Borel-Cantelli lemma: non-negative iid random variables I came across a claim in a paper on branching processes which says that the following is an immediate consequence of the B-C lemmas:

Let $X, X_1, X_2, \ldots$ be nonnegative iid random variables. Then $\limsup_{n \to \infty} X_n/n = 0$ if $EX<\infty$, and $\limsup_{n \to \infty} X_n/n = \infty$ if $EX=\infty$.

So to apply the BC lemmas to these, I want to essentially show that
$$(1) \; \textrm{If } EX<\infty, \textrm{ then } P(\limsup \{X_n/n > \epsilon\}) = 0 \quad \forall \epsilon>0$$
$$(2) \; \textrm{If } EX=\infty, \textrm{ then } P(\limsup \{X_n/n > \delta\}) = 1 \quad \forall \delta>0$$
But I keep getting stuck. For example if I want to apply the first BC lemma to (1), then using Markov's inequality only gives $P(X_n > n\epsilon) < EX/n\epsilon$, which isn't summable. Am I missing something right under my nose?
 A: (a): Say $\epsilon>0$. Let $E_n$ be the event $X_n/n > \epsilon$. Then $$\sum P(E_n)=\sum P(X_1>n\epsilon)<\infty$$because $EX_1<\infty$ (standard lemma). So BC says $$P(\limsup X_n/n>\epsilon)=P(E_n i.o.)=0.$$
(b): Same as (a).
Maybe what you're missing is that "standard lemma" referred to above?
Lemma. If $X\ge0$ then $\sum_{n=1}^\infty P(X>n)\le EX$.
Proof: (Notation inconsistent with the notation above.) Let $X_n=\min(X,n)$. Then $X_n$ increases to $X$ and $X_0=0$, so $EX=\sum_{n=0}^\infty E(X_{n+1}-X_n)$. But $X_{n+1}-X_n\ge0$ and $X_{n+1}-X_n=1$ on $E_n=\{X>n\}$, so $E(X_{n+1}-X_n)\ge P(E_n)$. QED.
Of course you need a slightly different version of the lemma for part (b).
A: Essentially what you need to show is that

If $X$ is a non-negative random variable, then $EX < \infty$ iff $\sum_n^{\infty} P(X > n\epsilon) < \infty$ for every $\epsilon > 0$.

The direct part is sufficient for part (a) obviously, but you will also want to establish the converse direction for part (b) because you will need to apply the 2nd BC lemma and show that, if $EX = \infty$ then $\sum P(X_n > n\epsilon) = \infty$ no matter how large of a $\epsilon>0$ you chose.
For the first direction $(\Rightarrow)$, note that for any $\epsilon>0$,
$$\sum_{n=1}^{\infty} \epsilon \cdot \mathbb{1}_{(X>n\epsilon)} < X$$
If you view the LHS as the limit of a sequence of partial sums then the MCT gives
$$\epsilon \sum_{n=1}^{\infty} P(X>n\epsilon) < EX$$
For the opposite direction $(\Leftarrow)$ just note that
$$\sum_{n=1}^{\infty} \mathbb{1}_{(X>n-1)} > X$$
Take expectations and limits as above and you will get
$$\sum_{n=1}^{\infty} P(X>n-1) > EX$$
And apply the assumption with $\epsilon=1$.
