what is the proof for $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $ Can someone provide a proof for the solution of this series
$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}} = {1 \over 4}:\
     {\large ?}}$
\begin{align}
{1 \over n\pars{n + 1}\pars{n + 2}}=
{1/2 \over n} + {-1 \over n + 1} + {1/2 \over n + 2}
=\half\pars{{1 \over n} - {1 \over n + 1}} -
\half\pars{{1 \over n + 1} - {1 \over n + 2}}
\end{align}
Now, you have 'telescoping' series:
\begin{align}
&\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}}}
=\half\sum_{n = \color{#f00}{\large 1}}^{\infty}\pars{{1 \over n} - {1 \over n + 1}}
-\half\sum_{n = \color{#f00}{\large 2}}^{\infty}\pars{{1 \over n} - {1 \over n + 1}}
\\[3mm]&=\half\pars{{1 \over 1} - {1 \over 1 + 1}}
=\color{#00f}{\large{1 \over 4}}
\end{align}
A: $$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$$
so
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)}&=\dfrac{1}{2}\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)\\
&=\dfrac{1}{2}\lim_{n\to\infty}\left(\dfrac{1}{1\times 2}-\dfrac{1}{(n+1)(n+2)}\right)\\
&=\dfrac{1}{4}
\end{align*}
