\text{Rezultat}=\sum _{n=1}^{\infty } \frac{1}{n(n+1)^2}=\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{1+n}-\frac{1}{(1+n)^2}\right)=\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{1+n}\right)-\sum _{n=1}^{\infty } \left(\frac{1}{(1+n)^2}\right)=1-(\zeta (2)-1)=2-\frac{\pi ^2}{6}