Let $W_0, W_1, W_2, \dots$ be random variables on a probability space $(\Omega, \mathscr{F}, \mathbb{P})$ where
$$\sum_{k=0}^{\infty}P(|W_k|>k) <\infty$$
Prove that $$\limsup \frac{|W_k|}{k} \le 1 \ \text{a.s.} $$
I initially thought the conclusion meant $(**)$ when it really means $(*)$:
$$P(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) = 1 \ \text{(*)}$$
$$P(\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}) = 1 \ \text{(**)}$$
What I tried:
By the first Borel-Cantelli Lemma, we have $P(\limsup (|W_k| > k)) = 0$
$\to P(\limsup (|W_k|/k > 1)) = 0$
$\to P(\liminf (|W_k|/k > 1)) = 0$
$\to P([\liminf (|W_k|/k > 1)]^C) = 1$
$\to P(\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}) = 1$
$\to P(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) = 1$ QED assuming $(**)$ implies $(*)$.
Is it really the case that $(**)$ implies $(*)$? Here is why I think such:
$\forall \omega \in \Omega, \omega \in (\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)})$
Then $\forall m \ge 1, \exists n \ge m$ s.t.
$\omega \in (\frac{|W_n(\omega)|}{n} \le 1)$
$\to \omega \in (\limsup \frac{|W_n(\omega)|}{n} \le \limsup 1 = 1)$
Now if $(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$, then it follows by monotonicity that
$(**)$ implies $(*)$ QED.
However $(*)$ does not imply $(**)$ because it does not hold that $(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \subseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$. Counterexample given in comments below. I guess the intuitive explanation is that
$\limsup \frac{|W_n(\omega)|}{n} \le \limsup 1$
--/--> $\frac{|W_n(\omega)|}{n} \le 1$
for similar reasons that
$x_n \le y_n \to \lim x_n \le \lim y_n$, if those limits exist (1)
but
$\lim x_n \le \lim y_n$ --/--> $x_n \le y_n$ (2), which makes sense:
(1) has an infinite number of statements which imply 3 statements (limsup, liminf, limsup=liminf)
(2) has 3 statements that claims to imply an infinite number of statements
Is that right?
Edit based on what Daniel Fischer said:
To prove $$\liminf \{\frac{W_k(\omega)}{k} \le 1\} \subseteq \{\limsup \frac{W_k(\omega)}{k} \le 1\}:$$
Suppose $$\omega \in \liminf \{\frac{W_k(\omega)}{k} \le 1\}$$
Then $\exists m \ge 1$ s.t. $\forall k \ge m$,
$$\omega \in \{\frac{|W_k(\omega)|}{k} \le 1\}$$
Since $\frac{|W_k(\omega)|}{k} \le 1 \to \limsup \frac{|W_k(\omega)|}{k} \le \limsup 1 = 1$, we have
$$\to \omega \in \{\limsup \frac{|W_k(\omega)|}{k} \le 1\}$$
Let's try to get things straight. We have two conditions,
\begin{align} P\left(\left\{\omega \in \Omega : \limsup_{k\to\infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}\right) &= 1,\text{ and} \tag{$\ast$}\\ P\left(\limsup_{k\to\infty} \left\{\omega\in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\}\right) &= 1. \tag{$\ast\ast$} \end{align}
The question is whether one condition implies the other. An inclusion between the sets $A := \left\{\omega \in \Omega : \limsup_{k\to\infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}$ and $B := \limsup_{k\to\infty} \left\{\omega\in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\}$ would give us such an implication.
However, in general we have neither $A \subset B$ nor $B \subset A$. We can use the verbal description of the limes superior of sets to describe $B$ as
$$B = \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \text{ for infinitely many } k \right\}.$$
This implies $\liminf\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1$ for all $\omega \in B$, but it doesn't say anything about $\limsup\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k}$. We could for example have $W_k \equiv 0$ for odd $k$, and $W_k \equiv 5k$ for even $k$. Then we have $B = \Omega$ and $A = \varnothing$, since for this example $\limsup\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k} = 5$ for all $\omega$.
On the other hand, we could have $W_k \equiv k(1+2^{-k})$ for all $k$, then we have $A = \Omega$ and $B = \varnothing$, since $\frac{\lvert w_k(\omega)\rvert}{k} \to 1$ uniformly on $\Omega$, but $\frac{\lvert W_k(\omega)\rvert}{k} = 1 + 2^{-k} > 1$ for all $k$ and $\omega$.
Thus, we have no implication whatsoever between the conditions $(\ast)$ and $(\ast\ast)$.
But the Borel-Cantelli lemma yields something stronger than $(\ast\ast)$. Namely, from
$$\sum_k P(\{\omega\in \Omega : \lvert W_k(\omega) \rvert > k\}) < +\infty$$
we conclude that the set of $\omega$ such that $\lvert W_k(\omega)\rvert > k$ for infinitely many $k$ is a null set. In other words, for almost all $\omega$, we have $\lvert W_k(\omega)\rvert \leqslant k$ for all but finitely many $k$. In other words, from the Borel-Cantelli lemma we obtain
$$P\left(\liminf_{k \to \infty}\left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}\right) = 1.\tag{$\ast{\ast}\ast$}$$
And now, we note that
$$C = \liminf_{k\to \infty} \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\} = \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \text{ for all but finitely many } k\right\}$$
is clearly a subset of $A$. Hence we have the implication $(\ast{\ast}\ast) \implies (\ast)$. Also it's clear that $C \subset B$, whence we have the implication $(\ast{\ast}\ast) \implies (\ast\ast)$.