TITLE: Prove that $n^{-1} \max\left(X_1,\cdots,X_n\right) \to 0$ in probability
QUESTION [5 upvotes]: Let $X_1,\cdots,X_n$ be sequence of positive, iid random variables such that $\mathbb{E} X_1 <\infty$. How can I show that

$$\frac{1}{n}\max\left(X_1,X_2,\cdots,X_n\right)\to 0 \text{ in probability}$$

I can prove it assuming $\mathbb{E}X_1^2 < \infty$, but I don't know how to prove it with the given assumption.
Note: While almost surely convergence implies convergence in probability, I am looking for a solution that does not take that route, and does not use Borel-Cantelli lemma.

REPLY [5 votes]: First,
\begin{eqnarray}
P\{|\max|\geq n\delta\}=1-[1-P(|X_1|\geq n\delta)]^n.
\end{eqnarray}
Next we estimate $P(|X_1|\geq n\delta)$. Since $E|X_1|<+\infty$, 
$$E\{|X_1|;|X_1|\geq n\delta\}:=\alpha(n)\to0,\text{as}\ n\to\infty.$$
So we have
$$P(|X_1|\geq n\delta)\leq\frac{\alpha(n)}{n\delta}.$$
Thus,
$$1-[1-P(|X_1|\geq n\delta)]^n\leq 1-[1-\frac{\alpha(n)}{n\delta}]^{\frac{n\delta}{\alpha(n)}\frac{\alpha(n)}{\delta}}\to 0.$$