Sum of Telescopic Series

$$\begin{eqnarray*}\frac{r^2-r-1}{(r+1)!}&=&\frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{(r-1)(r+1)}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{r-1}{r!}-\frac{r}{(r+1)!}\end{eqnarray*}$$

Therefore:

$$\begin{eqnarray*}\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}&=&\sum_{r=1}^n\left[\frac{r-1}{r!}-\frac{r}{(r+1)!}\right]\\&=&\frac{0}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{2}{3!}+\frac{2}{3!}-\frac{3}{4!}+...+\frac{n-1}{n!}-\frac{n}{(n+1)!}\\&=&\color{blue}{-\frac{n}{(n+1)!}}\end{eqnarray*}$$

For any $n\geq 2$ we have $$\begin{eqnarray*}\sum_{r=1}^{n}\frac{r^2-r-1}{(r+1)!}&=&\sum_{r=1}^{n}\frac{(r+1)r}{(r+1)!}-2\sum_{r=1}^{n}\frac{(r+1)}{(r+1)!}+\sum_{r=1}^{n}\frac{1}{(r+1)!}\\&=&\sum_{r=1}^{n}\frac{1}{(r-1)!}-2\sum_{r=1}^{n}\frac{1}{r!}+\sum_{r=1}^{n}\frac{1}{(r+1)!}\\ &=&\sum_{r=0}^{n-1}\frac{1}{r!}-2\sum_{r=1}^{n}\frac{1}{r!}+\sum_{r=2}^{n+1}\frac{1}{r!}\\&=&\frac{1}{0!}-\frac{1}{1!}-\frac{1}{n!}+\frac{1}{(n+1)!}=\color{red}{-\frac{n}{(n+1)!}}.\end{eqnarray*}$$