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

Anyone can give hint for this?

Thank you!

Answer: 
Hit : Compare the sum to the integral
$$\int \frac{\ln(x)^2}{x^2}dx=C-\frac{(\ln x)^2+2 \ln(x)+2}{x}$$
wich tends to a constant $C$ for $x$ tending to infinity. (Of course, $C$ is not the limit of the sum). I let you construct the full answer.

For information only : $\sum\_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}}=\zeta''(2) $ , second derivative of the Riemann zeta function. This is not necessary to be known, to answer to the question.
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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).