# Radius of convergence involving $z^{n^2}$

Consider the following complex power series: $\displaystyle\sum_{n=1}^\infty a_nz^{n^2}$ where , $\displaystyle a_n = \frac{1}{n!}$.

My approach is that by ratio test

$$\lim\limits_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \lim\limits_{n\to\infty} \frac{|z|^{2n+1}}{n+1} < 1$$

then by rearranging I have $$|z| < \lim\limits_{n\to\infty} (n+1)^{\frac{1}{2n+1}}$$

but I'm not sure how to evaluate this limit.

• Your first limit only converges when $|z|\leq 1$ – Kitegi Mar 11 '16 at 8:56
• Small comment: Limits don't converge or diverge; they either exist or don't. – zhw. Mar 11 '16 at 17:26

It is clear that the series converges absolutely when $|z|=1$. So by comparison it converges when $|z|\lt 1$.
To show divergence when $|z|\gt 1$, note that $$\frac{|z|^{n^2}}{n!}\ge \frac{|z^{n^2}|}{n^n}.$$ Now use the root test. The norm of the $n$-th root of $\frac{|z^{n^2}|}{n^n}$ is $\frac{|z|^n}{n}$, and this $\to\infty$ as $n\to\infty$ if $|z|\gt 1$.
Note that $$\sqrt{(n+1)^{1/(n+1)}}<(n+1)^{1/(2n+1)}<(n+1)^{1/(n+1)}$$ and $\sqrt[n]n\to 1$.
Alternatively, suppose $(n+1)^{1/(2n+1)}>1+\epsilon$. Then $$n+1>(1+\epsilon)^{2n+1}>1+(2n+1)\epsilon+(2n^2+n)\epsilon^2$$ is only possible for small $n$.