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

Anyone can give hint for this?

Thank you!

Answer: 
Note that $1+x\le e^x$ for all $x\in\mathbb{R}$. Thus, for $x\gt0$,
$$
1+\tfrac14\log(x)=1+\log\left(x^{1/4}\right)\le x^{1/4}
$$
Therefore,
$$
\log(x)\le4x^{1/4}
$$
Thus,
$$
\begin{align}
\sum\_{n=1}^\infty\frac{\log(n)^2}{n^2}
&\le\sum\_{n=1}^\infty\frac{16n^{1/2}}{n^2}\\
&=16\sum\_{n=1}^\infty\frac1{n^{3/2}}\\
\end{align}
$$
which converges by comparison to the [$p$-series](http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#P-series) where $p=\frac32\gt1$.

---

Using the [Euler-Maclaurin Sum Formula](http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula), we can compute
$$
\sum\_{n=1}^\infty\frac{\log(n)^2}{n^2}
\doteq1.98928023429890102342085868988
$$
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 $\text{Re}(s)>1$.