Find $\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}$

Question

Compute $$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$

My Approach

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

$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!} = \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{\left(k+3\right)!}\right)$$

But now I can't find this limit.

Answer 1

Good start!

$$ \begin{align} \lim_{n \to \infty}\sum_{k=1}^n\left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=1}^n\frac{1}{(k+3)!} \right) \\ &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=4}^{n+3}\frac{1}{k!} \right) \\ &= \lim_{n\to\infty}\left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - \frac{1}{(n+1)!} - \frac{1}{(n+2)!} - \frac{1}{(n+3)!} \right) \\ &= \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \\ &= \frac{5}{3} \end{align} $$

Answer 2

Use $$\sum_{k=1}^{\infty}\frac{1}{k!}=e-1$$ and $$\sum_{k=1}^{\infty}\frac{1}{(k+3)!}=e-2-\frac{1}{2}-\frac{1}{6}$$