Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{2^{n^2}}{n!}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{2^{(n+1)^2}}{(n+1)!}}{\dfrac{2^{n^2}}{n!}}|$
Thus, we have $l= \lim\limits_{n \to \infty}|\dfrac{2^{n^2+2n+1}}{(n+1)n!}|\lim\limits_{n \to \infty}|((\dfrac{n!}{2^{n^2}}))|=\lim\limits_{n \to \infty}|\dfrac{(2)(4^n)}{n+1}|$
So, $l=\dfrac{(2)(4^n)}{(1+\dfrac{1}{n})} =\infty \gt 1$
Hence, the series Diverges by the ratio test.