## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{\ln n}{n^3}$ Here, $\Sigma_{n=1}^\infty \dfrac{\ln n}{n^3} \leq \Sigma_{n=1}^\infty \dfrac{n}{n^3}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ we see that the series $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a convergent p-series. Hence, the given series converges by the direct comparison test.