# How do you find the radius of convergence of a power series?

Sep 13, 2014

I used to live in Hicksville too, when I was a kid!

To find the radius R of convergence of a power series
${\sum}_{n = 0}^{\infty} {c}_{n} {\left(x - a\right)}^{n} ,$ centered at $x = a$, use the Ratio Test,
and check that ${\lim}_{n \to \infty} | \frac{{c}_{n + 1} {\left(x - a\right)}^{n + 1}}{{c}_{n} {\left(x - a\right)}^{n}} | < 1 ,$ the same as
${\lim}_{n \to \infty} | \frac{{c}_{n + 1}}{{c}_{n}} | \cdot | x - a | < 1 ,$ or
$| x - a | < {\lim}_{n \to \infty} | \frac{{c}_{n}}{{c}_{n + 1}} |$

We wanted to find R such that our power series converged for
$a - R < x < a + R$, which is $| x - a | < R$, so we use the value
$R = {\lim}_{n \to \infty} | \frac{{c}_{n}}{{c}_{n + 1}} |$ for the radius of convergence.

Note: The word "radius" comes from the ability to use complex numbers for our variable x (and also the coefficients), and saying $| x - a | < R$ describes the inside of a circle of real radius R in the complex plane, centered at the complex number a.

| dansmath strikes again |