$\int_1^\infty \frac{\ln(t)}{t}dt$ is divergent?

Your solution is good, but perhaps this is simpler:

If $u = \ln t$ the $du = dt/t$ so $$ \int_1^\infty \frac{\ln t}{t} dt = \int_0^\infty udu = \left. \frac{u^2}{2} \right|_0^\infty $$

Your way by direct comparison is fine and we can simplify it further as follows

$$\int_1^\infty \frac{\ln(t)}{t}dt \ge \int_e^\infty \frac{\ln(t)}{t}dt \geq \int_e^\infty \frac{1}{t}dt = \infty$$