Power Series and Radius of Convergence

Power series are the local representation of holomorphic functions. Every holomorphic function is locally a convergent power series, and conversely, every convergent power series defines a holomorphic function.

Power Series

Definition5.1Power series

A power series centered at $z_0 \in \mathbb{C}$ is a formal series

$\sum_{n=0}^{\infty} a_n (z - z_0)^n$

where $\{a_n\}_{n=0}^\infty \subset \mathbb{C}$ are the coefficients. The series converges for some values of $z$ and diverges for others.

Theorem5.1Radius of convergence

For any power series $\sum a_n(z-z_0)^n$ , there exists a unique $R \in [0, \infty]$ called the radius of convergence such that the series converges absolutely for $|z - z_0| < R$ and diverges for $|z - z_0| > R$ . Furthermore,

$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}$

(Cauchy-Hadamard formula). If the limit exists, also $1/R = \lim_{n\to\infty} |a_{n+1}/a_n|$ .

ExampleComputing radii of convergence

$\sum z^n/n!$ : $|a_n|^{1/n} = (1/n!)^{1/n} \to 0$ , so $R = \infty$ (this is $e^z$ ).
$\sum n! z^n$ : $|a_n|^{1/n} = (n!)^{1/n} \to \infty$ , so $R = 0$ (converges only at $z = 0$ ).
$\sum z^n$ : $R = 1$ (geometric series).
$\sum z^n/n^2$ : $R = 1$ (converges on $|z| = 1$ as well).

Properties of Power Series

Theorem5.2Holomorphicity of power series

A power series $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n$ with radius of convergence $R > 0$ defines a holomorphic function on $D(z_0, R)$ . Moreover:

$f$ can be differentiated term by term: $f'(z) = \sum_{n=1}^\infty n a_n(z-z_0)^{n-1}$ .
The derived series has the same radius of convergence $R$ .
The coefficients are given by $a_n = f^{(n)}(z_0)/n!$ .

RemarkUniform convergence on compact subsets

The convergence of a power series is uniform on any compact subset of $D(z_0, R)$ — in particular, on every closed disk $\overline{D}(z_0, r)$ with $r < R$ . This uniform convergence justifies term-by-term differentiation and integration.

ExampleDifferentiating a power series

Starting from $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ for $|z| < 1$ , differentiation gives:

$\frac{1}{(1-z)^2} = \sum_{n=1}^\infty n z^{n-1} = \sum_{n=0}^\infty (n+1)z^n.$

A second differentiation: $\frac{2}{(1-z)^3} = \sum_{n=0}^\infty (n+1)(n+2)z^n / 1$ .

Abel's Theorem

Theorem5.3Abel's theorem

If $\sum_{n=0}^\infty a_n$ converges to $S$ , then $\lim_{r \to 1^-} \sum_{n=0}^\infty a_n r^n = S$ . More generally, if $f(z) = \sum a_n z^n$ has radius of convergence $1$ and $\sum a_n$ converges, then $\lim_{z \to 1} f(z) = \sum a_n$ as $z \to 1$ within any Stolz angle $\{z : |1-z| < C(1-|z|)\}$ .

ExampleAbel summation of $\ln 2$

The series $\sum_{n=1}^\infty (-1)^{n+1}/n = \ln 2$ (conditionally convergent). By Abel's theorem,

$\lim_{r \to 1^-} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} r^n = \lim_{r \to 1^-} \ln(1+r) = \ln 2.$

Multiplication of Power Series

RemarkCauchy product

If $f(z) = \sum a_n z^n$ and $g(z) = \sum b_n z^n$ both converge for $|z| < R$ , then $f(z)g(z) = \sum c_n z^n$ where $c_n = \sum_{k=0}^n a_k b_{n-k}$ (the Cauchy product), and this series also converges for $|z| < R$ .