Answer
(a) $\mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{{{\rm{e}}^k}}} = 0$; the series may either converge or diverge.
(b) $\mathop {\lim }\limits_{k \to \infty } \ln k = \infty $; the series diverges.
(c) $\mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\sqrt k }} = 0$; the series may either converge or diverge.
(d) $\mathop {\lim }\limits_{k \to \infty } \dfrac{{\sqrt k }}{{\sqrt k + 3}} \ne 0$; the series diverges.
Work Step by Step
(a) Let ${a_k} = \dfrac{k}{{{{\rm{e}}^k}}}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{{{\rm{e}}^k}}} = 0$
According to the divergence test, the series may either converge or diverge.
(b) Let ${a_k} = \ln k$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \ln k = \infty $
The limit does not exist. Since $\mathop {\lim }\limits_{k \to \infty } {a_k} \ne 0$, by the divergence test, the series diverges.
(c) Let ${a_k} = \dfrac{1}{{\sqrt k }}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\sqrt k }} = 0$
According to the divergence test, the series may either converge or diverge.
(d) Let ${a_k} = \dfrac{{\sqrt k }}{{\sqrt k + 3}}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\sqrt k }}{{\sqrt k + 3}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{1 + \dfrac{3}{{\sqrt k }}}} = 1$
Since $\mathop {\lim }\limits_{k \to \infty } {a_k} \ne 0$, by the divergence test, the series diverges.
