Answer
Convergent
Work Step by Step
$\Sigma_{k=1}^{\infty}\frac{1}{k\sqrt {k^{2}+1}}\lt \Sigma_{k=1}^{\infty}\frac{1}{k\sqrt {k^{2}}}$
Here, $\Sigma_{k=1}^{\infty}\frac{1}{k^{2}}$ is a p-series with $p=2$, hence convergent.
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