# Sum of Telescopic Series

Question :

$$\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}=?$$

I tried but I got $$(-1)!$$ in the end. This is what I tried:

$$\require{cancel} \frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\ \frac{\cancel{(r+1)}\cancel{(r-1)}}{\cancel{(r+1)}r\cancel{(r-1)}(r-2)!}-\frac{\cancel{r}}{(r+1)\cancel{r}(r-1)!}\\ T_r=\frac{1}{r(r-2)!}-\frac{1}{(r+1)(r-1)!}\\ \sum_{r=1}^{n} T_r = \frac{1}{1(1-2)!}-\frac{1}{(n+1)(n-1)!}\\ = \frac{1}{(-1)!}-\frac{1}{(n+1)(n-1)!}\\$$

• Please show what you have done. Sep 1, 2020 at 7:17
• how did you get (-1)! Sep 1, 2020 at 7:30
• Find $a,b,c$ such that $r^2-r-1=ar(r+1)+b(r+1)+c$, identically. Sep 1, 2020 at 7:36
• "I am taught in school that after splitting the fractions. Put the minima in least denominator and maxima in greater denominator." That is not a good way to teach. Sep 1, 2020 at 10:38
• Since $\frac1{(-1)!}=\frac0{0!}=0$, your answer is correct, in a sense.
– robjohn
Sep 1, 2020 at 18:37

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