Answer
Converges
Work Step by Step
Given
$$\sum_{n=1}^{\infty}\frac{n !}{n^{3}}$$
Compare with $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}} $, which is a convergent series ( $p-$series with $p=2$) and for $n\geq 1$
\begin{align*}
\frac{n !}{n^{3}}&=\frac{n \times(n-1) !}{n^{3}}\\
\frac{n !}{n^{3}}&=\frac{(n-1) !}{n^{2}}\\
& \geq \frac{1}{n^2}
\end{align*}
Then $\displaystyle\sum_{n=1}^{\infty} \frac{n !}{n^{3}}$ also converges.
