How to compute $\sum\limits_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$ I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!

The Heaviside Method gives the partial fraction decomposition
$$
\begin{align}
&\frac{1}{(n-2)(n-1)n(n+1)(n+2)}\\[6pt]
&=\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\tag{1}
\end{align}
$$
Notice that
$$
\frac{1}{24}-\frac{1}{6}+\frac{1}{4}-\frac{1}{6}+\frac{1}{24}=0\tag{2}
$$
Therefore,
$$
\begin{align}
&\sum_{n=3}^m\left(\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\right)\\
&=\frac{1}{24}\sum_{n=1}^{m-2}\frac1n-\frac{1}{6}\sum_{n=2}^{m-1}\frac1n+\frac{1}{4}\sum_{n=3}^{m}\frac1n-\frac{1}{6}\sum_{n=4}^{m+1}\frac1n+\frac{1}{24}\sum_{n=5}^{m+2}\frac1n\tag{$\ast$}\\
&=\frac{1}{24}\left(\frac11+\frac12+\frac13+\frac14\right)\\
&-\frac{1}{6}\left(\frac12+\frac13+\frac14+\frac{1}{m-1}\right)\\
&+\frac{1}{4}\left(\frac13+\frac14+\frac{1}{m-1}+\frac{1}{m}\right)\\
&-\frac{1}{6}\left(\frac14+\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}\right)\\
&+\frac{1}{24}\left(\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}+\frac{1}{m+2}\right)\\
&=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}\tag{3}
\end{align}
$$
where $\displaystyle\sum_{n=5}^{m-2}\frac1n$ is cancelled out of each summation in $(\ast)$ due to $(2)$.
As $m\to\infty$ the $(3)$ tends to $\dfrac{1}{96}$.
