Answer
(a) converges
(b) diverges
Work Step by Step
We show by graph that $\ln x \lt \sqrt x $ if $x \gt 0$ (please see the figure attached).
(a) For $k \ge 1$, we have $\ln k \lt \sqrt k $.
Thus,
$\dfrac{{\ln k}}{{{k^2}}} \lt \dfrac{{\sqrt k }}{{{k^2}}} = \dfrac{1}{{{k^{3/2}}}}$
$\mathop \sum \limits_{k = 1}^\infty \dfrac{{\ln k}}{{{k^2}}} \lt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^{3/2}}}}$
The bigger series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^{3/2}}}}$ is a convergent $p$-series (because $p = \dfrac{3}{2} \gt 1$). Thus, by the Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{\ln k}}{{{k^2}}}$ converges.
(b) For $k \ge 1$, we have $\ln k \lt \sqrt k $.
Thus,
$\dfrac{1}{{\ln k}} \gt \dfrac{1}{{\sqrt k }}$
$\dfrac{1}{{{{\left( {\ln k} \right)}^2}}} \gt \dfrac{1}{k}$
$\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{{\left( {\ln k} \right)}^2}}} \gt \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{k}$
The harmonic series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{k}$ diverges. Therefore, by the Comparison Test, the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{{\left( {\ln k} \right)}^2}}}$ diverges.
