$\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\dfrac{k^3+6k^2+11k+5}{(k+3)!}$ equals
Step-by-Step Solution
Key Concept: Factor the numerator as $(k+1)(k+2)(k+3)-1$ (since $(k+1)(k+2)(k+3)=k^3+6k^2+11k+6$) to split each term into $1/k!-1/(k+3)!$, which telescopes.
$(k+1)(k+2)(k+3)=k^3+6k^2+11k+6$, so numerator $=k^3+6k^2+11k+5=(k+1)(k+2)(k+3)-1$.
$$\frac{k^3+6k^2+11k+5}{(k+3)!}=\frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!}=\frac{1}{k!}-\frac{1}{(k+3)!}.$$
Summing from $k=1$ to $n$ telescopes to:
$$\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}-\frac{1}{(n+1)!}-\frac{1}{(n+2)!}-\frac{1}{(n+3)!}\xrightarrow{n\to\infty}1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3}.$$
Correct Answer: 4