Introduction

Welcome to this guide on radius and interval of convergence, this article will guide you through the following:

Radius and Interval of Convergence of Power Series

A Power Series takes in a form of

$\sum _{n=1}^{\infty }{a}_n(x-c{)}^n$

Where $c$ is the center of the series, since there is an extra $x$, for some $x$ the series might diverge, for example, consider a series where $a_n=1$ and $c=0$:

$\sum _{n=1}^{\infty }{x}^n$

If $x\ge 1$, this series will diverge to infinity, if $0<x<1$, this series will converge.

There are 3 possibilities of the convergence of a power series:

Interval of Convergence

Interval of Convergence is a fancy word for all the values that make the series converge, it is an important parameter of a power series.

To determine the radius of convergence, use the ratio test

If $\rho =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$

Where $a$ is the coefficient in front of the power term. Then radius of convergence of the series is $R=\frac{1}{\rho }$, if $\rho =\infty$, then $R=0$; if $\rho =0$, $R=\infty$

Example 1:

Find the radius of convergence of this power series:

$\sum _{n=0}^{\infty }(x+3{)}^n$

By ratio test:

$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{1}{1}\right|=1$

Thus the radius of convergence is $1$.

To find the interval of convergence for this series, notice that the center is at $x=-3$, and the radius of convergence is $1$, it means that for all values within $-2$ and $-4$, the series will converge.

To test for the endpoints, substitute the value to the series and test if it converge, for example, to determine the convergence of this series at $x=-2$, substitute $x=-2$ in

$\sum _{n=0}^{\infty }(-1{)}^n$

This is a geometric series with a common ratio of $r=-1$, which diverge, same logic applies to $x=-4$, therefore the interval of convergence for this series is $(-4,-2)$

Example 2:

Find the radius of convergence of this power series:

$\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{n+1}}{n^2}$

By ratio test:

$\underset{n\to \infty }{\lim }\left|\frac{n^2}{(n+1{)}^2}\right|=1$

Therefore the radius of convergence is $1$.

Practice

Determine the radius of convergence of the following power series.