The value of _n (_k=1^n (k^3+6 k^2+11 k+5)/((k+3)!)) is :

Question

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

Accepted Answer

$$\begin{aligned}
& \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \
& =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \
& =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \
& =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \
& =\lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right)
\end{aligned}$$

$$\begin{aligned}
& =\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{\mathrm{n!}}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots-\frac{1}{(\mathrm{n}+3)!}\right) \
& \quad=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}
\end{aligned}$$

Solutions · 1