Calculate $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$ 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
 A: 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 $f(1) = 1/4$.
A: Your partial sums are the same as those of $\frac 1 2 H_n-H_{n+1}+\frac 1 2 H_{n+2}$ minus some terms. Find those terms and use those harmonic sums tend to 0 in the limit. The answer is $1/4$. More generally, it happens that $$\binom{n+k-1}{k}^{-1}=\frac{k!}{n(n+1)\cdots (n+k)}=\sum_{j=0}^k \binom kj \frac{(-1)^j}{n+j}$$
This means that $$\sum_{n=1}^N\binom{n+k-1}k^{-1}=\sum_{j=0}^k\binom kj(-1)^j (H_{N+j}-H_j)$$
Using this and the fact that $\sum_{j=0}^k\binom kj(-1)^{j+1} H_j=\frac 1 k$ one gets $$\sum_{n\geqslant 1}\binom {n+k-1}k^{-1}=\frac 1 k$$
or $$\sum_{n\geqslant 1}\frac {1}{n(n+1)\cdots (n+k)}=\frac 1k\frac 1 {k!}$$
A: $$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}=\\\frac{1}{2n}-\frac{2}{2n+2}+\frac{1}{2n+4}=\\(\frac{1}{2n}-\frac{1}{2n+2})+(-\frac{1}{2n+2}+\frac{1}{2n+4})=\\(\frac{1}{2n}-\frac{1}{2n+2})-(\frac{1}{2n+2}-\frac{1}{2n+4})=\\\frac{1}{2}((\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2}))=\\ $$
A: Hint: $$\frac{1}{n(n+1)(n+2)}=\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$$
