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 $p&gt;1$ such that

$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$
Since

$\begin{array}\\
\ln(n)
&amp;=\int_1^n \frac{dt}{t}\\
&amp;&lt;\int_1^n \frac{dt}{t^{1/2}}\\
&amp;=\int_1^n t^{-1/2}dt\\
&amp;=2t^{1/2}|_1^n\\
&amp;&lt;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 $c &gt; 0$.
Of course the sum diverges
for $c = 0$.