Answer
The series converges
Work Step by Step
$ \frac{1}{n(n+2)} = \frac{1}{2} \frac{n+2-n}{n(n+2)}= \frac{1}{2} (\frac{1}{n}-\frac{1}{n+2})$
Therefore,
$\sum _{n=1}^{\infty \:} \frac{1}{n(n+2)}= \sum _{n=1}^{\infty \:}\frac{1}{2} (\frac{1}{n}-\frac{1}{n+2})$
$\:=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...$
We see that eventually, each term will cancel out except for $1+\frac{1}{2} = \frac{3}{2}$ to which this series converges