Test for convergence: $\int_1^\infty \frac{\ln x}{x^2} \, dx$ 
Is this improper integral convergent or divergent?$$\int_1^\infty \frac{\ln x}{x^2} \, dx$$

I tried $\int_1^\infty \frac{\ln x}{x^2} \, dx \le \int_1^\infty \frac{\ln x}x \, dx$ but the RHS diverges, which makes this relation inconclusive. I think the integral diverges.

The integral converges: $\log x / \sqrt{x} \rightarrow 0$ as $x\rightarrow +\infty$, and
$$\int_1^{+\infty} \sqrt{x}/x^2 dx <+\infty.$$ 
