TITLE: Prove the following identity
QUESTION [5 upvotes]: I am having some trouble proving following identity without use of induction, with which it is trivial.
$$\sum_{n=1}^{m}\frac{1}{n(n+1)(n+2)}=\frac{1}{4}-\frac{1}{2(m+1)(m+2)}$$
I did expand the expression:
$$\sum_{n=1}^{m}\left( \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \right)$$
I have no idea how to proceed further.

REPLY [6 votes]: This is a special case $(k=3)$ of the following
$$
\begin{align}
\sum_{n=k}^m\frac1{\binom{n}{k}}
&=\frac{k}{k-1}\sum_{n=k}^m\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n\vphantom{1}}{k-1}}\right)\\
&=\frac{k}{k-1}\left(1-\frac1{\binom{m}{k-1}}\right)
\end{align}
$$
Noting that $n(n+1)(n+2)\dots(n+k-1)=k!\binom{n+k-1}{k}$ yields
$$
\sum_{n=1}^m\frac1{k!\binom{n+k-1}{k}}=\frac1{k-1}\left(\frac1{(k-1)!}-\frac1{(k-1)!\binom{(m+1)+(k-2)}{k-1}}\right)
$$
that is,
$$
\sum_{n=1}^m\underbrace{\frac1{n(n+1)\dots(n+k-1)}}_{k\text{ factors}}=\frac1{k-1}\left(\frac1{(k-1)!}-\underbrace{\frac1{(m+1)(m+2)\dots(m+k-1)}}_{k-1\text{ factors}}\right)
$$
Substituting $k=3$ gives
$$
\sum_{n=1}^m\frac1{n(n+1)(n+2)}=\frac12\left(\frac12-\frac1{(m+1)(m+2)}\right)
$$