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?

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 $c = 0$.
