$X_1,X_2,\dots$ are i.i.d. and $X_1\thicksim\text{Exp}(\lambda)$, then $P\left(\limsup_{n\to\infty}\frac{X_n}{\log(n)}=\frac{1}{\lambda}\right)=1$ Problem: Let $X_1,X_2,\dots$ be i.i.d. random variables with $X_1\thicksim\text{Exp}(\lambda)$. Show that
$$P\left(\limsup_{n\to\infty}\frac{X_n}{\log(n)}=\frac{1}{\lambda}\right)=1.$$
My Approach: Let $\varepsilon>0$ be given and define the events
$$A_n=\left\{\frac{X_n}{\log(n)}>\frac{1+\varepsilon}{\lambda}\right\}\quad\text{for }n\in\mathbb N.$$
Then
$$\sum_{n=1}^\infty P(A_n)=\sum_{n=1}^\infty\int_{\lambda^{-1}(1+\varepsilon)\log(n)}^\infty \lambda e^{-\lambda x}\,dx=\sum_{n=1}^\infty\frac{1}{n^{1+\varepsilon}}<\infty.$$
It follows from the Borel-Cantelli lemma that
$$P\left(\frac{X_n}{\log(n)}\leq\frac{1+\varepsilon}{\lambda}\text{ for all but finitely many }n\right)=1.$$
Since $\varepsilon>0$ was arbitrary, we have
$$P\left(\limsup_{n\to\infty}\frac{X_n}{\log(n)}\leq\frac{1}{\lambda}\right)=1.$$
Next, we also have that
$$\sum_{n=1}^\infty P\left(\frac{X_n}{\log(n)}>\frac{1}{\lambda}\right)=\sum_{n=1}^\infty\frac{1}{n}=\infty.$$
Since the events $\left\{\frac{X_n}{\log(n)}>\frac{1}{\lambda}\right\}$ are independent due to the random variables being independent, the second Borel-Cantelli lemma implies that
$$P\left(\limsup_{n\to\infty}\frac{X_n}{\log(n)}>\frac{1}{\lambda}\right)=1.$$
It follows that
$$P\left(\limsup_{n\to\infty}\frac{X_n}{\log(n)}=\frac{1}{\lambda}\right)=1.$$

Do you agree with my proof above? Any feedback is much appreciated.
Thank you for your time.
 A: It looks very good. Just a few small comments. First of all, although it is easy, it is still not that obvious why $\mathbb{P}(\frac{X_n}{\log(n)}\leq\frac{1+\epsilon}{\lambda}$ eventually)$=1$ for every $\epsilon>0$ implies that $\mathbb{P}(\limsup_{n\to\infty}\frac{X_n}{\log(n)}\leq\frac{1}{\lambda})=1$. The reason is that if we let $B_{\epsilon}=\{\frac{X_n}{\log(n)}\leq\frac{1+\epsilon}{\lambda}$ eventually$\}$ then for every $k\in\mathbb{N}$ we have $\mathbb{P}(B_{\frac{1}{k}})=1$. Now let $B=\cap_{k=1}^\infty B_{\frac{1}{k}}$. A countable intersection of events with probability $1$ is an event with probability $1$, and so $\mathbb{P}(B)=1$. But note that if $\omega\in B$ then for every $k\in\mathbb{N}$ we have $\frac{X_n(\omega)}{\log(n)}\leq\frac{1+\frac{1}{k}}{\lambda}$ for all but finitely many $n$, and so $\limsup_{n\to\infty}\frac{X_n(\omega)}{\log(n)}\leq\frac{1+\frac{1}{k}}{\lambda}$. Since this is true for all $k\in\mathbb{N}$ it follows that $\limsup_{n\to\infty}\frac{X_n(\omega)}{\log(n)}\leq\frac{1}{\lambda}$. By monotonicity of probability it follows that:
$1=\mathbb{P}(B)\leq\mathbb{P}(\limsup_{n\to\infty}\frac{X_n}{\log(n)}\leq\frac{1}
{\lambda})\leq 1$
And so $\mathbb{P}(\limsup_{n\to\infty}\frac{X_n}{\log(n)}\leq\frac{1}{\lambda})=1$.
The second direction is very good, just in the end the second Borel-Cantelli implies that $\mathbb{P}(\limsup_{n\to\infty}\frac{X_n}{\log(n)}\geq\frac{1}{\lambda})=1$. It has to be $\geq$, not $>$.
Sorry if what I wrote is obvious to you. I just wrote here how would I look at it if I saw your answer in my exam.
