The value of $\lim_{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}\right)$ is:
(1) $4/3$
(2) $2$
(3) $7/3$
(4) $5/3$
The value of $\lim_{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}\right)$ is:\\
(1) $4/3$\\
(2) $2$\\
(3) $7/3$\\
(4) $5/3$