Answer
The Limit Comparison Test; diverges.
Work Step by Step
The leading term is $\dfrac{1}{{\sqrt k }}$. Thus, we try to use the limit comparison test with the series of $\dfrac{1}{{\sqrt k }}$.
Write ${a_k} = \dfrac{1}{{1 + \sqrt k }}$ and ${b_k} = \dfrac{1}{{\sqrt k }}$.
Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_k}}}{{{b_k}}} = \dfrac{{\sqrt k }}{{1 + \sqrt k }} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\dfrac{1}{{\sqrt k }} + 1}} = 1$
Since $\rho = 1 \gt 0$ and the $p$-series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\sqrt k }}$ diverges (because $p = \dfrac{1}{2} \lt 1$), by the limit comparison test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{1 + \sqrt k }}$ also diverges.
