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?
 A: 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
$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$
A: Since
$\begin{array}\\
\ln(n)
&=\int_1^n \frac{dt}{t}\\
&<\int_1^n \frac{dt}{t^{1/2}}\\
&=\int_1^n t^{-1/2}dt\\
&=2t^{1/2}|_1^n\\
&<2n^{1/2}\\
\end{array}
$,
$\dfrac{\ln(n)}{n^2}
\lt \dfrac{2n^{1/2}}{n^2}
= 2\dfrac1{n^{3/2}}
$
and the sum of this converges
by the integral test.
This can be modified
to show that
$\sum \frac{\ln(n)}{n^{1+c}}
$
converges for any $c > 0$.
Of course the sum diverges
for $c = 0$.
A: Denote $$x_n=\frac{\ln n}{n^2},~~~y_n=\frac{1}{n^{3/2}}.$$We have $$\lim_{n \to \infty}\frac{x_n}{y_n}=\lim_{n \to \infty}\frac{\ln n}{n^{1/2}}=\lim_{n \to \infty}\frac{\dfrac{1}{n}}{\dfrac{1}{2}n^{-1/2}}=2\lim_{n \to \infty}\frac{1}{n^{1/2}}=0.$$
Notice that $\sum y_n$ is convergent. By the comparison test of the limit form, $\sum x_n$ is also convergent.
