Answer
Converges
Work Step by Step
Let ${a_k} = \dfrac{{\ln k}}{k}$.
Condition 1. Examine if ${a_k} \gt {a_{k + 1}}$.
We plot the graph of $f\left( x \right) = \dfrac{{\ln x}}{x}$. The figure attached shows that $f$ is decreasing for $x \ge 3$.
Thus, for $k \ge 3$, we have ${a_k} - {a_{k + 1}} \gt 0$. Hence, ${a_k} \gt {a_{k + 1}}$.
Condition 2. Examine if $\mathop {\lim }\limits_{k \to \infty } {a_k} = 0$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\ln k}}{k}$
Applying L'Hôpital's Rule, we get
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\ln k}}{k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\dfrac{1}{k}}}{1} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{k} = 0$
Since the two conditions above are satisfied, by Theorem 9.6.1, the alternating series converges.
