# Prove that $M_n/\log n\to 1$ a.s. where $X_i$ are a sequence of i.i.d $\text{exp}(1)$ random variables and $M_n=\max_1^n X_i$

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.

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 $$y_n=\max \{x_1,x_2,\cdots, x_n\}$$.