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

We need to evaluate $$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}$$.

Observe that $$(k+1)(k+2)(k+3) = k^3 + 6k^2 + 11k + 6$$.

Therefore: $$k^3 + 6k^2 + 11k + 5 = (k+1)(k+2)(k+3) - 1$$.

$$\frac{k^3+6k^2+11k+5}{(k+3)!} = \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!}$$

$$= \frac{(k+1)(k+2)(k+3)}{(k+3)(k+2)(k+1) \cdot k!} - \frac{1}{(k+3)!}$$

$$= \frac{1}{k!} - \frac{1}{(k+3)!}$$

$$S_n = \sum_{k=1}^{n}\left(\frac{1}{k!} - \frac{1}{(k+3)!}\right)$$

Writing out the terms, most cancel in a telescoping fashion. The surviving terms are:

$$S_n = \left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}\right) - \left(\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!}\right)$$

As $$n \to \infty$$, the tail terms vanish:

$$S = 1 + \frac{1}{2} + \frac{1}{6} = \frac{6+3+1}{6} = \frac{10}{6} = \frac{5}{3}$$

The correct answer is Option 4: $$\frac{5}{3}$$.