Let $(X_n)_{n\in\mathbb{N}}$ be independent and \begin{equation} P(X_n=n)=P(X_n=-n)=\frac{1}{2}n^{-{\frac{5}{2}}}=\frac{1}{2}(1-P(X_n=0)), \ n\in\mathbb{N}. \end{equation} Show that $(X_n)_{n\in\mathbb{N}}$ satisfies the Strong Law of Large Numbers. My first guess was to use a Borel-Cantelli argument but I cant figure out how.

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    $\begingroup$ what exactly are you not able to figure out? Since $\sum_{n \geq 1} \mathbb{P}(|X_n|>0) < \infty$ it follows from the Borel Cantelli lemma that...... $\endgroup$ – saz Mar 2 at 15:29
  • $\begingroup$ then we know that $P(|X_n|>0 \ \text{infinitely often})=0$ so $P(X_n=0 \ \text{eventually})=1$. This means there is a $N\in \mathbb{N}$ such that for all $n\geq N$ we have $X_n=0$. So now one can split up $\frac{1}{n} \sum_{k=1}^n X_k = \frac{1}{n} \sum_{k=1}^{N-1} X_k + \frac{1}{n} \sum_{k=N}^{n} X_k $. And the first terms tends towards $0$ and in the second term $X_k=0 \ \forall k\geq N$ so we get the claim. Is this argumentation right? $\endgroup$ – Katakuri Mar 3 at 9:55
  • $\begingroup$ Yes, it is. You should keep in mind that $N$ depends, in general, on $\omega$. $\endgroup$ – saz Mar 3 at 15:38

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