Radius of convergence of $\sum_{n = 0}^\infty a_nx^n $ when $a_n = 1$ if n is a perfect square and $a_n = 0$ otherwise

Question

I'm studing the radius of convergence of $$\sum_{n = 0}^\infty a_nx^n $$

when

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

For $\phantom{2}a_n = 0\phantom{2}$ $\forall x \in \mathbb{R}$ the sum converges.

If $\phantom{2}a_n = 1\phantom{2} $ and $\phantom{2}|x| \le 1$ the sum converges. If $\phantom{2}a_n = 1\phantom{2} $ and $\phantom{2}|x| \gt 1$ the sum diverges.

But I'm confused when I try the ratio test.

Answer 1

Denoting $ R $ the radius of convergence of the power series $ \sum\limits_{n\geq 0}{a_{n}x^{n}} $, and $ R' $ the radius of convergence of $ \sum\limits_{n\geq 0}{x^{n}} $, we have the following : \begin{aligned}&\small \bullet \ \ \ a_{n}\leq 1\textrm{, thus }R\geq R'=1\cdot\\ &\small \bullet \ \ \ \textrm{Setting }x=1\textrm{, we get that }\sum_{n\geq 0}{a_{n}}\textrm{ diverges since }a_{n}\textrm{ does not go to }0\textrm{ as }n\textrm{ goes to }+\infty\textrm{, thus }R\leq 1 \cdot\end{aligned}

Hence, $ R=1 \cdot $

Answer 2

Why not just rewrite everything? Let $b_m=1$ for all $m$: then $$ \sum_{n=0}^{\infty}a_n x^n = \sum_{m=0}^{\infty} x^{m^2} $$Now do the Ratio Test.

Answer 3

We clearly have $\limsup_{n\to\infty}\sqrt[n]{|a_n|}=1$ as the term takes on the value $1$ infinitely often (and never a greater one).