My intended answer to a weak LLN problem on Math.SE.

Problem: Suppose $(X_n)$ is a sequence of r.v’s satisfying $P(X_n=\pm\ln (n))=\frac{1}{2}$ for each $n=1,2\dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.The idea is to show that $P(\overline{X_n}>\varepsilon)$ tends to 0, but I am unsure how to do so.

My solution: As in the accepted answer in OP’s previous question https://math.stackexchange.com/q/3021650/290189, I’ll assume the independence of $(X_n)$. By Chebylshev’s inequality,$$P(|\bar{X}_n|>\epsilon) = P(|\sum_{k=1}^n X_k|> n\epsilon) \le \frac{\sum_{k=1}^n var(X_k)}{n^2 \epsilon^2} = \frac{\sum_{k=1}^n (\ln(k))^2}{n^2 \epsilon^2} \le \frac{(\ln(n))^2}{n \epsilon} \to 0$$

Recall: $(\ln(n))^p/n \to 0$ whenever $p \ge 1$. To see this, make a change of variables $n = e^x$, so that it becomes $x^p / e^x$.