jmc

By partial fractions, $$\frac{1}{n(n+2)}=\frac{1/2}{n}-\frac{1/2}{n+2}.$$Hence, $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{1}{n(n+2)}& =\left(\frac{1/2}{1}-\frac{1/2}{3}\right)+\left(\frac{1/2}{2}-\frac{1/2}{4}\right)+\left(\frac{1/2}{3}-\frac{1/2}{5}\right)+\left(\frac{1/2}{4}-\frac{1/2}{6}\right)+\cdots \\ & =\frac{1/2}{1}+\frac{1/2}{2}\\ & =\boxed{\frac{3}{4}}.\end{array}$$