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.
If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(\log n)^2.$ Then $P(|\bar X_n|>\epsilon)\leq Var(\bar X_n)/\epsilon^2=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}Var(X_i)=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}(\log i)^2\leq \frac{(\log n)^2}{n\epsilon^2}\rightarrow 0.$