## Calculus (3rd Edition)

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.