Answer
Conditionally convergent.
Work Step by Step
We have $\mathop \sum \limits_{k = 3}^\infty \dfrac{{{{\left( { - 1} \right)}^k}\ln k}}{k}$. Let ${a_k} = \dfrac{{\ln k}}{k}$.
Step1. We show that the original series converges.
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.
Step 2. Here we test the series for absolute convergence by examining the series of absolute values:
$\mathop \sum \limits_{k = 3}^\infty \left| {\dfrac{{{{\left( { - 1} \right)}^k}\ln k}}{k}} \right| = \mathop \sum \limits_{k = 3}^\infty \dfrac{{\ln k}}{k}$
In Step 1, we have shown that $f$ is decreasing for $x \ge 3$. Since it is also continuous, we can use the Integral test to check the convergence of $\mathop \sum \limits_{k = 3}^\infty \dfrac{{\ln k}}{k}$.
Write $f\left( x \right) = \dfrac{{\ln x}}{x}$. Let $t = \ln x$. So, $dt = \dfrac{1}{x}dx$.
$\mathop \smallint \limits_3^\infty \dfrac{{\ln x}}{x}{\rm{d}}x = \mathop \smallint \limits_{\ln 3}^\infty t{\rm{d}}t = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{2}\left( {{t^2}} \right)|_{\ln 3}^p = \dfrac{1}{2}\mathop {\lim }\limits_{p \to \infty } \left[ {{p^2} - {{\left( {\ln 3} \right)}^2}} \right] = \infty $
Since the integral diverges, by the Integral Test, the series $\mathop \sum \limits_{k = 3}^\infty \dfrac{{\ln k}}{k}$ also diverges. Hence, the original series diverges absolutely.
Thus, the original series converges and also diverges absolutely. Hence, it converges conditionally.
