Answer
Divergent
Work Step by Step
Given $$
\sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !}
$$
By using the Ratio test where $a_n=\dfrac{2^{n^{2}}}{n !}$, $a_{n+1}=\dfrac{2^{(n+1)^{2}}}{(n+1) !}$
\begin{align*}
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|&=\lim _{n \rightarrow \infty}\left|\frac{2^{(n+1)^{2}}}{(n+1) !} \cdot \frac{n !}{2^{n^{2}}}\right|\\
&=\lim _{n \rightarrow \infty} \frac{2^{n^{2}+2 n+1}}{2^{n^{2}}(n+1)}\\
&=\lim _{n \rightarrow \infty} \frac{2^{2 n+1}}{n+1}\ \text{(L'Hopital Rule)}\\
&=\lim _{n \rightarrow \infty} \frac{(\ln 2)2^{2 n+2}}{ 1}\\
&=\infty
\end{align*}
Then $ \displaystyle \sum_{n=1}^{\infty} \frac{2^{n^{2}}}{n !}$ is divergent .
