Radius of Convergence - $\sum_{n=1}^{\infty}2^n x^{n^2}$

The radius of convergence is given by $$ \frac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}} $$ In this case $$ \frac1{\limsup\limits_{n\to\infty}\left(2^n\right)^{1/n^2}}=1 $$


Apparently, one can find the radius of convergence using the root test:

$$ \limsup_{n\to\infty} |a_n|^{1/n}=\lim_{n\to\infty}|2^{-n}|^{1/n^2}=1. $$

The radius of convergence is equal to $1$.