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) $5/3$