The value of \(\lim _{n\to \infty }\left(\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!}\right)\) is [JEE Main 2025…

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

\(\lim _{n\to \infty }\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!}\\ =\lim _{n\to \infty }\sum _{k=1}^{n}\frac{1}{k!}-\frac{1}{(k+3)!}\\ =\left(1+\frac{1}{2!}+\frac{1}{3!}+........\right)\\ -\left(\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+.........\right)\\ =1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3}\)