$\displaystyle \sum_{n=1}^\infty \frac{6}{n(n+2)}$

$\displaystyle \frac{6}{n(n+2)} = \frac{6}{n} - \frac{6}{n+2}$

$\displaystyle S_n = (6-\frac{6}{3}) + (\frac{6}{2}-\frac{6}{4}) + (\frac{6}{3}-\frac{6}{5})+$...$\displaystyle + (\frac{6}{n} - \frac{6}{n+2})$

not sure what to do next