Why $\limsup \left\{Y_n^+/n\le 0\right\} \text{ a.s.} \implies (max_{m\le n}Y_m)/n \rightarrow 0$ a.s? The original question is as follows:

$\left\{Y_n\right\}$ are i.i.d random variables. Find the sufficient and neccessary condition for $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.  

The answer is $EY_1^+ < \infty$. Then $\sum P(Y_n/n > \epsilon)\le EY_1^+< \infty \implies \limsup\left\{Y_n^+/n \le 0 \right\}$ a.s. I can't see why it implies  $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.  
My attempt is as follows:
Let $E=\left\{Y_n^+/n \le 0 \text{ i.o.}\right\}$. Then $P(E)=1$ and for $\omega \in E$, $Y_n^+(\omega)/n \le 0 \text{ i.o.}$ Why will this implies $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.? One counter-example I have in mind is let all even number satisfy this condition while all odd number don't.
 A: Let $M_n=\max\limits_{j\le n}Y_j$ and $F$ be the distribution function of $X_1$, then 
$$ P(M_n\le -n\epsilon)=[F(-n\epsilon)]^n<\alpha^n,\qquad \text{if n is large enough such that }F(-n\epsilon)<\alpha<1.$$
Therefore
\begin{eqnarray}
 P\Bigl(\Bigl\{\dfrac{M_n}n<-\epsilon\Bigr\}\quad \text{i.o.}\Bigr)=0,\qquad \forall \epsilon>0\quad
&\Rightarrow& P\Bigl(\liminf_{n\to\infty}\dfrac{M_n}n<-\epsilon\Bigr)=0\qquad \forall \epsilon>0\\
&\Rightarrow& \liminf_{n\to\infty}\dfrac{M_n}n\ge 0\qquad \text{a.s.}\tag{*}
\end{eqnarray}
Meanwhile, 
\begin{eqnarray}
EY_1^+<\infty &\Rightarrow &\sum_{n=1}^\infty P(Y^+_n>n\epsilon)<\infty\qquad \forall \epsilon>0\\
\text{by Borel-Cantelli Lemma}\quad&\Rightarrow & P\Bigl(\Bigl\{\dfrac{Y_n^+}n >\epsilon\Bigr\}\quad\text{i.o.}\Bigr)=0\qquad \forall \epsilon>0\\
&\Rightarrow & \limsup_{n\to\infty}\dfrac{Y_n^+}n \le \epsilon \quad\text{a.s.}\qquad\forall\epsilon>0\\
&\Rightarrow &\lim_{n\to\infty}\dfrac{Y_n^+}n=0 \quad\text{a.s.}\quad\Bigl(\Leftrightarrow \lim_{n\to\infty}\dfrac{\max\limits_{j\le n}Y_j^+}n=0\Bigr)\\
&\Rightarrow & \limsup_{n\to\infty}\dfrac{Y_n}n\le 0\quad\text{a.s.}\\
&\Rightarrow & \limsup_{n\to\infty}\dfrac{M_n}n\le 0\quad\text{a.s.}\\
\text{by} (*)\qquad &\Rightarrow & \lim_{n\to\infty}\dfrac{M_n}n=0  \quad\text{a.s.}
\end{eqnarray}
