limit/limsup of a sequence of random variables Suppose $X_i$'s are i.i.d, with the density distribution $f(x) = e^{-x}$, $x \geq 0$. I was able to show that
$$P(\limsup X_n/\log{n} =1)=1$$ using Borel-Cantelli.
Define $M_n=\max \{X_1,\ldots,X_n\}$, can I claim $M_n/\log{n} \rightarrow 1$ a.s. in this case? Is it still true in general without knowing the distribution of $X_i$?
 A: If $F(x) = 1 - e^{-x}$ for $x > 0$ is the CDF of each $X_i$, the CDF of $M_n$ is $F_{M_n}(x) = F(x)^n = (1 - e^{-x})^n$ for $x > 0$.  Note that $\ln(F_{M_n}(x)) = n \ln(1 - e^{-x})$ and since $- t - t^2 < \ln(1-t) < -t$ for $0 < t < .683$, for any $c>0$ we have $-n^{1-c} - n^{1-2c} \le \ln(F_(M_n)(c \ln n)) \le -n^{1-c}$ for $n$ large enough.  If $c < 1$, this says
$P\left( \frac{M_n}{\ln n} \le c\right) \le e^{-n^{1-c}}$, and $\sum_n e^{-n^{1-c}} < \infty$ so almost surely only finitely many $\frac{M_n}{\ln n} \le c$.  If $c > 1$, 
$P(\left( \frac{M_n}{\ln n} \le c \right) \ge e^{-n^{1-c} - n^{1-2c}} \to 1$ as $n \to \infty$, so almost surely infinitely many $\frac{M_n}{\ln n} \le c$.  Thus $\lim \sup_n \frac{M_n}{\ln n} = 1$ almost surely.  However, it's not so clear to me that $\lim \inf_n \frac{M_n}{\ln n} = 1$ almost surely. 
A: this is a small observation not an answer: 
the distribution is in fact important, for example if the random variables are bounded almost surely the limit is zero a.s.
For the unbounded case, (that more likely you are thinking about), I just got the trivial lower bound 
$$
1\leq \liminf_{n\to\infty} \frac{M_n}{\log n}
$$
in the i.i.d case, by the Borel-Cantelli Lemma, supposing that 
$$\sum_{n=1}^{\infty} \mathbb{P}(X_1\leq \ln n)^n<+\infty.$$ 
 Edition. I replaced the limsup by liminf, which is better in this case, based on the comments made by  Didier Piau. 
