When trying to find the sum of these types of series, try looking for a 'telescoping' sum. If expanded, it is easier to see. When telescoping, most terms will cancel one another out and what remains is the sum.
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+3)}=\frac{1}{3}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\displaystyle =\frac{1}{3}\left[\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+....\right]\)
Add a few more terms if need be. Note the cancelling terms. What are you left with?. Let me know what you get.
