Show convergence/divergence for $\sum_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}} $ $$\sum_{n=1}^{\infty} \frac{{(\ln n)}^{2}}{{n}^{2}} $$
Anyone can give hint for this?
Thank you!
 A: 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 where $p=\frac32\gt1$.

Using the Euler-Maclaurin Sum Formula, we can compute
$$
\sum_{n=1}^\infty\frac{\log(n)^2}{n^2}
\doteq1.98928023429890102342085868988
$$
A: Hint: Try with Cauchy's condensation theorem; you will get a series of the form
$$
\sum_{n=1}^{\infty} \frac{n^{\alpha}}{2^{n\beta}}.
$$
Now, this series converges for any $\beta>0$.
A: 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.
A: 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 $\beta>1$).
A: I guess it's nowhere exciting as other answers, but since $\frac{\log^2 x}{x^2}$ is monotone decreasing for $x>1$, you can compare it to the integral
$$
\lim_{n \to \infty} \int_{1}^{n}\frac{\log ^2 x dx}{x^2}
$$
and IBP twice, both times with $\int f' = \int \frac{dx}{x^2}$.
A: 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 $\text{Re}(s)>1$.
A: 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 $(log x)^2/x^2 = O(1/x^{1+\varepsilon})$).
