The value of $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{k^{3}+6k^{2}+11k+5}{(k+3)!}$ is:
Step-by-Step Solution
Key Concept: $k^{3}+6k^{2}+11k+6=(k+1)(k+2)(k+3)$, so the numerator $=(k+1)(k+2)(k+3)-1.$ This breaks the term into $\dfrac{1}{k!}-\dfrac{1}{(k+3)!}$, which telescopes (lag of $3$).
Identify $(k+1)(k+2)(k+3)=k^{3}+6k^{2}+11k+6.$ So
$$\dfrac{k^{3}+6k^{2}+11k+5}{(k+3)!}=\dfrac{(k+1)(k+2)(k+3)-1}{(k+3)!}=\dfrac{1}{k!}-\dfrac{1}{(k+3)!}.$$
Sum from $k=1$ to $n$:
$$\sum_{k=1}^{n}\frac{1}{k!}-\sum_{k=4}^{n+3}\frac{1}{k!}=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}-\frac{1}{(n+1)!}-\frac{1}{(n+2)!}-\frac{1}{(n+3)!}.$$
As $n\to\infty$:
$$1+\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{6+3+1}{6}=\dfrac{10}{6}=\dfrac{5}{3}.$$
Correct Answer: 4