Answer
See proof
Work Step by Step
Write
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 2} \right)}} = \dfrac{1}{{1\cdot3}} + \dfrac{1}{{2\cdot4}} + \dfrac{1}{{3\cdot5}} + \cdot\cdot\cdot$
We can write $\dfrac{1}{{k\left( {k + 2} \right)}} = \dfrac{1}{2}\left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right)$.
So,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 2} \right)}} = \dfrac{1}{2}\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right)$
According to Exercise 35: $\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \dfrac{3}{2}$. Therefore,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 2} \right)}} = \dfrac{1}{2}\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \dfrac{1}{2}\cdot\dfrac{3}{2} = \dfrac{3}{4}$
Hence,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k\left( {k + 2} \right)}} = \dfrac{1}{{1\cdot3}} + \dfrac{1}{{2\cdot4}} + \dfrac{1}{{3\cdot5}} + \cdot\cdot\cdot = \dfrac{3}{4}$
