The value of $\lim _{n \rightarrow \infty}\left(\sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}\right)$ is:

Question

Aicharya · Accepted Answer

$\begin{aligned} & \lim _{n ightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \ & =\lim _{n ightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \ & =\lim _{n ightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \ & =\lim _{n ightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \ & =\lim _{k=1} \sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}ight)\end{aligned}$ $\begin{aligned} & =\lim _{k=1}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{n!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots .-\frac{1}{(n+3)!}ight) \ & \quad=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}\end{aligned}$ ~

The value of \(\lim _{n \rightarrow \infty}\left(\sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}\right)\) is:

Solution