Answer
Converges
Work Step by Step
Given
$$\sum_{n=1}^{\infty} \frac{1}{(n+1) !} $$
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*}
(n+1) ! &\geq n^{2}\\
\frac{1}{(n+1) !} &\leq \frac{1}{n^{2}}
\end{align*}
Then $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1) !} $ also converges.
