I have a question that
$\sum \_{k =1}^n\frac{1} {k(k+3)} $
I resolved this into partial fraction 
$\sum \left(\frac{1} {3k} - \frac{1} {3(k+3)} \right) $

I had a geometric series which converges to $1/2$ but the other has a different pattern

Where exactly is the way to go?