[tex]\it \dfrac{1}{k(k+2)} =\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)\\ \\ \\ S=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\ ...\ +\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{1}{2}\left( \dfrac{1}{1}-\dfrac{1}{15}\right) =\\ \\ \\ =\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}[/tex]
