Example of i.d. random variables s.t. $\frac{X_n}{n} \not\rightarrow 0 ~~~a.s.$

Question

I am looking for a counterexample to the following statement:

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of identically distributed random variables, then

$$ \frac{X_n}{n} \rightarrow 0 ~~~a.s.$$

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The statement is true when the $(X_n)_{n \in \mathbb{N}}$ are i.i.d. and all $X_n \in L^1$. Also, the statement is true for convergence in probability since

$$\forall \varepsilon > 0: \mathbb{P}\left(\left\vert \frac{X_n}{n} \right\vert > \varepsilon\right) = \mathbb{P}(\vert X_n \vert > \varepsilon n) \rightarrow 0$$

My first try was to modify the usual example of a sequence converging in probability but not almost surely (i.e., typewriter sequence). However, that forces the random variables to have different distributions.

Any tips or pointers would be greatly appreciated.

Answer 1

Consider an i.i.d. sequence $\{X_n\}$. Since $$ \sum_{n\ge 1}\mathsf{P}(|X_1|>n)\ge \int_{1}^{\infty} \mathsf{P}(|X_1|>x)\, dx \ge \mathsf{E}|X_1|-1, $$ and the events $\{|X_n|>n\}$, $n\ge 1$, are independent, the 2nd Borel-Cantelli lemma implies that $$ \mathsf{P}(|X_n|>n\text{ i.o.})=1 $$ whenever $\mathsf{E}|X_1|=\infty$.