1} \leadsto 1 + z + z^2 + \cdots + z^n = \frac{1 - z^{n + 1}}{1 - z}, \forall z \ne 1$.

Inside the unit disk ($|z| < 1$): If ∣z∣ < 1, then $\lim_{n \to \infin} z^{n + 1} = 0$. Therefore, the geometric series $\sum_{n=0}^\infty z^n$ converges and its sum is the function $\frac{1}{1-z}$ inside the unit disk.

On the unit circle, ∣z∣ = 1, there are two options:

z = 1, the series becomes $1 + 1 + \cdots + 1$ which clearly diverges to ∞.
z = $e^{i\theta} \text{ where } \theta \ne 0$ (any other point on the unit circle). The magnitude of each term is$|z^n| = |e^{in\theta}| = 1$ for all $n$. The terms do not shrink (approach zero, i.e., the series diverges because the terms don’t satisfy the necessary condition for convergence: $\lim_{n \to \infin} a_n = 0$), they keep rotating (endlessly) around the unit circle. Therefore, the series diverges for |z| = 1.

Outside the unit disk, ∣z∣ > 1, we have $|z^{n + 1}| = |z|^{n+1} \to \infin$. The terms grow without bond, “blow up”, so the series diverges.

The unit circle is the event horizon for $\sum_{n=0}^\infty z^n$, everything inside shrink (gets sucked into 0) and the series converges; everything on or outside stays large (or blow up) and the series diverges.

This example reveals a principle: the convergence region of a power series is a disk, and there is a sharp boundary —the circle of convergence— separating convergence from divergence. The next theorem shows this is universal.

The Radius of Convergence

Theorem. . Given a power series $\sum_{n=0}^\infty a_n z^n$, there exists a unique value $R$ with $0 \le R \le \infty$ (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 the “green light” zone).
For any z with $|z| > R$, the series diverges (this is the “red light” zone).
On the Circle (|z| = R), this theorem gives no information. This is the yellow light zone —the series may converge absolutely, converge conditionally, or diverge, depending on the specific series and the specific point on the circle. We shall see all three behaviors in the examples below.

For a series centered at $z_0$, the same result holds with $|z|$ replaced by $|z - z_0|$. The convergence region is a disk of radius $R$ centered at $z_0$, called the disk of convergence: $D(z_0, R) = \{z \in \mathbb{C} : |z-z_0| < R\}.$

Proof.

Defining $R$. Let $S := \bigl\{r \ge 0 : \sup_{n \ge 0}\, |a_n|\,r^n < \infty \bigr\}$ be the set of non-negative real numbers $r$ for which the sequence $\{ |a_n|\,r^n \}_{n\ge 0}$ is bounded. Note that $0 \in S$ (since $|a_n| \cdot 0^n$ is bounded), so $S$ is non-empty.

We define R as the suppremum (the least upper bound) S := { $r\ge 0 : \sup_{n\ge 0} |a_n|r^n < \infty$. Intuitively, $R$ is the largest radius at which the terms of the series remain “manageable” (bounded). Beyond $R$, the terms inevitably escape all bounds.

Absolute convergence for $|z| < R$.

Pick any complex number z such that |z| < R (inside the circle). Since |z| is less than the supremum R, we can always find another number, $r_1 \ge 0$, that is “sacrificed” to be between them (it sits between |z| and R). Let’s choose an $r_1$ such that $|z| \lt r_1 \le R$, so the sequence of terms $|a_n|r_1^n$ is bounded.

Since $r_1 \in S$, the sequence $|a_n|\,r_1^n$ is bounded by some constant $M$: $|a_n|\,r_1^n \le M$ for all $n \ge 0$.

Now we look at the absolute value of the terms of our original series, $|a_n z^n| = |a_n|\,|z|^n = \bigl(|a_n|\,r_1^n\bigr)\cdot\left(\frac{|z|}{r_1}\right)^n \le M\cdot\rho^n,$ where we define $\rho := |z|/r_1$. Since $|z| < r_1$, we have $0 \le \rho < 1$.

The series $\sum_{n=0}^\infty M\rho^n$ is a convergent geometric series (ratio $\rho < 1$). By the Comparison Test, since every term $|a_n z^n|$ is less than or equal to the corresponding term of a convergent series, our series $\sum |a_n z^n|$ must also converge. This proves that the original series converges absolutely.

Proving Divergence for |z| > R.

We aim to show that if z is outside the circle, |z| > R, the series $\sum a_n z^n$ diverges.

Since |z| exceeds the supremum R, |z| cannot be in our set (the set of radii that keep the terms bounded, $|z|\notin S$). This means the sequence $|a_n| |z|^n$ is unbounded, hence cannot tend to 0. A series whose terms do not tend to 0 diverges.

