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

Anyone can give hint for this?

Thank you!

Answer: 
I guess it's nowhere exciting as other answers, but since $\frac{\log^2 x}{x^2}$ is monotone decreasing for $x>1$, you can compare it to the integral
$$
\lim\_{n \to \infty} \int\_{1}^{n}\frac{\log ^2 x dx}{x^2}
$$
and IBP twice, both times with $\int f' = \int \frac{dx}{x^2}$.
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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).