# Prove that ${S_n/n}$ does not converge a.s.

This is an old qualifying exam question of probability theory.

Let $$\{X_n\}$$ be a sequence of independent random variables with $$X_1=0$$ and for $$k\geq 2$$, define $$X_k$$ as $$\mathbf{P}(X_k=k)=\mathbf{P}(X_k=-k)=\frac{1}{2k\log k},\mathbf{P}(X_k=0)=1-\frac{1}{k\log k}.$$

Let $$S_n=\sum_{k=1}^nX_n$$. Show that $$\dfrac{S_n}{n}$$ converges to 0 in probability but not almost surely.

My attempt: For given $$\epsilon>0$$, \begin{align*}\mathbf{P}(|S_n|\geq n\epsilon) & \leq \frac{\text{Var}(S_n)}{n^2\epsilon^2}=\frac{1}{n^2\epsilon^2}\sum_{k=1}^n\text{Var}(X_k)\end{align*}\\ =\frac{1}{n^2\epsilon^2}\sum_{k=2}^n \frac{k}{\log k}\leq \frac{1}{n^2\epsilon^2}(n-1)\frac{n}{\log n}\to0 as $$n\to\infty$$. This proves that $$\dfrac{S_n}{n}\to 0$$ in probability.

However, I'm stuck with proving that $$\dfrac{S_n}{n}$$ does not converge to 0 a.s. I'd like to apply Borel-Cantelli lemma here, but each $$S_n$$ is not independent, so the fact that $$\sum_{k=1}^\infty \frac{1}{k\log k}=\infty$$ cannot tell anything about convergence. Does anyone have ideas?

• The second Borel-Cantelli lemma is sufficient to prove that : en.wikipedia.org/wiki/… You then need to use the fact that $\sum \frac{1}{n\log(n)}$ diverges Oct 1, 2018 at 10:22
• @P.Quinton The second Borel-Cantelli lemma works only when each of the event $(|S_n/n|\geq \epsilon$ is independent. This is not the case. Oct 1, 2018 at 10:25
If $$\frac {S_n} n$$ converges a.s. it can only converge to $$0$$ a.s.. In this case $$\frac {X_n} n =\frac {S_n-S_{n-1}} n \to 0$$ a.s. But $$P\{X_n =n\ i.o. \}=1$$ because $$\sum P\{X_n =n \} =\sum \frac 1 {n log \, n} =\infty$$ and the events here are independent.