Mathematics - Sequence & Series Question with Solution | TestHub

The general term of the series is $T_n=\frac{n^2+2n}{(n+1)!}$. We can rewrite the numerator as$n(n+2)$.$T_n=\frac{n(n+2)}{(n+1)!}=\frac{(n+1{)}^2-1}{(n+1)!}=\frac{(n+1{)}^2}{(n+1)!}-\frac{1}{(n+1)!}=\frac{n+1}{n!}-\frac{1}{(n+1)!}$. Further,$\frac{n+1}{n!}=\frac{n}{n!}+\frac{1}{n!}=\frac{1}{(n-1)!}+\frac{1}{n!}$. So,$T_n=\frac{1}{(n-1)!}+\frac{1}{n!}-\frac{1}{(n+1)!}$. The sum$S={\sum }_{n=1}^{\infty }{T}_n={\sum }_{n=1}^{\infty }\left(\frac{1}{(n-1)!}+\frac{1}{n!}-\frac{1}{(n+1)!}\right)$. This can be split into three separate sums:$S_1={\sum }_{n=1}^{\infty }\frac{1}{(n-1)!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots =e$.$S_2={\sum }_{n=1}^{\infty }\frac{1}{n!}=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots =e-\frac{1}{0!}=e-1$.$S_3={\sum }_{n=1}^{\infty }\frac{1}{(n+1)!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots =e-\left(\frac{1}{0!}+\frac{1}{1!}\right)=e-2$. Therefore,$S=S_1+S_2-S_3=e+(e-1)-(e-2)=e+1$.