Answer
See proof
Work Step by Step
The $n$th partial sums:
$\mathop \sum \limits_{k = 1}^n \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \left( {1 - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{6}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{n - 2}} - \dfrac{1}{n}} \right) + \left( {\dfrac{1}{{n - 1}} - \dfrac{1}{{n + 1}}} \right) + \left( {\dfrac{1}{n} - \dfrac{1}{{n + 2}}} \right)$
$ = 1 + \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{4} + \dfrac{1}{4}} \right) + \cdot\cdot\cdot + \left( { - \dfrac{1}{n} + \dfrac{1}{n}} \right) - \dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}$
So, $\mathop \sum \limits_{k = 1}^n \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \dfrac{3}{2} - \dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}$.
Taking the limit:
$\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{3}{2} - \dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}} \right) = \dfrac{3}{2} - \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{n + 1}} - \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{n + 2}} = \dfrac{3}{2}$
Hence, $\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{k} - \dfrac{1}{{k + 2}}} \right) = \dfrac{3}{2}$.
