# Radius of convergence of a power series whose coeffecients are “discontinuous”

I have a power series:

$s(x)=\sum_0^\infty a_n x^n$

with

$a_n= \begin{cases} 1, & \text{if$n$is a square number} \\ 0, & \text{otherwise} \end{cases}$

What is the radius of convergence of this series?

I tried to use ratio test, but as$\lim\limits_{n \to \infty} \frac{a_n x^n}{a_{n+1} x^{n+1}}$does not exist, I don't know how to apply the ratio test on the problem.

• wouldn't this be the same as $s(x) = \sum\limits_{n=0}^\infty x^{n^2}$? – Brent Mar 15 '15 at 5:53
• @Brent You are right. I did not realize that. Thank you! – fanmingyu212 Mar 15 '15 at 5:56
• One approach is to omit all the zero terms from the series. The ratio test may be considered a form of series comparison to a geometric series. – hardmath Mar 15 '15 at 5:56
• You can use the root test. – science Mar 15 '15 at 6:02

Although $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ and $\lim_{n\to\infty}\sqrt[n]{a_n}$ don't exist, you can still find $\limsup_{n\to\infty} \sqrt[n]{a_n}$. Since $$\sup_{m\ge n} \sqrt[m]{a_m}=1$$ for all $n\in\mathbb{N}$, $\limsup_{n\to\infty}\sqrt[n]{a_n}=1$ and so the radius of convergence is $1$.