Finding the infinite Sum of a series: $\sum\frac1{n(n+1)(n+2)}$ Find the infinite Sum of the series with general term $\frac{1}{n(n+1)(n+2)}$.
I decomposed the fraction upto this $1/(2n)-1/(n+1)+1/(2n+4)$. But I find no link about cancelling terms. So how should I find the infinite Sum? 
Please don't say that this is off topic or homework.  if wanna say that this is homework, at least give a hint if I am going the right way? 

Notice, use partial fractions in form of difference of two terms as follows $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$$$=\sum_{n=1}^{\infty}\frac 12\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$$
$$=\small  \frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+\left(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}\right)+\ldots+\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\right)$$
$$=\frac 12\lim_{n\to \infty}\left(\frac{1}{1\cdot 2}-\frac{1}{(n+1)(n+2)}\right)$$
$$=\frac 12\left(\frac{1}{2}-0\right)=\color{red}{\frac 14}$$
