I'm trying to prove that the probability distribution $P(X=k) \sim \dfrac{1}{k^2}, k \in \mathbb{Z} \setminus \{0\}$ does not hold the weak law of large numbers. it means if $\overline{X}$ is the average of $n$ iid samples from this distribution, then for any $\epsilon>0$ we have a $\delta >0$ in which $P(|\overline{X}|>\epsilon) > \delta$ for any $n \in N$.
I don't even know if this fact is right or not, so I'd also be happy to see why is this fact wrong. I was just thinking that maybe it's true because this distribution is similar to $\text{Cauchy}(0,1)\sim \dfrac{1}{1+x^2}$ and $\text{Cauchy}(0,1)$ is stable on average which means it's average is not convergent to zero in probability (the average of $n$ iid samples from $\text{Cauchy}(0,1)$ also has the $\text{Cauchy}(0,1)$ distribution).
It is known that if $(X_i)_i$ is an i.i.d. sequence and $S_n=\sum_{i=1}^n X_i$, a necessary and sufficient condition for the existence of a sequence $(a_n)$ for which $S_n/n-a_n\to 0$ in probability is that $n\mathbb P(\lvert X_1\rvert>n)\to 0$ as $n\to\infty$ (actually, when such a sequence exits, then $a_n=\mathbb E\left[X_1\mathbf{1}_{\{\lvert X_1\rvert\leqslant n\}}\right]$ is the natural choice).
Here, $\mathbb P(\lvert X_1\rvert>n)$ is of order $1/n$ hence we cannot have the weak law of large numbers.