Why does $|a_nz^n| \not\to 0$ imply $a_nz^n \not\to 0$? Convergence to $0$ in $\mathbb{C}$ is equivalent to the modulus tending to $0$: $a_nz^n \to 0$ if and only if $|a_nz^n| \to 0$. So if the moduli fail to vanish, the terms themselves fail to vanish.

Uniqueness of $R$.

Take any z with |z| < R’. Property (1) for R’ says the series converges at this z. But R is defined as the supremum of all radii for which convergence holds. So if convergence happens at |z| < R’, then all such |z| must lie within the convergence disk of radius R. Thus, $|z| \lt R'\quad \Rightarrow \quad |z|\leq R.$
Taking the supremum over all such |z| gives: $R'\leq R.$
Take any z with |z| > R. Property (2) for R says the series diverges at this z. But property (2) for R’ says divergence happens for all |z| > R’. If divergence already occurs at |z| > R, then the “divergence zone” for R’ cannot start later than that of R.
Thus, $|z| > R \quad \Rightarrow \quad |z|\geq R'.$ Taking the infimum over all such |z| gives: $R\leq R'$.
Conclusion: Both inequalities force equality $R'\leq R\quad \mathrm{and}\quad R\leq R'\quad \Rightarrow \quad R = R'.$
So the radius of convergence is unique.

Geometric Interpretation

Theorem. For a complex power series $\sum a_n (z - z_0)^n$ with radius of convergence $R$, the set of points $z \in \mathbb{C}$ where the series converges absolutely is the open disk $D(z_0, R) = \{z \in \mathbb{C} : |z - z_0| < R\}$, and the series diverges on the exterior $\{z : |z - z_0| > R\}$.

There are three extreme cases:

$R = 0$: The series converges only at the center $z_0$. It is “trapped at a point.”
$R = \infty$: The series converges absolutely for every $z \in \mathbb{C}$. There is no boundary. Such series define entire functions.
$0 < R < \infty$: The generic case. The boundary circle $|z-z_0| = R$ separates convergence from divergence, and behavior on this circle must be analyzed separately.

Computing the Radius of Convergence

For a power series $\sum a_n (z - z_0)^n$, there are two common ways to determine the radius of convergence R:

Theorem .The Cauchy-Hadamard Formula (Root Test). The radius of convergence of $\sum a_n z^n$ is given by $R = \frac{1}{\displaystyle\limsup_{n\to\infty} |a_n|^{1/n}},$ with the conventions: if $\limsup = 0$ then $R = \infty$ (the series converges everywhere); if $\limsup = \infty$ then $R = 0$ (the series converges only at z = z₀).

Theorem (Ratio Test for Power Series). If the limit $\ell := \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}$ exists (and is finite or $+\infty$), then $R = \frac{1}{\ell} = \lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|,$ with the conventions $1/0 = \infty$ and $1/\infty = 0$.

The Cauchy-Hadamard Formula (Root Test)

This formula always works —the $\limsup$ exists in $[0, \infty]$ for every sequence, so the Cauchy–Hadamard formula always yields a definite answer.

Proof:

For simplicity, we’ll analyze the proof for $\sum |a_nz^n|$ centered at $z_0 = 0$.

Convergence when ∣z∣ < R = 1/L where $L = \limsup_{n \to \infty} |a_n|^{1/n}$. Choose a number q such that $L < q < \frac{1}{|z|}$ (possible since $|z| < 1/L$ implies $L < 1/|z|$). By the definition of limsup, eventually (for all large n): $|a_n|^{1/n} \le q \leadsto |a_n| \le q^n \leadsto |a_nz^n| \le (q|z|)^n$

Since q∣z∣<1 by construction, $\sum(q ∣z∣)^n$ is a convergent geometric series. By the Comparison Test, $\sum |a_nz^n|$ converges, then $\sum |a_nz^n|$ converges absolutely.

Divergence when ∣z∣ > 1/L. Choose a number p such that $\frac{1}{|z|} \lt p \lt L$ (possible since $|z| > 1/L$ implies $1/|z| < L$). By the definition of limsup, infinitely many n satisfy: $|a_n|^{1/n} \ge p \leadsto |a_n| \ge p^n \leadsto |a_nz^n| \ge (p|z|)^n$

Since $p > 1/|z|$, we have $p|z| > 1$, so $(p|z|)^n \to \infty$. The terms $|a_n z^n|$ exceed $1$ for infinitely many $n$, so $a_n z^n \not\to 0$, and the series diverges. $\square$

$|a_nz^n| ↛ 0$, which direly implies $a_nz^n ↛ 0$. That’s because convergence to $0 \in \mathbb{C}$ requires both the real and imaginary parts to vanish, which is equivalent to the modulus tending to 0.

