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
$$
\sum\_{n=1}^\infty\frac{\log(n)^2}{n^2}
\doteq1.98928023429890102342085868988
$$
*With asymptotic analysis:*

It is easy to check:
$$\frac{\log^2 n}{n^2}=o\biggl(\frac1{n^{3/2}}\biggr)$$
hence it converges.

Indeed the little-oh assertion means that
$$n^{3/2}\frac{\log^2 n}{n^2}=\frac{\log^2x}{n^{1/2}}\xrightarrow[n\to\infty]{}0.$$

*Note:*
This series is but a particular case of a *Bertrand's series* : $\,u\_n=\dfrac1{n^{\alpha}\log^{\beta}n}\enspace\alpha,\beta\in\mathbf R$, hich converges if $\alpha>1$ or if ($\alpha=1$ a $\beta>1$).