Answer
Converges
Work Step by Step
$\sum ^{n}_{1}\left( \dfrac {1}{n}-\dfrac {1}{n+2}\right) =\left( \dfrac {1}{1}-\dfrac {1}{1+2}\right) +\left( \dfrac {1}{2}-\dfrac {1}{2+2}\right) +\left( \dfrac {1}{3}-\dfrac {1}{3+2}\right) +\ldots \left( \dfrac {1}{n-1}-\dfrac {1}{n+1}\right) +\left( \dfrac {1}{n}-\dfrac {1}{n+2}\right) =\dfrac {3}{2}-\dfrac {1}{n+1}-\dfrac {1}{n+2}$
$\lim _{n\rightarrow \infty }\left( \dfrac {3}{2}-\dfrac {1}{n+1}-\dfrac {1}{n+2}\right) =\dfrac {3}{2}\Rightarrow \sum ^{\infty }_{n=1}\left( \dfrac {1}{n}-\dfrac {1}{n+2}\right) $
So series converges