Why $\limsup$ and not $\lim$? The ordinary limit $\lim_{n\to\infty} |a_n|^{1/n}$ may not exist (the sequence might oscillate). The $\limsup$ captures the worst-case growth rate by tracking the largest cluster point. This is exactly the right quantity: if even a subsequence of terms grows too fast, the series diverges.

For a general center $z_0$, replace $z^n$ by $(z-z_0)^n$; the same proof goes through with $|z-z_0|$ in place of |z|.

The Ratio Test

Derivation. Apply the classical ratio test for absolute convergence to $\sum |a_n z^n|$:

$\frac{|a_{n+1} z^{n+1}|}{|a_n z^n|} = \frac{|a_{n+1}|}{|a_n|}\,|z| \;\;\xrightarrow{n\to\infty}\;\; \ell\,|z|.$

The series converges absolutely when $\ell|z| < 1$ (i.e., $|z| < 1/\ell$) and diverges when $\ell|z| > 1$ (i.e., $|z| > 1/\ell$). Hence $R = 1/\ell$.

Cauchy–Hadamard vs. Ratio Test. The Cauchy–Hadamard formula always works. The Ratio Test works only when $\lim |a_{n+1}/a_n|$ exists. When both apply, they give the same $R$. In practice, the Ratio Test is often easier to compute.

Some examples

Geometric Series $\sum_{n=0}^\infty z^n$ with coefficients $ a_n = 1$, so using Ratio Test $R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{1}{1} \right| = 1$. It converges for |z| < 1, diverges for |z| > 1. On the boundary |z| = 1, every term has modulus $|z^n| = 1 \ne 0$ (so they cannot approach zero), so the series diverges at every point of the unit circle.
Exponential Function $e^z$: Power series: $ \sum_{n=0}^\infty \frac{z^n}{n!}$. Ratio Test, $\left| \frac{a_n}{a_{n+1}} \right| = \left| \frac{1/n!}{1/(n+1)!} \right| = \frac{(n+1)!}{n!} = \frac{(n+1) \cdot n!}{n!} = n+1 \;\xrightarrow{n\to\infty}\; \infty$. Radius $R = \lim_{n \to \infty} (n+1) = \infty $. The series converges absolutely for every $z \in \mathbb{C}$. There is no boundary circle.
The exponential function $e^z$ is entire (analytic on all of $\mathbb{C}$).
Convergence Nowhere but the Origin. $\sum_{n=0}^\infty n!\, z^n$. Coefficients: $a_n = n!$.
Ratio Test: $\displaystyle \frac{|a_{n+1}|}{|a_n|} = \frac{(n+1)!}{n!} = n+1 \;\xrightarrow{n\to\infty}\; \infty.$
Hence $\ell = \infty$ and $R = 1/\infty = 0$. The series converges only at $z = 0$ (where every term but the constant vanishes). This is the extreme case where the coefficients grow so fast that no non-zero $z$ can tame them.
Convergence Everywhere on the Boundary. The Series $\sum_{n=1}^\infty \frac{z^n}{n^2}$.
Coefficients: $a_n = 1/n^2$. Root Test: $\displaystyle |a_n|^{1/n} = \frac{1}{n^{2/n}} = \frac{1}{(n^{1/n})^2}.$
Since $\lim_{n\to\infty} n^{1/n} = 1$, we get $L = \limsup |a_n|^{1/n} = 1$, so $R = 1$.
Boundary ($|z| = 1$): For $z = e^{i\theta}$, $|a_n z^n| = 1/n^2$. Since $\sum 1/n^2 < \infty$, the series converges absolutely at every point of the unit circle. This shows that boundary convergence everywhere is possible.
Almost-Everywhere Convergence on the Boundary. The Series $\sum_{n=1}^\infty \frac{z^n}{n}$.
Coefficients: $a_n = 1/n$. Root Test: $|a_n|^{1/n} = 1/n^{1/n} \to 1$, so $R = 1$. (Alternatively, the Ratio Test also gives $|a_n/a_{n+1}| = (n+1)/n \to 1$).
This series represents $-\log(1 - z)$ for $|z| < 1$.

Boundary ($|z| = 1$):

At $z = 1$: $\sum 1/n$ is the harmonic series—diverges.
At $z = -1$: $\sum (-1)^n/n$ is the alternating harmonic series—converges (conditionally, to $-\ln 2$).
At $z = e^{i\theta}$, $\theta \ne 0$: By Dirichlet’s test, the series converges.

