## Idea

The radius of convergence of a power series tells how far out the series will converge.

## Definition

Let $a = (a_0, a_1, \ldots)$ be an infinite sequence of complex numbers, let $\zeta$ be a particular complex number, and consider the power series

(1)$\sum_n a_n (z - \zeta)^n .$

The radius of convergence is the supremum of the positive numbers $\epsilon$ such that (1) converges for ${|z - \zeta|} \lt \epsilon$. (This supremum is a nonnegative lower real in $[0,\infty]$.)

The phrasing above is ambiguous. If $\epsilon$ is less than the radius of convergence, does this mean that (1) converges for every $z$ with ${|z - \zeta|} \leq \epsilon$ or for some such $z$? It doesn't matter:

###### Theorem

If (1) converges for some $z$ with ${|z - \zeta|} = \epsilon$, then (1) converges for every $z$ with ${|z - \zeta|} \lt \epsilon$.

(We do not say that (1) converges for $z$ with ${|z = \zeta|} = \epsilon$, but this has no effect on the supremum.)

## Properties

The radius of convergence is clearly independent of $\zeta$. It can be calculated quite easily from the coefficients $a_i$:

The radius of converge of (1) is

$R = \liminf_n {|a_n|}^{-1/n} .$

For ${|z - \zeta|} \lt R$, (1) is (pretty much by definition) an analytic function of $z$. There is a partial converse:

###### Theorem

If a function $f$ is analytic at all $z$ with ${|z - \zeta|} \lt R$, then there is a power series (1) that converges to $f(z)$ for all $z$ with ${|z - \zeta|} \lt R$.

(Specifically, $a_n = f^{(n)}(\zeta)/n!$.)

###### Remark

Over the real numbers, Theorem  fails. I'm sure (says one of this pages authors) that there are interesting things to say about series in adic numbers, matrices, and things like that, but I don't know them.

Last revised on July 5, 2013 at 11:36:49. See the history of this page for a list of all contributions to it.