Exercise about Convergence in Probability I am preparing myself for an exam in probabilities and I have some trouble answering the following question. Can someone help me out please?
Let $X_1,X_2,...$ be independent, identically distributed random variables.
Prove that $\frac{MAX\{|X_1|,...,|X_n|\}}{\sqrt{n}}\rightarrow 0$ in probability as $n\rightarrow \infty$ if and only if $\forall c>0:n\mathbb{P}(|X_1|>c\sqrt{n})\rightarrow 0$, as $n\rightarrow \infty$.
I have approached it in the following way: 
$(\Rightarrow)$
Let $|X_i|:=MAX\{|X_1|,...,|X_n|\}$. Since $\frac{|X_i|}{\sqrt{n}}$ converges to $0$ in probability, we get that: $\mathbb{P}(|\frac{|X_i|}{\sqrt{n}}-0|>\epsilon)\rightarrow 0,\forall \epsilon>0$.
Therefore, we get that: $\mathbb{P}(|X_i|>\epsilon\sqrt{n})\rightarrow 0$.
Now since $|X_i|:=MAX\{|X_1|,...,|X_n|\}$, we get that: $|X_1|\leq |X_i|\forall \omega\in\Omega$ and therefore: $\mathbb{P}(|X_1|>\epsilon\sqrt{n})\leq \mathbb{P}(|X_i|>\epsilon\sqrt{n})\rightarrow 0, \forall \epsilon>0$, getting the one-way result.
$(\Leftarrow)$
Now for the other direction, if $|X_1|=MAX\{|X_1|,...,|X_n|\}$, then we can get the desired result using an analogous argument. However I am out of ideas for the case that $|X_1|\neq MAX\{|X_1|,...,|X_n|\}$.
 A: In what follows, let
$$
\rho_{i,n}:=P\left(\left|X_i\right|{}>{}\epsilon\sqrt{n}\right)\,.
$$
We need to show,
$$
\begin{eqnarray*}
&\lim\limits_{n\to\infty}P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right){}={}0\mbox{, for all }\epsilon>0& \newline
&& \newline
&\iff&\newline
&&\newline
&\lim\limits_{n\to\infty}n\rho_{1,n}{}={}0\mbox{, for all }\epsilon>0\,.& 
\end{eqnarray*}
$$
Note: For each $n$ and $\epsilon{}>{}0$, observe that for $\omega{}\in{}\left\{\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right\}$ it is necessary, and sufficient, that there exist some $i$ such that $X_i(\omega){}>{}\epsilon\sqrt{n}$ . In other words, $\omega\in\bigcup\limits_{1\leq i\leq n}\left\{X_i{}>{}\epsilon\sqrt{n}\right\}$ equivalently. 
So, choose $n$ and $\epsilon{}>{}0$. $\color{brown}{\textit{Let's assume that }\lim\limits_{k\to\infty}k\rho_{1,k}{}={}0\textit{, for all }\epsilon>0}$. Then, using the fact that the $X_i$ are identically distributed,
$$
\begin{eqnarray*}
P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right)&{}={}&P\left(\bigcup\limits_{1\leq i \leq n}\left\{\left|X_i\right|{}>{}\epsilon\sqrt{n}\right\}\right) \newline
&{}\leq{}&\sum\limits_{i{}={}1}^{n} \rho_{i,n}\newline
&{}={}&n\rho_{1,n}\,.
\end{eqnarray*}
$$
Hence, $0\leq{}\lim\limits_{n\to\infty}P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right){}\leq{}\lim\limits_{n\to\infty}n\rho_{1,n}{}={}0$ .
Conversely, choose $n$ and $\epsilon>0$. $\color{brown}{\textit{Assume } \lim\limits_{n\to\infty}P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right){}={}0,\textit{ for all }\epsilon>0}$. In passing, note that this means 
$$
\lim\limits_{n\to\infty}\rho_{1,n}{}\leq{}\lim\limits_{n\to\infty}P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right)=0.
$$
By using the "even" maximum power Bonferri Inequalities (make sure you convince yourself of this) and the i.i.d $X_i$'s, we have
$$
\begin{eqnarray*}
n\rho_{1,n}&{}={}&\sum\limits_{1\leq i\leq n}\rho_{i,n}\newline
&{}\leq{}&P\left(\bigcup\limits_{1\leq i \leq n}\left\{\left|X_i\right|{}>{}\epsilon\sqrt{n}\right\}\right){}+{}\sum\limits_{i<j}\rho_{i,n}\rho_{j,n}\newline
&{}={}&P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right){}+{}\dfrac{n(n-1)}{2}\rho_{1,n}^2\,,
\end{eqnarray*}
$$ 
which means
$$
\begin{eqnarray*}
0{}\leq{}n\rho_{1,n} {}-{} \dfrac{1}{2}\bigg(n\rho_{1,n}\bigg)^2{}+{}\dfrac{1}{2}\bigg(n\rho_{1,n}\bigg)\rho_{1,n}{}\leq{}P\left(\max\limits_{1\leq i \leq n}\left\{\left|X_i\right|\right\}{}>{}\epsilon\sqrt{n}\right)\,.
\end{eqnarray*}
$$
Therefore, upon taking the limit, as $n\to\infty$, of the terms involved in these inequalities, we deduce that
$$
\lim\limits_{n\to\infty}n\rho_{1,n}{}={}0\mbox{ or }2\,.
$$
But, by applying ever higher even power Bonferri inequalities in a similar manner, we can convince ourselves that the only value of the limit that holds for all $n$ is $0$. Since $\epsilon > 0$ was arbitrary in the foregoing, we are done.
A: Hint: show the inequality 
$$nt-\frac{n(n+1)}2t^2\leqslant 1-(1-t)^n\leqslant nt,$$
valid for any positive integer $n$ and any $t\in (0,1)$. 
Then try to simplify the quantity $\mathbb P\left\{\max_{1\leqslant j\leqslant n}|X_j|/\sqrt n\gt\varepsilon     \right\}$ for a fixed $\varepsilon$.
