lim n→∞ ∑ k^3+6k^2+11k+5/(k+3)! k ∈ to (k=1, n) is equal to

Question

\(\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}\) is equal to

(1) \(\frac{5}{3}\)

(2) \(\frac{8}{3}\)

(3) 3

(4) \(\frac{7}{3}\)

Answer 1

Correct option is (1) \(\frac{5}{3}\)    

\(\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 x^2+11 k+6-1}{(k+3)!} \)

\(=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \)

\(=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{k!}-\frac{1}{(k+3)!} \)

\(=\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots \infty\right)-\left(\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\cdots \infty\right) \)

\(=(e-1)-\left(e-1-\frac{1}{1!}-\frac{1}{2!}-\frac{1}{3!}\right) \)

\(=1+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}\)