We know that $\frac{1}{n^{2}+2n}=\frac{1}{n(n+2)}=\frac{\frac{1}{n}-\frac{1}{n+2}}{2}$. So, if we sum this from 1 to $\infty$, all terms except for $\frac{1}{2}+\frac{\frac{1}{2}}{2}$ will cancel out (a 'telescoping' series). Therefore, the sum will be $\frac{3}{4}$.