Radius of Convergence

We can apply the ratio test to get some information about where a power series converges absolutely. Trying the ratio test on

$$\sum_{n=0}^\infty \vert a_n x^n\vert$$

leads us to look at

$$\lim_{n\to \infty} \frac{\vert a_{n+1}\vert}{\vert a_n\vert}\vert x\vert$$

and ask for which x the limit is strictly less than 1. To this end we take $${\lim_{n\to \infty}\vert\frac{a_{n+1}}{a_n}\vert = L}$$ and ask that L|x|<1, or equivalently, $-\frac1L< x < \frac1L$.

Definition 4.2   The number $\frac1L$ is called the radius of convergence of the power series.

Inside its radius of convergence a power series is absolutely convergent and has a little wiggle room before you get outside the radius of convergence.

It is possible for L=0 in which case we will get convergence for all x. If the limit gives $+\infty$ then any x we use other than 0 will give a divergent series.

Examples:

1.

$\sum_{n=0}^\infty \frac{x^n}{n!}$ converges for all x. We actually knew this already from Taylor's theorem, since this is the series for e^x, but it is instructive to see how the ratio test gives it:

$$\lim_{n\to \infty}\frac{\left(\frac{1}{(n+1)!}\right)}{\left(\frac{1}{n!}\right)} = \lim_{n\to \infty}\frac1{n+1} = 0$$

The ratio test then gives us convergence for all x.

2.

Turning the previous example upside down gives $${\sum_{n=0}^\infty n!x^n}$$. Here the ratio of test gives an infinite limit, so only x=0 works.

3.

$${\sum_{n=0}^\infty \frac{n x^n}{2^n}}$$ gives a radius of convergence 1/L where

$$L=\lim_{n\to \infty} \frac{(n+1)^2 2^n}{n^2 2^{n+1}}= \frac12$$

Thus the radius of convergence is 2.

Exercises: Find the radius of convergence for the following power series:

1.

$${\sum \frac{2^n x^n}{n}}$$

2.

$${\sum \frac{nx^n}{n+1}}$$

3.

$${\sum \frac{4^nx^n}{n!}}$$

4.

$${\sum \frac{n^nx^n}{n!}}$$