Answer
Please see the explanations below.
Work Step by Step
(a) For $k \ge 3$, we have $\ln k > 1$ (as is shown in the figure attached).
Thus,
$\dfrac{{\ln k}}{k} \gt \dfrac{1}{k}$
$\mathop \sum \limits_{k = 3}^\infty \dfrac{{\ln k}}{k} \gt \mathop \sum \limits_{k = 3}^\infty \dfrac{1}{k}$
Since the harmonic series $\mathop \sum \limits_{k = 3}^\infty \dfrac{1}{k}$ diverges, by the Comparison Test, the bigger series $\mathop \sum \limits_{k = 3}^\infty \dfrac{{\ln k}}{k}$ also diverges.
By Theorem 9.4.3, divergence is unaffected by deleting a finite number of terms from a series, hence $\mathop \sum \limits_{k = 1}^\infty \dfrac{{\ln k}}{k}$ diverges.
(b) For $k \ge 1$, we have
$\dfrac{k}{{{k^{3/2}} - \dfrac{1}{2}}} \gt \dfrac{k}{{{k^{3/2}}}} = \dfrac{1}{{{k^{1/2}}}}$
Thus,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{k}{{{k^{3/2}} - \dfrac{1}{2}}} \gt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^{1/2}}}}$
The series on the right-hand side is a divergent $p$-series because $p = \dfrac{1}{2} \lt 1$. Thus, by the Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{k}{{{k^{3/2}} - \dfrac{1}{2}}}$ is divergent.
