[tex]\dfrac{1}{k(k+1)} =\dfrac{1}{k}-\dfrac{1}{k+1} \\ \\ \\ \\a_{10}=\sum_{k=1}^{10}\dfrac{1}{k(k+1)}=\sum_{k=1}^{10}(\dfrac{1}{k}-\dfrac{1}{k+1} ) \\ \\ \\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3} +...+\dfrac{1}{10}-\dfrac{1}{11} =1-\dfrac{1}{11} =\dfrac{10}{11}[/tex]
