TITLE: Calculate $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$
QUESTION [11 upvotes]: I am trying to calculate the following series:
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$
and I managed to reduce it to this term
$$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$
How should I go about solving this question

REPLY [8 votes]: Consider 
$$f(x) = \sum_{n=1}^{\infty} \frac{x^{n+2}}{n (n+1)(n+2)} $$
Then
$$f''(x) = -\log{(1-x)}$$
$$f'(x) = (1-x) \log{(1-x)} +x $$
$$f(x) = -\frac14 [x (2-x) - 2 (1-x)^2 \log{(1-x)}] + \frac12 x^2$$
The sum is then $f(1) = 1/4$.