Answer
Diverges
Work Step by Step
Given
$$ \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !}$$
We use the comparison test with:
$$\frac{2^{n^{2}}}{n !}\geq \frac{\left(2^{n}\right)^n}{n ^n} $$
Then, by using the $ n$th Root Test, we get:
\begin{align*}
\rho&= \lim_{n\to \infty} \sqrt[n]{ a_n}\\
&= \lim_{n\to \infty} \sqrt[n]{\frac{\left(2^{n}\right)^n}{n ^n} }\\
&= \lim_{n\to \infty} \frac{2^n}{n}\\
&=\lim_{n\to \infty} \frac{\ln 22^n}{1}\\
&=\infty
\end{align*}
Thus, by the comparison test, the series $ \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !} $ also diverges.
