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

Anyone can give hint for this?

Thank you!

Answer: 
Hint: Try with Cauchy's condensation theorem; you will get a series of the form
$$
\sum\_{n=1}^{\infty} \frac{n^{\alpha}}{2^{n\beta}}.
$$
Now, this series converges for any $\beta>0$.
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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).