Method of Differences/Partial fractions with factorials

Question

By first expressing $\frac{1}{r!(r+2)}$ in the form $\frac{A}{(r+2)!} + \frac{B}{(r+1)!}$, find $\sum\limits_{r}^n \frac{1}{r!(r+2)}$. Struggling to do the partial fractions to begin.

Answer 1

One way is: $$\frac{1}{r!(r+2)}=\frac{r+1}{(r+2)!}=\frac{r+2-1}{(r+2)!}=\frac1{(r+1)!}-\frac1{(r+2)!}.$$ And it results in telescoping difference.