Prove that $n^{-1} \max\left(X_1,\cdots,X_n\right) \to 0$ in probability 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.
 A: 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,
$$1-[1-P(|X_1|\geq n\delta)]^n\leq 1-[1-\frac{\alpha(n)}{n\delta}]^{\frac{n\delta}{\alpha(n)}\frac{\alpha(n)}{\delta}}\to 0.$$
A: Hint:
Each random variable has an identical cumulative distribution function $\phi(x)$. The probability that
$$\max(X_1, ..., X_n) > x$$
is equal to 1 minus the probability that all $X_i$ are less than $x$, which is $\phi(x)^n$.
That is, the cdf for the max is $1-\phi(x)^n$. Now realize that $0<\phi(x)<1$ and think about for a fixed $x$ what the limit is. Can you now use the cdf to show convergence to zero in probability?
A: Take some $\lambda > 0$. Then $P(\max(X_1,\dots,X_n) > n\lambda) = P(\cup_j \{X_j > \lambda n\}) = 1-P(\cap_j \{X_j \le \lambda n\})$
$= 1-(P(X_1 \le \lambda n))^n = 1-(1-P(X_1 > \lambda n))^n$. So we just need to show that $(1-P(X_1 > \lambda n))^n \to 1$. But the LHS is $e^{n\log(1-P(X_1 > \lambda n))}$, so it suffices to show $n\log(1-P(X_1 > \lambda n)) \to 0$. Note $-n\log(1-P(X_1 > \lambda n)) = n\log(\frac{1}{1-P(X_1 > \lambda n)}) \asymp nP(X_1 > \lambda n)$. But since $X_1 \in L^1$, $nP(X_1 > \lambda n) \to 0$, as desired.
