Applying the Law of Large Numbers? $X_k$, $k \geq 1$ are iid random variables such that
$$\limsup_{n\rightarrow\infty} \frac{X_n}{n} < \infty$$
with probability 1. We want to show that
$$\limsup_{n\rightarrow\infty} \frac{\sum_{i=1}^n X_i}{n} < \infty$$
with probability 1.
The hint says to apply the law of large numbers to the sequence $\max(X_k,0), k \geq 1$. SLLN gives that
$$\frac{\sum_{i=1}^n \max(X_i,0)}{n} \rightarrow \mathbb{E}\max(X,0) = \mathbb{E}(X; X>0)$$
almost surely. I feel that the idea here is that $\limsup X_n/n < \infty$ a.s. implies that $\mathbb{E}(X;X>0)$, but I am not really sure how to approach this...
 A: Starting where you stopped, assume that $E(X^+)$ is infinite, then $\sum\limits_{n=1}^\infty P(X_n\geqslant xn)=\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges for every $x\gt0$ hence Borel-Canteli lemma implies that $X_n\geqslant xn$ for infinitely many $n$, almost surely, thus $\limsup\limits_{n\to\infty}\frac{X_n}n\geqslant x$, almost surely. This holds for every $x\gt0$ hence you are done.
To show that the series $\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges, note that for every $x\gt0$ and every random variable $X$, one has $x\sum\limits_{n=1}^\infty \mathbf 1_{X\geqslant xn}\geqslant X^+$ almost surely hence $x\sum\limits_{n=1}^\infty P(X\geqslant xn)\geqslant E(X^+)$.
A: Consider $X_k^+ := \max(X_k,0)$. Then,
\begin{align*}
P\left(\limsup \frac{X_n}{n} < \infty\right)=1 &\Rightarrow P\left(\limsup \frac{X_n^+}{n} < \infty\right)=1 \\
&\Rightarrow \exists A: P\left(\frac{X_n^+}{n} > A \text{ i.o.}\right)=0 \text{ a.s.}\\
&\Rightarrow \sum_{i=1}^n P\left(\frac{X_i^+}{i} > A\right) < \infty \text{ a.s.} \\
&\Rightarrow \sum_{i=1}^n P\left( X^+ > iA \right) < \infty \text{ a.s.} \\
&\Rightarrow \mathbb{E}X^+ < \infty \text{ a.s.}
\end{align*}
By the Strong Law of Large Numbers,
$$\frac{\sum_{i=1}^n X_i^+}{n} \rightarrow \mathbb{E}X^+ < \infty \text{ a.s.},$$
and so,
$$\limsup \frac{\sum_{i=1}^n X_i}{n} \leq \frac{\sum_{i=1}^n X_i^+}{n} < \infty \text{ a.s.}$$
Lemma. $\limsup_n X_n < \infty$ a.s. if and only if $\sum P(X_n > A) < \infty$ for some $A$.
Proof of Lemma. We write $Y = \limsup_n X_n$ for notational simplicity. Since $X_n$ are independent, Borel-Cantelli Lemmas show that
\begin{align*}
\sum_{n=1}^\infty P(X_n > A) < \infty &\iff P(X_n > A \text{ i.o.}) = 0 \\
\sum_{n=1}^\infty P(X_n > A) = \infty &\iff P(X_n > A \text{ i.o.}) = 1.
\end{align*}
To relate this with the finiteness of $Y$, note that
$\bullet$ If $a_n > A$ i.o., then $\limsup_n a_n \geq A$.
$\bullet$ If $\limsup_n a_n \geq A$, then for any $\epsilon > 0$, we have $a_n \geq A-\epsilon$ i.o.
This yields the inequality
$$P(Y \geq A+\epsilon) \leq P(X_n > A \text{ i.o.}) \leq P(Y \geq A)$$
This shows that
$\bullet$ If $P(Y < \infty) = 0$, then $P(Y \geq A) < 1$ for some constant $A$. Then, $P(X_n > A \text{ i.o.}) < 1$ and hence, $P(X_n > A \text{ i.o.}) = 0$.
$\bullet$ If $P(X_n > A \text{ i.o.}) = 0$, then $P(Y \geq A + \epsilon) = 0$, and thus, $P(Y= \infty) = 0$ as well.
These combine to show that $limsup_n X_n < \infty $ a.s. $\iff \sum_n P(X_n > A) < \infty$ for some $A$.
