Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{1}{k^2 \ln k}$ and $b_k=\dfrac{1}{k^{2}}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{1}{k^2 \ln k}}{1/k^2}\\=\lim\limits_{k \to \infty} \dfrac{1}{\ln k } \\=0$
Therefore, the series converges by the limit comparison test.
