For non-negative iid random variables, show that the max converges ip From Resnick's A Probability Path, exercise 6.7.11.  (Studying for my comprehensive exam.)

Suppose $\{X_n, n \ge1\}$ are iid and non-negative and define $M_n=\bigvee^n_{i=1}X_i$.
a) Check that $P[M_n > x] \le nP[X_1 > x].$
b) Show that if $E(X^p_1) < \infty$, then $\frac{M_n}{n^{1/p}} \stackrel{p}{\to} 0.$

I did part a:
\begin{equation}
[M_n > x] \subseteq \bigcup^n_{i=1}[X_i>x] \\
P[M_n > x] \le P\left(\bigcup^n_{i=1}[X_i>x]\right) \le \sum^n_{i=1}P[X_i>x] = nP[X_1>x]
\end{equation}
Part b is giving me trouble--I must be missing something!  I assumed that part (a) was meant to be used in part b:
\begin{align}
P\left(\left|\frac{M_n}{n^{1/p}}-0\right|>\epsilon\right)&=P[M_n > n^{1/p}\epsilon]\\
&\le nP[X_1>n^{1/p}\epsilon]\\
&=nP[X_1^p > n\epsilon^p]\\
&\le \frac{E(X_1^p)}{\epsilon^p} \nrightarrow0.
\end{align}
Now, I recognize that although Markov's inequality does not push the probability to zero doesn't mean that the probability actually fails to converge to zero, but it seems as if the author is implying I should use the inequality from part (a), which naturally seems to lead to Markov.  What am I missing here?

Use this instead in the very last inequality.
$$ \frac{E(X_1^pI_{X_1 > n^{1/p}\epsilon})}{\epsilon^p} .$$
You should have a theorem somewhere that if $Y \ge 0$ and $EY < \infty$, then $E(YI_{Y>\alpha}) \to 0$ as $\alpha \to \infty$.
