Given Xr be independent, non-negtive and iid with infinite mean. Show that limsup $\frac{X_r}{r} = \infty $
Now the solution assess that: $$ \sum_{r=1} (\frac{X_r}{r} \geq x) = \sum_{r=1} (\frac{X_r}{x} \geq r) = \frac{E(X_r)}{x} = \infty $$
Then by Borel Cantelli, the proof is completed.
I'm wondering what's the procedure that has been applied. I was wondering about Markov's inequality but I suppose that since the mean is infinite, this cannot be applied. So what else has been made?
The following inequality applies for any nonnegative random variable $X_1$ and any real number $x > 0$:
$$\tag{*}\sum_{r=1}^\infty \mathbb{P}(X_1 > rx) < \frac{1}{x} \mathbb{E}(X_1) \leqslant 1 + \sum_{r=1}^\infty \mathbb{P}(X_1 > rx) $$
Given non-negative iid random variables $X_r$ where $\mathbb{E}(X_r)= \mathbb{E}(X_1) = +\infty$, it then follows from the second Borel-Cantelli lemma that $\mathbb{P}([X_r/r > x] \,\,\, \text{i.o.})= 1$ for any $x > 0$ and $\limsup_{r \to \infty} \frac{X_r}{r} = +\infty $ almost surely.
To prove (*), note that
$$\sum_{r=1}^\infty \mathbb{P}(X_1 > rx) = \sum_{r=1}^\infty \mathbb{E}(\mathbb{1}_{\{X_1 > rx\}})= \mathbb{E} \left(\sum_{r=1}^\infty \mathbb{1}_{\{X_1 > rx\}} \right),$$
where the expectation operator and the summation can be switched by the monotone convergence theorem, and
$$\sum_{r=1}^\infty \mathbb{1}_{\{X_1 > rx\}} = \begin{cases}0, &X_1(\omega) = 0\\j, &X_1(\omega) \in (jx,(j+1)x] \end{cases}, $$
which implies that $\sum_{r=1}^\infty \mathbb{1}_{\{X_1 > rx\}} < \frac{1}{x} X_1 \leqslant 1 + \sum_{r=1}^\infty \mathbb{1}_{\{X_1 > rx\}}$ . Taking expected values yields (*).