Answer
(a) diverges
(b) converges
Work Step by Step
(a) Here, the leading term is $\dfrac{k}{{{k^2}}} = \dfrac{1}{k}$. Thus, we can apply the Comparison Test with the divergent $\dfrac{1}{k}$ series.
We check the following:
$\dfrac{{k + 1}}{{{k^2} - k}} - \dfrac{1}{k} = \dfrac{{{k^2} + k - {k^2} + k}}{{{k^3} - {k^2}}} = \dfrac{{2k}}{{{k^3} - {k^2}}} = \dfrac{2}{{{k^2} - k}}$
For $k \ge 2$, we have $\dfrac{2}{{{k^2} - k}} \gt 0$. Therefore,
$\dfrac{{k + 1}}{{{k^2} - k}} - \dfrac{1}{k} \gt 0$
Thus,
$\mathop \sum \limits_{k = 2}^\infty \dfrac{{k + 1}}{{{k^2} - k}} \gt \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{k}$
The smaller series is divergent since it is the harmonic series. It follows from the Comparison Test, the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{{k + 1}}{{{k^2} - k}}$ diverges.
(b) The leading term is $\dfrac{2}{{{k^4}}}$. We guess we can apply the Comparison Test with the convergent $\dfrac{2}{{{k^4}}}$ $p$-series.
For $k \ge 1$, we have
$\dfrac{2}{{{k^4} + k}} \lt \dfrac{2}{{{k^4}}}$
Thus,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{2}{{{k^4} + k}} \lt \mathop \sum \limits_{k = 1}^\infty \dfrac{2}{{{k^4}}}$
The series on the right-hand side converges since it is a constant times a convergent $p$-series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^4}}}$. Thus, by the comparison test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{2}{{{k^4} + k}}$ also converges.
