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

$$\begin{gathered} \begin{array}{l}\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{k}=1}^{\mathrm{n}} \frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+11\mathrm{k}+5}{(\mathrm{k}+3)!}\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{k}=1}^{\mathrm{n}} \frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+11\mathrm{k}+6-1}{(\mathrm{k}+3)!}\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{k}=1}^{\mathrm{n}} \frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)-1}{(\mathrm{k}+3)!}\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \sum _{\mathrm{k}=1}^{\mathrm{n}} \frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{(\mathrm{k}+3)!}-\frac{1}{(\mathrm{k}+3)!}\\ =\underset{\mathrm{k}=1}{lim} \sum _{\mathrm{k}=1}^{\mathrm{n}} \left(\frac{1}{\mathrm{k}!}-\frac{1}{(\mathrm{k}+3)!}\right)\end{array} \\ \begin{array}{l}=\underset{\mathrm{k}-1}{lim} \left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}\dots +\frac{1}{\mathrm{n}!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!}\dots -\frac{1}{(\mathrm{n}+3)!}\right)\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}\end{array} \end{gathered}$$