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
$$
\sum\_{n=1}^\infty\frac{\log(n)^2}{n^2}
\doteq1.98928023429890102342085868988
$$
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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).