Answer
\[\text{The integral test does not apply}\]
Work Step by Step
\[\begin{align}
& \sum\limits_{k=1}^{\infty }{\frac{k}{\sqrt{{{k}^{2}}+4}}} \\
& \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{k}{\sqrt{{{k}^{2}}+4}},\text{ then let }f\left( x \right)=\frac{x}{\sqrt{{{x}^{2}}+4}} \\
& \text{The domain of }f\left( x \right)\text{is }\left( -\infty ,\infty \right)\text{, so }f\left( x \right)\,\text{is continuous and positive } \\
& \text{for }x\ge 1, \\
& \text{Now we will determine if }f\left( x \right)\text{ is decreasing}\text{, then } \\
& f\left( x \right)=\frac{x}{\sqrt{{{x}^{2}}+4}} \\
& \text{Differentiating} \\
& f'\left( x \right)=\frac{\sqrt{{{x}^{2}}+4}-x\left( \frac{2x}{2\sqrt{{{x}^{2}}+4}} \right)}{{{\left( \sqrt{{{x}^{2}}+4} \right)}^{2}}} \\
& f'\left( x \right)=\frac{{{x}^{2}}+4-{{x}^{2}}}{{{\left( \sqrt{{{x}^{2}}+4} \right)}^{3/2}}} \\
& f'\left( x \right)=\frac{4}{{{\left( \sqrt{{{x}^{2}}+4} \right)}^{3/2}}} \\
& f'\left( x \right)>0\text{ for the interval }\left( 1,\infty \right) \\
& \text{The function is not decreasing for }x\ge 1,\text{ then} \\
& \text{The integral test does not apply} \\
\end{align}\]
