<s>[INST] Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper

Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
  and find the sum if it exists.

I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
 [/INST] There is an alternate method and is as follows.
Notice that 
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\
&= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\
&= \frac{1}{4}
\end{align}
This leads to the known result
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}.
\end{align}
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