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.

First, we solve the integral using integration by parts (let me know if I should elaborate more on this) 
$$\lim_{a\to\infty}\int_1^a \frac{\ln x}{x^2}\,dx=\lim_{a\to\infty}\left.\frac{-\ln x-1}{x}\right|_1^a=\Big(\lim_{a\to\infty}\frac{-\ln a-1}{a}\Big)+1$$
Now, we have to solve the limit
$$\Big(\lim_{a\to\infty}\frac{-\ln a-1}{a}\Big)+1$$
Apply L'Hôpitals Rule
$$=\lim_{a\to\infty}-\frac1a+1=1$$
Therefore, the integral converges to $1$.
