<s>[INST] Evaluating an infinite series using partial fractions... I am having trouble evaluating an infinite series that uses partial fractions.  The problem is as follows:
$$
\sum_{n = 1}^{\infty}{1 \over n\left(n + 1\right)\left(n + 2\right)}
$$
I realize that this is a telescoping series, but I am unable to find a general formula for the Sn.  
After the partial fraction decomposition the problem looks like
$1/\left(2n\right) - 1/\left(n + 1\right) + 1/\left(2n + 4\right)$. I input values for $n = 1, n = 2,$ etc, and some cancel out but I am unable to determine a pattern to write to find Sn...  Thanks.
I found this answer to the problem online which makes sense, up to the step where it makes Sn equal to
$1/2(1/2 - 1/\left(n+1\right) + 1/\left(n+2\right)$.  I get the prior steps but I am not sure how they make that leap.

 [/INST] $$\frac{1}{n(n+1)(n+2)} = \frac{1}{(n+1)} \cdot \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right) = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$$
So
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{1(1+1)}-\frac{1}{(1+1)(1+2)}+\frac{1}{2(2+1)}-\frac{1}{(2+1)(2+2)}+\cdots\right)$$
$$=\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$
</s>