The Radius of Convergence of a Power Series

# The Radius of Convergence of a Power Series

Recall from the Power Series page that we saw that a power series will converge at it's center of convergence $c$, and that it is possible that a power series can converge for all $x \in \mathbb{R}$ or on some interval centered at the center of convergence. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below.

 Definition: The Radius of Convergence, $R$ is a non-negative number or $\infty$ such that the interval of convergence for the power series $\sum_{n=0}^{\infty} a_n(x - c)^n$ is $[c - R, c + R]$, $(c - R, c + R)$, $[c - R, c + R)$, $(c - R, c + R]$.

For example, in the case that a power series $\sum_{n=0}^{\infty} a_n(x - c)^n$ is convergent only at $x = c$, then the radius of convergence for this power series is $R = 0$ since the interval of convergence is $[c - 0, c + 0] = [c, c]$. Similarly, if the power series is convergent for all $x \in \mathbb{R}$ then the radius of convergence of the power series is $R = \infty$ since the interval of convergence is $(-\infty, \infty)$.

# Determining the Radius of Convergence of a Power Series

We will now look at a technique for determining the radius of convergence of a power series using The Ratio Test for Positive Series

 Theorem 1: If $\lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = L$ where $L$ is a positive real number or $L = 0$ or $L = \infty$, then the power series $\sum_{n=0}^{\infty} a_n(x - c)^n$ has a radius of convergence $R = \frac{1}{L}$ where if $L = 0$ then $R = \infty$ and if $L = \infty$ then $R = 0$.

Let's now look at some examples of finding the radius of convergence of a power series.

## Example 1

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{1}{1 + n^3} (x + 6)^n$.

We note that the center of convergence is $c = -6$. Now we want to find the radius of convergence using the ratio test. Let $a_n = \frac{1}{1 + n^3}$. Thus:

(1)
\begin{align} \quad L = \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{1}{1 + (n+1)^3}}{\frac{1}{1 + n^3}} = \lim_{n \to \infty} \frac{1 + n^3}{1 + (n+1)^3} = \lim_{n \to \infty} \frac{1 + n^3}{2 + 3n + 3n^2 + n^3} = \lim_{n \to \infty} \frac{\frac{1}{n^3} + 1}{\frac{2}{n^3} + \frac{3}{n^2} + \frac{3}{n} + 1} = 1 \end{align}

So $L = 1$, and so the radius of convergence is $R = \frac{1}{L} = 1$.

## Example 2

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n}{n!}(x - 3)^n$.

Once again we note that the center of convergence is $c = 3$. We now want to find the radius of convergence using the ratio test once again.

(2)
\begin{align} \quad L = \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{ \frac{(n+1)}{(n+1)!}}{ \frac{n}{n!}} = \lim_{n \to \infty} \frac{(n + 1)n!}{n(n + 1)!} = \lim_{n \to \infty} \frac{(n+1)!}{n(n+1)!} = \lim_{n \to \infty} \frac{1}{n} = 0 \end{align}

Since $L = 0$ we get that our radius of convergence $R = \infty$.