I have a power series:

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


$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.

Thanks in advance!

  • 4
    $\begingroup$ wouldn't this be the same as $s(x) = \sum\limits_{n=0}^\infty x^{n^2}$? $\endgroup$
    – Brent
    Commented Mar 15, 2015 at 5:53
  • $\begingroup$ @Brent You are right. I did not realize that. Thank you! $\endgroup$ Commented Mar 15, 2015 at 5:56
  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Commented Mar 15, 2015 at 5:56
  • 3
    $\begingroup$ You can use the root test. $\endgroup$
    – science
    Commented Mar 15, 2015 at 6:02

1 Answer 1


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 $1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .