Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{1}{\sqrt n \ln n}$
When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase.
Thus, we have $a_n \geq \dfrac{1}{\sqrt n \sqrt n}=\dfrac{1}{n}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent series. Thus the series diverges by the direct comparison test.