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

Hint: $$\frac{1}{n(n+1)(n+2)}=\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$$

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 $f(1) = 1/4$.

$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}=\\\frac{1}{2n}-\frac{2}{2n+2}+\frac{1}{2n+4}=\\(\frac{1}{2n}-\frac{1}{2n+2})+(-\frac{1}{2n+2}+\frac{1}{2n+4})=\\(\frac{1}{2n}-\frac{1}{2n+2})-(\frac{1}{2n+2}-\frac{1}{2n+4})=\\\frac{1}{2}((\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2}))=\\ $$