Answer
(a) converges
(b) diverges
Work Step by Step
(a) The leading term is $\dfrac{1}{{5{k^2}}}$. Thus, we guess we can apply the comparison test with the convergent $\dfrac{1}{{{k^2}}}$ series.
We check the following:
$\dfrac{1}{{5{k^2} - k}} - \dfrac{1}{{{k^2}}} = \dfrac{{{k^2} - 5{k^2} + k}}{{5{k^4} - {k^3}}} = \dfrac{{ - 4{k^2} + k}}{{5{k^4} - {k^3}}} = \dfrac{{ - 4k + 1}}{{5{k^3} - {k^2}}}$
For $k \ge 1$, we have
$ - 4k + 1 \lt 0$, ${\ \ \ \ \ \ \ }$ $5{k^3} - {k^2} \gt 0$
Therefore,
$\dfrac{1}{{5{k^2} - k}} - \dfrac{1}{{{k^2}}} \lt 0$
Hence,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{5{k^2} - k}} \lt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^2}}}$
Since the bigger series is a convergent $p$-series, it follows from the Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{5{k^2} - k}}$ converges.
(b) The leading term is $\dfrac{3}{k}$. We guess we can apply the Comparison Test with the divergent $\dfrac{1}{k}$ series.
We have,
$\dfrac{3}{{k - \dfrac{1}{4}}} \gt \dfrac{1}{k}$
Thus,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{3}{{k - \dfrac{1}{4}}} \gt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$
The harmonic series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$ diverges. Therefore, by the Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{3}{{k - \dfrac{1}{4}}}$ also diverges.