So $\sum z^n/n$ converges at every point on the unit circle except $z = 1$ —the location of the singularity of $-\log(1 - z)$.

Logarithm Series $\log(1 + z)$: Power series: $\sum_{n=1}^\infty (-1)^{n+1} \frac{z^n}{n}$. Root Test, $|a_n|^{1/n} = \left| \frac{1}{n} \right|^{1/n} = \frac{1}{n^{1/n}} = \frac{1}{\sqrt[n]{n}}$. Take the limit (L): $L = \limsup_{n \to \infty} \frac{1}{\sqrt[n]{n}} = \frac{1}{1} = 1$ (Recall $\lim_{n \to \infty} \sqrt[n]{n} = 1$). Radius R = 1/L = 1/1 = 1. Converges for |z| < 1.

Boundary Behavior (when $|z|=1$):

Case: z = 1. The series becomes $\sum_{n=1}^\infty (-1)^{n+1} \frac{(1)^n}{n} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$. This is the alternating harmonic series ($1 - \frac{1}{2} + \frac{1}{3} - \dots$). By the Alternating Series Test, this series converges (it converges conditionally to $\ln 2$).
Alternating Series Test (Leibniz Criterion). A series $\sum_{n=1}^\infty (-1)^{n+1} b_n$ with $b_n \ge 0$ converges if: (i) Monotonic decrease: $b_{n+1} \le b_n$ for all sufficiently large $n$ (monotonic decrease), and (ii) $\lim_{n\to\infty} b_n = 0$.
Here $b_n = 1/n$: both conditions hold ($1/(n+1) \le 1/n$ and $1/n \to 0$). The series converges conditionally to $\ln 2$.
Case: z = -1. The series becomes $\sum_{n=1}^\infty (-1)^{n+1} \frac{(-1)^n}{n} = \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n}$. Since 2n+1 is always odd, $(-1)^{2n+1} = -1 \leadsto \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n} = \sum_{n=1}^\infty \frac{-1}{n} = -1 \cdot \sum_{n=1}^\infty \frac{1}{n}$. This is the negative of the harmonic series, which famously diverges.
Case $z = e^{i\theta}$, $\theta \ne \pi$ (all other boundary points): By Dirichlet’s test, the series converges.

The logarithm series $\sum (-1)^{n+1}z^n/n$ converges for every $|z| \le 1$ except at the single point $z = -1$ —precisely the singularity of $\log(1 + z)$.

Theorem (Dirichlet’s Test). Let $\{ a_n \}$ be a sequence of complex numbers whose partial sums $A_N = \sum_{n=1}^N a_n$ are bounded: $|A_N| \le C$ for all $N$. Let $\{ b_n \}$ be a sequence of positive reals that is monotonically decreasing to $0$. Then $\sum_{n=1}^\infty a_n b_n$ converges.

Example: $\sum z^n/n$ on $|z| = 1$, $z \ne 1$: Set $a_n = z^n$ and $b_n = 1/n$. The partial sums of the geometric series $\sum z^n$ are:

$A_N = \sum _{n=1}^Nz^n = z\, \frac{1-z^N}{1-z}, \quad |A_N| \le \frac{2}{|1 - z|} < \infty \text{ (since } z \ne 1\text{)}$, the denominator is nonzero, so this a finite constant. Hypothesis (1) is satisfied.

We have |z| = 1: $|z|=1 \implies |z^N| = 1.$ So $|1-z^N| \leq 2$. Thus, $|A_N|=\left| z\frac{1-z^N}{1-z}\right| =\frac{|1-z^N|}{|1-z|}\leq \frac{2}{|1-z|}.$

$b_n=1/n$ is positive, decreasing, and tends to zero (Hypothesis is (2) satisfied), Dirichlet’s test guarantees convergence.

Definition. A series of functions $\sum f_n(z)$ converges uniformly on a set $K$ if $\sup_{z \in K} \left|\sum_{n=0}^\infty f_n(z) - \sum_{n=0}^N f_n(z)\right| \to 0$ as $N \to \infty$.

In other words, the “tail” of the series can be made uniformly small across all of $K$ simultaneously.

Theorem. Weierstrass $M$-Test. If $|f_n(z)| \le M_n$ for all $z \in K$ and $\sum M_n < \infty$, then $\sum f_n(z)$ converges absolutely and uniformly on $K$.

Theorem. A power series $\sum a_n(z-z_0)^n$ with radius of convergence $R > 0$ converges uniformly and absolutely on every compact subset of the open disk $D(z_0, R)$.

Complex Power Series: A Comprehensive Guide