How to find the radius of convergence of $\sum_{n=0}^{\infty}[2^{n}z^{n!}]$ I tried with $$1/R = \lim_{n\to\infty}{\sup({\sqrt[n]{2^n}})} = \lim_{n\to\infty}{2} = 2$$
But that don't seem correct.
Thank you for your help!

We have 
$$\sqrt[n]{2^n\left|z^{n!}\right|}= 2|z|^{(n-1)!} \to \begin{cases}0&,|z|<1\\\\2&,|z|=1\\\\\infty&,|z|>1\end{cases}$$
and the series converges when $|z|<1$ and diverges otherwise.
