Prove that almost surely $\lim\sup_{n\to\infty}\frac{X_n}{\ln n}=\frac{1}{\lambda}$

Question

Let ${(X_n)_{n \in \mathbb{N}}}$, be independent random variables distributed $Exp(\lambda)$.

Prove that almost surely $$\limsup_{n\to\infty}\frac{X_n}{\ln n}=\frac{1}{\lambda}$$ My idea was to look at the $$\sum_{k=1}^{\infty}P\left(\left|\frac{X_k}{\ln k}-\frac{1}{\lambda}\right|>\epsilon\right)$$ and prove it is finite. But I think it is not. $$P\left(X_k<\left(-\epsilon+\frac{1}{\lambda}\right)\ln k\right)=\frac{1}{k}(1-k^{\lambda\epsilon})$$ and we can take such espilon value so it is divergent as $k\to\infty$, which means sum does not converge.

Answer 1

Observe that for $c>0$, we have that $$ \sum_{n=0}^\infty P(X_n>c\log n)=\sum_{n=0}^\infty\frac{1}{n^{\lambda c}}=\begin{cases} <\infty&\text{if }c>1/\lambda\\ \infty&\text{if }c<1/\lambda \end{cases}\quad (1) $$ where we used the fact that $P(X>x)=e^{-\lambda x}$ for $x\geq 0$ if $X\sim \text{exp}(\lambda)$. It follows by Borel-Cantelli that for each $\varepsilon>0$ $$ P\left(\frac{X_n}{\log n}>\lambda^{-1}+\varepsilon\quad \text{i.o}\right)=0 $$ whence $$ \limsup \frac{X_n}{\log n}\leq \lambda^{-1}. $$ with probability $1$. Similarly for each $\varepsilon>0$, $$ P\left(\frac{X_n}{\log n}>\lambda^{-1}-\varepsilon\quad \text{i.o}\right)=1 $$ whence $$ \limsup \frac{X_n}{\log n}\geq \lambda^{-1}.$$ with probability $1$.