How to tell if $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges using Integral Test? So I have this problem:
Determine whether or not the following infinite series converge or diverge. State what test you used.

$$\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$$

So I decided to do the Integral Test for $$\int^\infty_1\frac{ln(n)}{n^2}dx$$
I made $u = \ln(n)$ and $du = \frac{1}{n}dx$ and got:
$$\int^\infty_1\frac{u}{n}*du$$
which would become:
$$u[\ln(n)]|^\infty_1$$
This is incomplete, but does this look right so far? If not, how do I fix it?

By integral test we should obtain
$$\int_1^\infty \frac{\ln x}{x^2} dx=\left[-\frac{1+\log x}{x}\right]_1^\infty$$
or by $\ln x=u \implies \frac1x dx=du$
$$\int_0^\infty \frac{u}{e^u} du=\left[-\frac{u+1}{e^u}\right]_0^\infty$$
If you are not forced to use integral test, as an effective alternative, we can use limit comparison test with
$$\sum_{n=1}^\infty\frac{1}{n^p}$$
with $p>1$ such that
$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$
