Does $\sum \limits_{n=1}^{\infty}{\frac{(\ln n)^2}{n^2}}$ converge? I'm trying to find out whether
\begin{equation}
\sum \limits_{n=1}^{\infty}{\dfrac{(\ln n)^2}{n^2}}
\end{equation}
is convergent or divergent, but I can only use Comparison test or Ratio test.
I gave it some thought, but logarithms confused me. Regards and thanks a lot.
 A: Compare with
$$\sum_{n=1}^{\infty} \dfrac1{n^s}$$where $s \in (1,2)$ and make use of the fact that $\dfrac{\ln(n)}{n^s} \to 0$ as $n \to \infty$ for $s > 0$.
For instance, compare with $$\sum_{n=1}^{\infty} \dfrac1{n^{1.5}}$$
A: There's one more way. Compare it to the integral:
$$
I=\int_{1}^{\infty}\bigg(\frac{\log x}{x}\bigg)^2dx
$$
Also note $\frac{\log x}{x^2}<\big(\frac{\log x}{x}\big)^2<\frac{\log x}{x}<\frac{\log^2 x}{x}$and the limit of all these sequences is $0$ as $x \to \infty$.
Now integrate I by parts in the following way:
$$
I=\int_{0}^{\infty}1 \cdot \bigg(\frac{\log x}{x}\bigg)^2dx=\frac{\log^2 x}{x}\bigg|^{\infty}_{1}-2 \int_{1}^{\infty}\frac{\log x}{x^2}dx+2 \int_{1}^{\infty}\bigg(\frac{\log x}{x}\bigg)^2dx \\
=-2\int_{1}^{\infty}\frac{\log x}{x^2}dx+2I
$$
so you get 
$$
I=2\int_{1}^{\infty}\frac{\log x}{x^2}dx
$$
Again integrate the expression on RHS by parts to get 
$$
I=2(0+\int_{1}^{\infty}\frac{dx}{x^2})=2
$$
Hence the original sum converges too.
A: To prove $\ln n < C(\epsilon) n^{\epsilon}$
where $ C(\epsilon)$ depends only on $\epsilon$:
Since $e^x \ge 1+x > x$,
$x > \ln x$.
Now watch: The fingers never leave the hands!
Substitute $x^{\epsilon}$ for $x$.
Then 
$x^{\epsilon} > \ln (x^{\epsilon})
= {\epsilon} \ln x
$
so $\ln x < \dfrac{x^{\epsilon}}{{\epsilon}}$.
Setting ${\epsilon}=\frac{1}{m}$,
$\ln x < m x^{1/m}$.
In particular,
for $m=2$,
$\ln x < 2 x^{1/2}$.
A: Since for all $\,n\,$ big enough we have that $\,\log n<n^\epsilon\;$ , with arbitrary $\,\epsilon>0\;$ , we get
$$\frac{\left(\log n\right)^2}{n^2}<\frac{\left(n^{1/4}\right)^2}{n^2}=\frac1{n^{3/2}}\ldots$$
