Answer
The series $\Sigma \dfrac{1}{\sqrt k}$ diverges by using the comparison test .
Work Step by Step
Here, we can see that the series $\dfrac{1}{k} \leq \dfrac{1}{\sqrt k}$
where, $k=1,2,3,.....$
But the series $\Sigma \dfrac{1}{k}$ diverges.
Thus, we can say that the series $\Sigma \dfrac{1}{\sqrt k}$ diverges by using the comparison test .
