I'm trying to solve the following exercise from Resnick's book
Suppose ${X_n, n > 1}$ are identically distributed with finite variance. Show that $\frac { { \bigvee }_{ i=1 }^{ n }|{ X }_{ i }| }{ \sqrt { n } } \rightarrow 0$ in probability.
I already showed in this same exercise the following result
$nP[|{X_1} \ge \varepsilon \sqrt n ] \to 0$
If the maximum of the $|X_i|$'s is greater than $c$, then at least one of the $|X_i|$'s is greater than $c$, so:
$$ \Big\{\max_{i=1,\ldots,n} |X_i| \geq c \Big\} \; \subset \; \bigcup_{i=1}^n \Big\{ |X_i| \geq c \Big\} $$
Then using the union bound and setting $c = \epsilon \sqrt{n}$ for some $\epsilon>0$: $$ P(\max_{i=1,\ldots,n} |X_i| \geq \epsilon \sqrt{n}) \; \leq \; \sum_{i=1}^n P(|X_i| \geq \epsilon\sqrt{n}) = n P(|X_1| \geq \epsilon \sqrt{n})$$ which goes to 0 as $n \to \infty$, as you have already shown.