_n(_k=1^n(k^3+6k^2+11k+5)/((k+3)!)) is:

Question

$\displaystyle\lim_{n	o\infty}\left(\sum_{k=1}^{n}\frac{k^3+6k^2+11k+5}{(k+3)!}ight)$ is:

Accepted Answer

Note $k^3+6k^2+11k+5=(k+1)(k+2)(k+3)-1$, so
$$\sum_{k=1}^{n}\frac{k^3+6k^2+11k+5}{(k+3)!}=\sum_{k=1}^{n}\left(\frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!}\right)=\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right)$$

This telescopes: as $n\to\infty$,
$$\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right)\to\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+\frac12+\frac16=\frac53$$

Solutions · 1