Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{1}{k^{\ln k}}$ and $b_k=\dfrac{1}{k^2}$
We can see that for $\dfrac{1}{k^{\ln k}} \lt \dfrac{1}{k^2}$ on $(e^2, \infty)$.
But $b_k=\dfrac{1}{k^2}$ corresponds to a convergent p-series.
Therefore, the series converges by the comparison test.
