Borel-Cantelli argument for maximum of random variables Let $(X_n)_{n \geq 1}$ be a sequence of random variables taking non-integer values, such that for each $n$ and each $i$, $\mathbb{P}(X_n \geq i) = 1/i$. By Borel-Cantelli, I have managed to show that (almost surely) 
\begin{equation}
\limsup_{n \to \infty} \dfrac{\log X_n}{\log n} = 1.
\end{equation}
Next I need to prove that, for $M_n = \max_{1 \leq k \leq n} X_k$, we have almost surely:
\begin{equation}
\lim_{n \to \infty} \dfrac{\log M_n}{\log n} = 1.
\end{equation}
My idea was to fix an $\epsilon > 0 $, then show that:
\begin{equation}
\mathbb{P} \left(\limsup_{n \to \infty} \dfrac{\log M_n}{\log n} \leq 1+ \epsilon \right) = 1,
\end{equation}
\begin{equation}
\mathbb{P} \left(\liminf_{n \to \infty} \dfrac{\log M_n}{\log n} \geq 1 - \epsilon \right) = 1,
\end{equation}
then take a monotonic intersection over all $\epsilon \in \mathbb{Q}^+$ and conclude. However, I have been unable to prove the above equalities, and I am not sure how (and if) the first property that I proved can be useful. Any help appreciated!
 A: Without independence this is obviously wrong. If $X_n=X_1$ for all $n$ then the result you claim to have proved is clearly false since the $\lim\ sup$ is $0$, not $1$. 
Assume independence.  The  proof  for limsup does not require any probability theory. 
If $\lim \sup \frac {a_n} {b_n}=1$ for some positive increasing sequence $b_n \to \infty$ then $\lim \sup \frac {c_n} {b_n}=1$ where $c_n =\max \{a_1,a_2,...,a_n\}$. 
A: I think I have finally found a (probabilistic) solution to this question. 
For any $\alpha > 0$, we have:
\begin{equation}
\mathbb{P}[M_n < n^{\alpha}] = \mathbb{P}[X_k < n^{\alpha}]^n \leq (1-n^\alpha)^n \leq \exp(-n^{1-\alpha}).
\end{equation}
Next, we use the fact that $\sum_n e^{-n^{\epsilon}}$ converges for any $\epsilon > 0$. This means that for any $\alpha <1$, the sum of  the above probabilities is  finite. Therefore, by the  first Borel-Cantelli lemma, 
\begin{equation}
\mathbb{P}[M_n \geq n^{\alpha} \text{ eventually}] = \mathbb{P}\left[\liminf_{n  \to \infty} \frac{\log M_n}{\log n} \geq \alpha\right] = 1.
\end{equation}
So  taking a  countable  union over rational $\alpha \in (0,1)$, we get
\begin{equation}
\mathbb{P}\left[\liminf_{n  \to \infty} \frac{\log M_n}{\log n} \geq 1 \right] = 1.
\end{equation}
On the other hand, $\limsup \frac{\log M_n}{\log n} \leq \limsup \frac{\log X_n}{\log n}$. This is because if $\exists N_1 \in \mathbb{N}$ such that $X_n \leq n^\alpha$ whenever $n \geq N_1$, then $\exists N_2$ s.t. $M_n \leq n^\alpha$ whenever $n \geq N_2$. This is accomplished by taking $M$ such that $M^\alpha > \max(X_1, X_2,...,X_{N_1})$. This shows the $\limsup$ is at most $1$, so we're done. 
