Answer
Converges
Work Step by Step
Given
$$\sum_{n=1}^{\infty} \frac{e^{n}}{n !} $$
By using the Ratio test
\begin{align*}
\rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim _{n \to \infty}\frac{e^{n+1}}{(n+1) !} \frac{n !} {e^{n}}\\
&=\lim _{n \to \infty}\frac{ee^{n}}{(n+1)n !} \frac{n !} {e^{n}}\\
&=\lim _{n \to \infty}\frac{e}{(n+1)}\\
&=0\lt 1
\end{align*}
Thus the given series converges.
