Computing the sum $\sum_\limits{n=2}^\infty \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{n+1}$ I have come across an infinite series, but I have no clue on how to compute its sum.

$$\sum_{n=2}^\infty \left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{n+1}$$

It should have something to do with the Taylor expansion of $e^x$, but I could not figure out how to do this.

\begin{align*}
\sum_{n=2}^\infty \left (\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{(n+1)}
&=\sum_{n=2}^\infty \left (\frac{n}{(n+1)!}-\frac{1}{(n+1)!}\right)\\
&=\sum_{n=2}^\infty \left (\frac{n+1-1}{(n+1)!}-\frac{1}{(n+1)!}\right)\\
&=\sum_{n=2}^\infty \frac{1}{n!}-2\sum_{n=2}^\infty \frac{1}{(n+1)!}\\
&=-2+\sum_{n=0}^\infty \frac{1}{n!}-2\sum_{n=3}^\infty \frac{1}{n!}\\
&=-2+e-2(e- \frac 5 2 )=-e+3
\end{align*}
