Show $\max_{i\leq n} |X_i|^p/n$ converges to $0$ in probability $X_i$'s are i.i.d random variables and  $\mathbb E(|X_i|^p)=k<\infty$ for some $p,k$. I want to show $Z= \max_{i\leq n} |X_i|^p/n$ converges to $0$ in probability.
Here is what I have tried.
$$\mathbb P(Z\geq \epsilon)=1-P(Z< \epsilon)=1-\prod_{i\leq n}\mathbb P(|X_i|^p/n<\epsilon),$$by indepdence. That is $$1-\prod_{i\leq n}(1-\mathbb P(|X_i|^p/n\geq\epsilon))=1-(1-P(|X_1|^p/n\geq\epsilon))^n.$$ Now apply Chebyshev we have $$\mathbb P(Z\geq \epsilon)\leq1-(1-\mathbb E(|X_1|^p/n\epsilon))^n=1-(1-k/n\epsilon))^n$$ which does not converges to 0 as $n \to\infty$. I don't know If I can find any stronger inequality to make this proof work . Any suggestions will be appreicated.

Let $Y_n=\max_{1\le i\le n}|X_i|^p$. Then $Y_n/n\xrightarrow{p}0$ iff $n\mathsf{P}(|X_1|^p>n\epsilon)\to 0$ as $n\to \infty$ for all $\epsilon>0$. Since $\mathsf{E}|X_1|^p<\infty$,
$$
n\mathsf{P}(|X_1|^p>n\epsilon)\le \epsilon^{-1}\mathsf{E}[|X_1|^p 1\{|X_1|^p>n\epsilon\}]\to 0
$$
as $n\to\infty$ by the dominated convergence theorem.
