Let $X_1, X_2, \ldots$ be i.i.d. nonnegative random variables. I want to use the Borel–Cantelli lemma, to show that \begin{equation*} \limsup_{n\rightarrow\infty} \frac{1}{n} X_n = \begin{cases} 0 \text{ a.s.}, & \text{ if } \mathbb{E}[X_1] < \infty,\\ \infty \text{ a.s.}, & \text{ if } \mathbb{E}[X_1] = \infty. \end{cases} \end{equation*}
The second case actually works quite nicely when you use that $\mathbb E[X] \le \sum_{n=0}^{\infty} \mathbb P[X \ge n]$. But I don't know how the first case works.
On a different thread I saw this estimation:
$ \sum_{n=1}^\infty \mathbb{P}\bigl[X_n/n \geq \varepsilon \bigr] = \frac{1}{\varepsilon} \sum_{n=1}^\infty \mathbb{P}\bigl[X_1 \geq n \varepsilon\bigr] \cdot \varepsilon \leq \frac{1}{\varepsilon}\cdot \underbrace{\mathbb{E}[X_1]}_{<\infty} < \infty$
but I don't see why I can get a lower bound for my expected value here. Additionally it also resembles the estimation I used for the second case, but the other way around.
$\sum_{n=1}^{\infty} P\{\frac {X_n} n \geq \epsilon\} =\sum_{n=1}^{\infty} P\{X_1 \geq n\epsilon\}=\sum_{n=1}^{\infty} P\{\frac {X_1} {\epsilon} \geq n\}\leq E\frac {X_1} {\epsilon}$ since $EY\geq \sum_{n=1}^{\infty} P\{Y\geq n\}$ for any non-negative random variable $Y$. Hence $\sum_{n=1}^{\infty} P\{\frac {X_n} n \geq \epsilon\} <\infty$ which, by Borel - Cantelli Lemma, shows that $\frac {X_n} n <\epsilon$ for all $n$ suficiently large, with probabilty 1. Letting $\epsilon \to 0$ through a sequence we get $\limsup \frac {X_n} n =0$ almost surely.
Proof of $EY\geq \sum_{n=1}^{\infty} P\{Y\geq n\}$: $EY=\sum EYI_{n\leq Y <n+1}\geq \sum nP\{n\leq Y <n+1\}$. Now $\sum_{n=1}^{\infty} P\{Y\geq n\}=\sum_n \sum_{j=n}^{\infty} P\{j\leq Y <j+1\}=\sum_j \sum_{n=1}^{j} P\{j\leq Y <j+1\}=\sum_j j P\{j\leq Y <j+1\}\leq EY.$