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*}

Hint. You may write, for $n\ge1$, $$\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)\frac{1}{n+1}=\frac{1}{(n-1)!}\left(1-\frac{1}{n}\right)\frac{1}{n+1}=\frac{1}{n!}-\frac{2}{(n+1)!}$$ then apply $$\sum_{n=0}^\infty\frac{x^n}{n!}=e^x,\quad x \in \mathbb{R}.$$

Hint:

$$\dfrac1{n+1}\left(\dfrac1{(n-1)!}-\dfrac1{n!}\right)=\dfrac n{(n+1)!}-\dfrac1{(n+1)!}$$

Now $\dfrac n{(n+1)!}=\dfrac{n+1-1}{(n+1)!}=\dfrac1{n!}-\dfrac1{(n+1)!}$

Now $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$