Sum of probabilities is infinite I'm stucked solving this problem:

Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of i.i.d. random variables
  with exponential distribution and $\lambda=1$. Show that
  $$\limsup_{n\to\infty}\frac{X_n}{\ln(n)}=1\ a.s$$

So 
$$\limsup_{n\to\infty}\frac{X_n}{\ln(n)}=1\ a.s\Leftrightarrow P(\limsup_{n\to\infty}\frac{X_n}{\ln(n)}=1)=1$$
Since we are looking at i.i.d. random variables Borel-Cantelli say that this can only be true if
$$\sum\limits_{n=0}^\infty P(\frac{X_n}{\ln(n)}=1)=\infty$$
How do I show this last equality?
 A: Here is a guideline. Some details are left as smaller exercises, but I really encourage you to fill them up. I'll first recall how to prove that a sequence of random variables converges almost surely, before adapting the argument to the limsups.
Convergence of sequences of random variables
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and let $\ell \in \mathbb{R}$. We say that $\lim_{n \to + \infty} a_n = \ell$ if and only if:
$$\forall \varepsilon \in \mathbb{R}_+^*, \ \exists N \geq 0 : \ \forall n \geq N, \ |a_n-\ell| \leq \varepsilon.$$
Note that, in the definition, we may require without loss of generality that $\varepsilon$ be a positive rational (or even that $\varepsilon = 1/M$ for some integer $M$).
Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of real-valued random variables, and let $\ell \in \mathbb{R}$. Then $\lim_{n \to + \infty} Y_n = \ell$ almost surely if and only if
$$\left[ \forall \varepsilon \in \mathbb{R}_+^*, \ \exists N \geq 0 : \ \forall n \geq N, \ |Y_n-\ell| \leq \varepsilon \right] \ \text{almost surely}.$$
For $\varepsilon > 0$, let $A_\varepsilon$ be the set of sequences $(Y_n)_{n \in \mathbb{N}}$ such that $\exists N \geq 0 : \ \forall n \geq N, \ |Y_n-\ell| \leq \varepsilon$. Then 
$$\lim_{n \to + \infty} Y_n = \ell \Leftrightarrow (Y_n)_{n \in \mathbb{N}} \in \bigcap_{\varepsilon > 0} A_\varepsilon = \bigcap_{\varepsilon \in \mathbb{Q}_+^*} A_\varepsilon.$$
Since the RHS is a countable intersection of measurable sets, it has full measure if and only if each set has full measure. Hence:
$$\lim_{n \to + \infty} Y_n = \ell \ \text{a.s.} \Leftrightarrow \forall \varepsilon \in \mathbb{Q}_+^*, \ \left[ (Y_n)_{n \in \mathbb{N}} \in A_\varepsilon \ \text{a.s.} \right].$$
Finally, I can replace the condition $\forall \varepsilon \in \mathbb{Q}_+^*, \ \left[ (Y_n)_{n \in \mathbb{N}} \in A_\varepsilon \ \text{a.s.} \right]$ by the condition $\forall \varepsilon >0, \ \left[ (Y_n)_{n \in \mathbb{N}} \in A_\varepsilon \ \text{a.s.} \right]$, which is implied by the LHS above, and implies the RHS.
Hence, is you want to prove that a sequence of random variables almost surely converges to some constant $\ell$, the most fundamental method is to fix an arbitrary $\varepsilon >0$, and then to prove that $A_\varepsilon$ has full measure, or in other words that almost surely, there exists $N \geq 0$ such that $|X_n-\ell| \leq \varepsilon$ for all $n \geq N$. We have managed to get the "almost surely" inside the "for all $\varepsilon > 0$, which is very important and not completely obvious (until you get used to it).
Convergence of sequences of random variables
To goal of the section above was to remind you of something you should have already seen, so that what follows does not drops from nowhere.
You want to compute a limsup instead of a mere limit. It shall be quite convenient to use the following characterization of the limsup (left as an exercise). Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and let $\ell \in \mathbb{R}$. Then $\limsup_{n \to + \infty} a_n = \ell$ if and only if:
$$\forall \varepsilon >0, \ \{n: \ a_n >\ell+\varepsilon\} \text{ is finite and } \{n: \ a_n >\ell-\varepsilon\} \text{ is infinite.}$$
Let $(Y_n)_{n \in \mathbb{N}}$. Using the same trick as in the first section (again, the details are left as an exercise), 
$$\limsup_{n \to + \infty} Y_n = \ell \ \text{a.s.} \Leftrightarrow \forall \varepsilon >0, \ \left[ \{n: \ Y_n >\ell+\varepsilon\} \text{ is finite and } \{n: \ X_n >\ell-\varepsilon\} \text{ is infinite a.s.} \right].$$
Finally, by a usual characterization of the limsup and liminf of sets, $\{n: \ Y_n >\ell+\varepsilon\}$ is finite if and only if 
$$(Y_n)_{n \geq 0} \notin \limsup_{n \geq 0} \ \{Y_n >\ell+\varepsilon\},$$
and $\{n: \ Y_n >\ell-\varepsilon\}$ is infinite if and only if 
$$(Y_n)_{n \geq 0} \in \limsup_{n \geq 0} \ \{Y_n >\ell-\varepsilon\}.$$
So what you need to prove is that, for all $\varepsilon > 0$,
$$\mathbb{P} \left( \limsup_{n \geq 0} \ \{Y_n >\ell+\varepsilon\}\right) = 0 \text{ and } \mathbb{P} \left( \limsup_{n \geq 0} \ \{Y_n >\ell-\varepsilon\}\right) = 1.$$
Choose wisely the sequence $(Y_n)_{n \geq 0}$ and the constant $\ell$, use Borel-Cantelli's lemma, and you should be able to prove the claim.
