Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3+6k^2+11k+5}{(k+3)!}\\ & =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3+6k^2+11k+6-1}{(k+3)!}\\ & =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{(k+1)(k+2)(k+3)-1}{(k+3)!}\\ & =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!}\\ & =\underset{k=1}{\lim }\sum _{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right)\end{array}$


$\begin{array}{rl}& =\underset{k=1}{\lim }\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}\dots +\frac{1}{n!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!}\dots .-\frac{1}{(n+3)!}\right)\\ & =\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}\end{array}$