We have independent random variables such that
$$\mathbb{P}(X_n=n)=\mathbb{P}(X_n=-n)=\frac{1}{2(n+1)\ln(n+1)}$$ and
$$\mathbb{P}(X_n=0)=1-\frac{1}{(n+1)\ln(n+1)}$$
I am trying to show that $\frac{S_n}{n}$ does not converge to $0$ almost surely. I'm thinking about Borel-Cantelli Lemmas, so I'd like to show that $\sum_{n=0}^{\infty} \mathbb{P}(S_n=0)\lt\infty$.
I have tried coming up with a recurrence relation for $\mathbb{P}(S_n=0)$. I have also tried finding out whether $X_n$ could take the value $0$/$n$/$-n$ for infinitely many $n$'s. But these two did not help much.Do you see a hint you could give me?
Best Answer
We have $$\sum_{n=1}^\infty \mathbb P(X_n\geqslant n) = \sum_{n=1}^\infty\frac1{2(n+1)\log (n+1)}=\infty $$ which implies that $$\mathbb P\left(\limsup_{n\to\infty} X_n \geqslant n\right)=1 $$ by the second Borel-Cantelli lemma. Now consider the difference in consecutive partial sums, as @Kika: $$\frac{S_{n+1}}{n+1}-\frac{S_n}n = \frac{X_{n+1}}{n+1} - \frac1{n(n+1)}\sum_{k=1}^n X_k.$$ Since $$\sum_{k=1}^n X_k \leqslant \sum_{k=1}^n k = \frac{n(n+1)}2, $$ it follows that $$\frac{S_{n+1}}{n+1} - \frac{S_n}n \geqslant 1 - \frac{n(n+1)}{2n(n+1)} = \frac12, $$ for infinitely many $n$, and hence $$\mathbb P\left(\limsup_{n\to\infty} \frac{S_n}n\geqslant\frac12\right) =1. $$ We conclude that $\frac{S_n}n$ does not converge to $0$ a.s.