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!
 A: Hint: There exists some $c_k$ independent of $n$ such that
$$
\frac1{(n-2)(n-1)n(n+1)(n+2)}=\sum_{k=-2}^2\frac{c_k}{n+k}.
$$
To find $c_k$, multiply both sides by $n+k$ and evaluate the result at $n=-k$. For example,
$$
c_{-2}=\left.\frac1{(n-1)n(n+1)(n+2)}\right|_{n=2}=\frac1{24}.
$$
Sanity check: $\sum\limits_{k=-2}^2c_k=0$ and every $c_k$ should be a multiple of $\frac1{24}$ with the sign of $(-1)^k$ and depending only on $|k|$.
Once this is done, note that the value $S$ of the series you are looking for is
$$
S=c_{-2}\cdot\left(\frac11+\frac12\right)+c_{-1}\cdot\left(\frac12\right)+c_{1}\cdot\left(-\frac13\right)+c_{2}\cdot\left(-\frac13-\frac14\right),
$$
which yields the value you got thanks to W|A, namely, $S=\dfrac1{96}$.
A: 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 $\dfrac{1}{96}$.
A: If you know partial fractions, this should be 
$$\frac{(n-3)!}{(n+2)!}=\frac{1}{4n}+\frac{1}{24(n-2)}+\frac{1}{24(n+2)}-\frac{1}{6(n-1)}-\frac{1}{6(n+1)}$$
And you might have to simplify the finite sum to get an expression like
$$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
A: (n−3)!/(n+2)! = 1/[(n+2)(n+1)n(n-1)(n-2)] and you can easily solve by using fractional part.
