Answer
(a) $p = 3$; the series converges.
(b) $p = \dfrac{1}{2}$; the series diverges.
(c) $p = 1$; the series diverges.
(d) $p = \dfrac{2}{3}$; the series diverges.
Work Step by Step
(a) $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^3}}}$
$p = 3$
We have $p = 3 \gt 1$, by Theorem 9.4.5, the series converges.
(b) $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\sqrt k }} = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^{1/2}}}}$
$p = \dfrac{1}{2}$
Since $p = \dfrac{1}{2} \lt 1$, by Theorem 9.4.5, the series diverges.
(c) $\mathop \sum \limits_{k = 1}^\infty {k^{ - 1}} = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$
$p = 1$
This is the harmonic series, which diverges.
(d) $\mathop \sum \limits_{k = 1}^\infty {k^{ - 2/3}} = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{k^{2/3}}}}$
$p = \dfrac{2}{3}$
Since $p = \dfrac{2}{3} \lt 1$, by Theorem 9.4.5, the series diverges.
