Originally Posted by

**arbolis** Find the region of convergence of the power series $\displaystyle \sum _{n=0}^{+\infty} a^{n^2} z^n$ where $\displaystyle a \in \mathbb{C}$.

According to my class notes, if I have a power series of the form $\displaystyle \sum _{n=0}^{+\infty} a_n z^n$, the radius of convergence is equal to $\displaystyle R=\lim _{n \to \infty} \frac{|a_n|}{|a_{n+1}|}$. So I applied the formula and I got that $\displaystyle R= \lim _{n \to \infty} \frac{1}{|a^{2n+1}|}$.

I wrote $\displaystyle a=re^{i \theta} \Rightarrow a^{2n+1}=r^{(2n+1)} e^{i(2n+1)\theta}$. So I think that $\displaystyle \lim _{n \to \infty} |a^{2n+1}|$ depends on $\displaystyle |a|=r$. If $\displaystyle |a|<1$ the series converges while if $\displaystyle |a|>1$ the series diverges, $\displaystyle \forall z \in \mathbb{C}$. So the radius of convergence would be 1, although it seems totally wrong to me since I think I should have had a condition on z, not only on a. For instance if $\displaystyle a=1$ I fall in the case when I have $\displaystyle \sum _{n=0}^{+\infty} z^n$ and the radius of convergence is indeed 1 but because of a condition on z.

So I'd like to know what I'm doing wrong.