Sum the series (real analysis) $$\sum_{n=1}^\infty {1 \over n(n+1)(n+2)(n+3)(n+4)}$$
I tried to sum the above term as they way I can solve the term $\sum_{n=1}^\infty {1 \over (n+3)}$ by transforming into ${3\over n(n+3)} ={1\over n}-{1\over(n+3)}$ but I got stuck while trying to transform $12\over n(n+1)(n+2)(n+3)(n+4)$ into something solvable. 
 A: Hint
Notice that
$${1 \over n(n+1)(n+2)(n+3)(n+4)}=\frac14{(n+4 )-n\over n(n+1)(n+2)(n+3)(n+4)}\\=\frac14\left({1\over n(n+1)(n+2)(n+3)}-{1\over (n+1)(n+2)(n+3)(n+4)}\right)$$
and then telescope.
A: By working it out, by various methods, it is seen that
\begin{align}
\frac{1}{n(n+1)(n+2)(n+3)(n+4)} = \frac{1}{4!} \, \sum_{r=0}^{4} (-1)^{r} \binom{4}{r} \, \frac{1}{n+r}
\end{align}
Now,
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)(n+4)} \\
&= \frac{1}{4!} \, \sum_{r=0}^{4} (-r)^{r} \binom{4}{r} \, \sum_{n=1}^{\infty} \frac{1}{n+r} \\
&= \frac{1}{4!} \left[ \zeta(1) - 4 \left( \zeta(1) - 1\right) + 6 \left( \zeta(1) - 1 - \frac{1}{2} \right) - 4 \left( \zeta(1) - 1 - \frac{1}{2} - \frac{1}{3} \right) + \left(\zeta(1) - 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} \right) \right] \\
&= \frac{1}{4!} \left[ 4 - 6 \left( 1 + \frac{1}{2} \right) + 4 \left( 1 + \frac{1}{2} + \frac{1}{3} \right) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) \right] \\
S &= \frac{1}{4 \cdot 4!}  
\end{align}
