Limes superior of $X_n>\alpha \ln(n)$

Question

Given the random variable $X_n$, to be exponential distributed with parameter $1$, we define $$A_n:=\left\{X_n>\alpha \ln(n)\right\}.$$ We've shown in the classes that for $\alpha \leq 1$, $A_n$ occurs infinitely times, and for $\alpha>1$ finitely times. Now calculating the limes superior $\overline{\lim}_{n\rightarrow\infty}\frac{X_n}{\ln(n)}$, we have said that $\frac{X_n}{\ln(n)}\geq 1$ is true for infinitely many $n$.
Why is that true? Is this fact connected with the case $\alpha \leq 1$?

Answer 1

$P(X_n >\alpha \log\, n) =\int_{\alpha \log\, n}^{\infty} e^{-x}\, dx=\frac 1 {n^{\alpha}}$. Assuming independence of $X_n$'s we can conclude from Borel Cantelli Lemma and convergence of $\sum \frac 1 {n^{\alpha}}$ that $X_n \leq \alpha \log\, n$ for all $n$ sufficently large with probability $1$ if $\alpha >1$. If $\alpha \leq 1$ the same argument shows that $X_n >\alpha \log\, n$ for inifintely many $n$ with probability $1$. This implies that $\lim \sup \frac {X_n} {\log\, n} \geq 1$ with probability $1$.