Question: $$\sum\_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}} $$

Anyone can give hint for this?

Thank you!

Answer: 
*With asymptotic analysis:*

It is easy to check:
$$\frac{\log^2 n}{n^2}=o\biggl(\frac1{n^{3/2}}\biggr)$$
hence it converges.

Indeed the little-oh assertion means that
$$n^{3/2}\frac{\log^2 n}{n^2}=\frac{\log^2x}{n^{1/2}}\xrightarrow[n\to\infty]{}0.$$

*Note:*
This series is but a particular case of a *Bertrand's series* : $\,u\_n=\dfrac1{n^{\alpha}\log^{\beta}n}\enspace\alpha,\beta\in\mathbf R$, hich converges if $\alpha>1$ or if ($\alpha=1$ a $\beta>1$).
As usual for this type of series, you should compare to an integral. In this case, to $\int \frac{\log^2 x}{x^2} dx$. This primitive has a simple expression through usual functions, and by carefully writing your series-integral comparison you should be able to prove easily that this does converge.

(Or, but this is more specific, you could also directly use that $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).