[tex] \frac{2+3^2}{1!+2!+3!+4!}+ \frac{3+4^2}{2!+3!+4!+5!} +........+ \frac{2013+2014^2}{2012!+2013!+2014!+2015!} [/tex]

Sesuaikan bentuk pada satu suku:
[tex]$\begin{align}&\frac{(n+1)+(n+2)^2}{n!+(n+1)!+(n+2)!+(n+3)!}\\&=\frac{n^2+5n+5}{n!(1+(n+1)+(n+1)(n+2)+(n+1)(n+2)(n+3))} \\ &=\frac{n^2+5n+5}{n!(10+15n+7n^2+n^3)}=\frac{n^2+5n+5}{n!(n+2)(n^2+5n+5)} \\ &=\frac{1}{n!(n+2)}=\frac{n+1}{(n+2)(n+1)\times n!} \\ &=\frac{n+1}{(n+2)!}\end{align}[/tex]

Cek penjumlahan dari 2012 suku pertama:
[tex]$\begin{align}\sum_{i=1}^{2012}\frac{i+1}{(i+2)!}&=\sum_{i=2}^{2013}\frac{i}{(i+1)!}=\sum_{i=2}^{2013}\frac{i+1-1}{(i+1)!}\\&=\sum_{i=2}^{2013}\left(\frac1{i!}-\frac{1}{(i+1)!}\right) \\ &=\frac1{2!}+\left(\frac1{3!}-\frac1{3!}+\dots+\frac1{2013!}-\frac1{2013!}\right)-\frac1{2014!} \\ &=\frac12-\frac1{2014!}\end{align}[/tex]