Probability of $\limsup_n\left\{|\frac{\max_{1\leq k \leq n}X_k}{\ln(n)}-1| >\epsilon\right\}$ Let $(X_n)_n$ be a sequence of independent random variables and identically distributed, following the exponential distribution with parameter 1.
Let $0<\epsilon<1.$ I want to compute $P\left(\limsup_n\left\{\left|\frac{\max_{1\leq k \leq n}X_k}{\ln(n)}-1\right| >\epsilon\right\}\right)$
I am thankful for any idea.
 A: Claim
Let $\{X_k\}_{k=1}^{\infty}$ be i.i.d. exponentially distributed random variables with parameter $\lambda=1$. Define $M_n = \max_{\{k : 1 \leq k \leq n\}}X_k$.  Then:
$$\lim_{n\rightarrow\infty} \frac{M_n}{\log(n)} = 1 \quad \mbox{(with prob 1)} $$
Derivation
Fix $\epsilon>0$.  Then
$$ \{|M_n/\log(n) - 1| > \epsilon\} = \underbrace{\{\{M_n < \log(n^{1-\epsilon})\}}_{\mbox{type 1}} \cup \underbrace{\{M_n> \log(n^{1+\epsilon})\}}_{\mbox{type 2}}   $$
Set Type 1:
Indeed $P[M_n < \log(n^{1-\epsilon})]= (1-\frac{1}{n^{1-\epsilon}})^n \approx exp(-n^\epsilon)$. This is summable (as Clement C notes above) and so by Borel-Cantelli we can conclude that, with prob 1, at most finitely many of the events $\{M_n < \log(n^{1-\epsilon})\}$ occur.
Set Type 2:
Indeed $P[M_n > \log(n^{1+\epsilon})] = 1-(1-\frac{1}{n^{1+\epsilon}})^n$.  This case is more tricky since indeed these are not summable over all $n$.  Nevertheless there is a nice technique of summing over the sparse subsequence of indices of the form $2^k$: 
$$ \sum_{k=1}^{\infty} P\left[M_{2^k}>\log\left((2^k)^{1+\epsilon}\right)\right] = \sum_{k=1}^{\infty}\left[1 - \left(1 - \frac{1}{(2^k)^{1+\epsilon}}\right)^{2^k}\right] $$
This sum is not trivial to evaluate, but indeed it is finite.
Wait a minute, you say, what about the $M_n$ values for indices $n$ that are not powers of 2? Well,  for any $n$ that satisfies $2^k \leq  n \leq 2^{k+1}$ we can say:
$$ \frac{M_n}{\log(n)} \leq \frac{M_{2^{k+1}}}{\log(2^k)}$$
So 
$$ \cup_{\{n : 2^k\leq n \leq 2^{k+1}\}}\left\{\frac{M_n}{\log(n)}>1+\epsilon\right\} \subseteq \left\{ \frac{M_{2^{k+1}}}{\log(2^k)} > 1+\epsilon\right\} $$
so it remains to compute 
$$ \sum_{k=1}^{\infty} P\left[\frac{M_{2^{k+1}}}{\log(2^k)}>1+\epsilon \right] = \sum_{k=1}^{\infty} \left[1 - \left(1-\frac{1}{2^{k(1+\epsilon)}}\right)^{2^{k+1}} \right]$$
This sum is similar to the previous one and is also finite.
So, with prob 1, at most finitely many type 2 events occur.
