Although $a_n\=\ln \left(n\right)\/n\to 0$ as $n\to \infty$, the divergence of the given series can be shown by both the Limit-Comparison and Integral tests (Table 8.3.1).

Since $\frac{d}{\mathrm{dx}}\left(\frac{\ln \left(x\right)}{x}\right)\=\frac{1-\ln \left(x\right)}{x^2}$, which is negative for $x\>e\doteq 2.7$, the sequence $\left\{\ln \left(n\right)\/n\right\}$ is eventually monotone decreasing to zero, so the Integral test applies, and gives

${\int }_3^{\infty }\frac{\ln \left(x\right)}{x} \mathit{\ⅆ}x$ = $\mathrm{\∞}$$$

Since this integral diverges, so too does the given series.

Alternatively, since $\underset{n\to \infty }{lim}\frac{ \ln \left(n\right)\/n}{1\/n}\=\underset{n\to \infty }{lim}\ln \left(n\right)$ = $\mathrm{\∞}$, by part (3) of the Limit-Comparison test the given series diverges because the comparison series is the divergent harmonic series.