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Question

Let $(X_n)_{n\geq 1}$ be an i.i.d sequence of random variables with $P(X_1>x)=e^{-x}$ and put $M_n=\max_1^n X_i$. Then $M_n/\log n\to 1$ a.s.

My attempt

I have shown that $\limsup X_n/\log n=1$ a.s and $\lim\inf M_n/\log n\geq 1$ a.s.

For the first claim it suffices to note that $$ \sum _{n=1}^\infty P(X_n\geq c \log n)=\sum _n n^{-c} $$ which is finite if $c>1$ and equal to $\infty$ if $c<1$. In particular for any $\epsilon >0$, $X_n/\log n\ge 1+\epsilon $ finitely many times with probability one which implies that $\lim \sup X_n/\log _n\leq 1$. Similarly, for any $\epsilon >0$, $X_n/\log n\geq 1-\epsilon$ infinitely often with probability one, so $\limsup X_n/\log n \geq 1$,

For the other claim it suffices to show that $P(M_n/\log n <1-\epsilon \quad \text{i.o})=0$. This can be shown by an application of Borel Cantelli. Indeed, $$ P(M_n<(1-\epsilon)\log n)=P(X_1<(1-\epsilon)\log n)^n=(1-n^{-(1-\epsilon)})^n\leq \exp(-n\times n^{-(1-\epsilon)}) $$ But $\sum \exp(-n^{\epsilon})<\infty$.

My Problem

I am unable to show that $\limsup M_n/\log n\leq 1$ a.s. I tried to use Borel cantelli by showing that $P(M_n>(1+\epsilon)\log n \quad \text{i.o})=0$. To this end, put $c_n=(1+\epsilon)\log n)$ and $$ P(M_n>c_n)=1-P(M_n\le c_n)=1-(1-e^{-c_n})^n\leq \exp (-(1-e^{-c_n})^n) $$ but I am not sure if this sequence is summable. Any help is appreciated and other methods are welcome.

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You don't any probbaility theory to complete that argument. Try to prove the following lemma:

if $a_n$ increases to $\infty$ and $\lim \sup \frac {x_n} {a_n} \leq 1$ then $\lim \sup \frac {y_n} {a_n} \leq 1$ where $y_n=\max \{x_1,x_2,\cdots, x_n\}$.

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