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

Anyone can give hint for this?

Thank you!

Answer: 
If you like an overkill, the Kummer's Fourier series for the $\log\Gamma$ function gives:

> 
> $$\log\Gamma(t) = \frac{1}{2}\,\log\left(\frac{\pi}{\sin(\pi t)}\right)+(\gamma+\log(2\pi))\left(\frac{1}{2}-t\right)+\frac{1}{\pi}\sum\_{n\geq 1}\frac{\log n}{n}\,\sin(2\pi n t) $$
> 
> 
> 

hence your series is converging by Parseval's theorem, since both
$$ \log\Gamma(t),\quad \log\left(\frac{\pi}{\sin(\pi t)}\right),\quad \left(\frac{1}{2}-t\right) $$
belong to $L^2(0,1)$. Namely:
$$ \int\_{0}^{1} \log^2\left(\frac{\pi}{\sin(\pi t)}\right)\,dt = \frac{\pi^2}{12}+\frac{1}{4}\,\log^2(4\pi^2). $$
Moreover:
$$ \sum\_{n\geq 1}\frac{\log^2 n}{n^2}=\zeta''(2),$$
where the $\zeta(s)$ function is analytic over $\text{Re}(s)>1$.
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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).