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

Consider $a_n=\dfrac{\ln n}{n}$ Since, we know that $\ln n \geq 1$ for the all value of $n$. Thus, $\Sigma_{n=1}^\infty \dfrac{\ln n}{n} \geq \Sigma_{n=1}^\infty \dfrac{1}{n}$ we see that the series $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent harmonic series. Hence, the given series Diverges by the direct comparison test.