Answer
See proof
Work Step by Step
Write
$\dfrac{{\sqrt {k + 1} - \sqrt k }}{{\sqrt {{k^2} + k} }} = \dfrac{{\sqrt {k + 1} }}{{\sqrt {{k^2} + k} }} - \dfrac{{\sqrt k }}{{\sqrt {{k^2} + k} }} = \dfrac{{\sqrt {k + 1} }}{{\sqrt k \sqrt {k + 1} }} - \dfrac{{\sqrt k }}{{\sqrt k \sqrt {k + 1} }} = \dfrac{1}{{\sqrt k }} - \dfrac{1}{{\sqrt {k + 1} }}$
Thus, the $n$th partial sums of the series can be written as
$\mathop \sum \limits_{k = 1}^n \dfrac{{\sqrt {k + 1} - \sqrt k }}{{\sqrt {{k^2} + k} }} = \mathop \sum \limits_{k = 1}^n \left( {\dfrac{1}{{\sqrt k }} - \dfrac{1}{{\sqrt {k + 1} }}} \right)$
$ = \left( {1 - \dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }}} \right) + \left( {\dfrac{1}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 4 }}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right)$
$ = 1 + \left( { - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}} \right) + \left( { - \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }}} \right) + \cdot\cdot\cdot + \left( { - \dfrac{1}{{\sqrt n }} + \dfrac{1}{{\sqrt n }}} \right) - \dfrac{1}{{\sqrt {n + 1} }}$
$\mathop \sum \limits_{k = 1}^n \dfrac{{\sqrt {k + 1} - \sqrt k }}{{\sqrt {{k^2} + k} }} = 1 - \dfrac{1}{{\sqrt {n + 1} }}$
Taking the limit:
$\mathop \sum \limits_{k = 1}^\infty \dfrac{{\sqrt {k + 1} - \sqrt k }}{{\sqrt {{k^2} + k} }} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{1}{{\sqrt {n + 1} }}} \right) = 1$
Hence, $\mathop \sum \limits_{k = 1}^\infty \dfrac{{\sqrt {k + 1} - \sqrt k }}{{\sqrt {{k^2} + k} }} = 1$.
