I can't seem to write a solid proof for the following problem:
Consider an independent sequence $\left(X_n\right)_{n \in \mathbb{N}}$ of random variables with $$ P\left(X_n=0\right)=1-\frac{1}{n \log (n+1)}, \quad P\left(X_n=n\right)=P\left(X_n=-n\right)=\frac{1}{2 n \log (n+1)} . $$
Show that $\frac{1}{n}\left(X_1+\cdots+X_n\right)$ does not converge almost surely. Hint: Use the Borel-Cantelli Lemma.
So, from Borel-Cantelli I simply state that
$\sum_{n \in \mathbb{N}} P\left(A_n\right)= \sum_{n \in \mathbb{N}} P(\frac{1}{2 n \log (n+1)}) < \infty$, then $$ P\left(\limsup _{n \rightarrow \infty} A_n\right) \neq 1 $$ and thus the seuqnece doesn ot converge a.s. I know my solution is not rigorous enough (and could be worng), so I would love to see sugestions or solutions to the problem. Thank you !
Let $E_n = \{|X_n|/n = 1\}$, then \begin{align} \sum_{n=1}^\infty \mathbb P(E_n) &= \sum_{n=1}^\infty \mathbb P(|X_n|/n=1)\\ &=\sum_{n=1}^\infty (\mathbb P(X_n=n)+\mathbb P(X_n=-n))\\ &= \sum_{n=1}^\infty \frac1{n\log(n+1)}\\ &= +\infty, \end{align} so by independence of the $E_n$ the second Borel-Cantelli lemma implies that $\mathbb P(\limsup_{n\to\infty} E_n)=1$.
Let $S_n=\frac1n\sum_{i=1}^n X_i$, then it follows that $$ \mathbb P\left(\limsup_{n\to\infty} |S_n|>\frac12 \right) \geqslant \mathbb P\left(\limsup_{n\to\infty} |X_n|/n>\frac12 \right)\geqslant \mathbb P(\limsup_{n\to\infty} E_n)=1, $$ and hence $S_n$ does not converge almost surely (to zero).