Prove Sn of $Ur ={1/n!} - {2/(n+1)!} +{ 1/(n+2)!}$

Question

$Ur ={1/r!} - {2/(r+1)!} +{ 1/(r+2)!}$ Prove that $\sum_{r=1}^n Ur= {1/2} - { (n+1)/ (n+2)! }$

Answer 1

Much more generally, if $u_r =v_r-2v_{r+1}+v_{r+2} $ then

$\begin{array}\\ \sum_{r=1}^n u_r &=\sum_{r=1}^n (v_r-2v_{r+1}+v_{r+2})\\ &=\sum_{r=1}^n v_r-2\sum_{r=1}^nv_{r+1}+\sum_{r=1}^nv_{r+2}\\ &=\sum_{r=1}^n v_r-2\sum_{r=2}^{n+1}v_{r}+\sum_{r=3}^{n+2}v_{r}\\ &=v_1+v_2+\sum_{r=3}^n v_r-2(v_2+v_{n+1}+\sum_{r=3}^{n}v_{r})+\sum_{r=3}^{n}v_{r}+v_{n+1}+v_{n+2}\\ &=v_1-v_2-v_{n+1}+v_{n+2}\\ \end{array} $

If $v_r = \frac1{r!}$, this gives $\sum_{r=1}^n u_r =\frac1{1}-\frac1{2} -\frac1{(n+1)!} +\frac1{(n+2)!} =\frac1{2} +\frac{1-(n+2)}{(n+2)!} =\frac1{2} -\frac{n+1}{(n+2)!} $.

This generalizes to $u_r =\sum_{k=0}^m (-1)^k \binom{m}{k}v_{r+k} $. This is the case $m=2$.