Hi I came up with a proof and I'm not really sure about it. Can someone help me to check whether is correct or not?
Statement:
Consider a sequence $(Xn)_{n\geq1}$ of exponentially-distributed random variables with mean $λ>0 \ (i.e.EX_1 =λ)$. Prove or disprove wether
$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg)=1$$
Proof:
Let us define a sequence of events $(E_n)_{n \geq 1}$ where
$$E_n = \{ \omega \in \Omega: X_n(\omega)< \lambda \log n \}$$
Let us note moreover that $(X_n)_{n \geq 1}$ is a sequence of iid exponentially distributed random variable and hence $(E_n)_{n \geq 1}$ is a sequence of independent events. Let us now compute
$$\forall n\geq 1 \qquad P(E_n)=P(X_n< \lambda \log n)=F_{X_n}( \lambda \log n)=1-e^{-\lambda^2 \log n}$$
Hence we have that
$$\sum_{n=1}^\infty P(E_N)= \infty$$
So by the second Borel-Cantelli lemma we have that
$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)=1$$
We now observe that
$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)=1 \rightarrow P \bigg( \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)^c \bigg)=0$$
Now, since
$$\bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg) \subseteq \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)^c$$
we conclude by monotonicity that
$$P\bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg)=0$$
disproving the initial statement.