Chapter 10: Infinite Sequences and Series - Section 10.5 - Absolute Convergence; The Ratio and Root Tests - Exercises 10.5 - Page 598: 27

Converges

Let us consider $a_n=\dfrac{\ln n}{n^3}$ Apply direct comparison test. This implies that $\Sigma_{n=1}^\infty \dfrac{\ln n}{n^3} \leq \Sigma_{n=1}^\infty \dfrac{(n)}{(n^3)}=\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent $p$-series with $p=2 \gt 1$ Thus, the series converges by the direct comparison test.