convergence radius

Radius of convergence

Radius of convergence


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


Let a=(a 0,a 1,)a = (a_0, a_1, \ldots) be an infinite sequence of complex numbers, let ζ\zeta be a particular complex number, and consider the power series

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

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

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


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

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


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

Theorem (Cauchy–Hadamard)

The radius of converge of (1) is

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

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


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

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


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.