Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{1}{\ln (\ln n)}$
When the numerator increases, the value of fraction is always increasing whereas when the denominator decreases, the value of the fraction is also increasing.
$\implies u_n \geq \dfrac{1}{n}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent series by the limit comparison test.
Thus, the series diverges.