<s>[INST] 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. 
 [/INST] 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.
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