Upperbound of tail event

Question

Let $X_n\sim \mathcal{E}(n)$ I need to show that $P (\{X_2 + \ldots + X_n \geq 3 \log{n} \text{ infinitely often}\}) = 0$.

I also have the following hint: with $n \geq 2 \text{ and } 0 \leq x \leq \frac{1}{n}, \frac{1}{1-x} \leq ( \exp(\frac{n}{n-1} x))$.

I started upper-bounding: \begin{align}P (\{X_2 + \ldots + X_n \geq 3 \log{n}\}) &\leq E(\prod_{i=2}^n e^{X_i}) n^{-3} \\ &= \prod_{i=2}^n\bigg(\frac{i}{i-1}\bigg) n^{-3}, \end{align}

but then I can not use the hint, because $\frac{i}{i-1} = \frac{1}{1-\frac{1}{i}}, \frac{1}{i}\geq \frac{1}{n}$, although, $\frac{1}{n} \leq \frac{1}{i}\leq \frac{1}{2}$, can I use the bound and say that $\frac{i}{i-1} \leq \exp(2\cdot\frac{1}{i})$? Am I doing something wrong?

Answer 1

I do not understand the role of the hint. We have a telescopic product and we can simplify it by $$ \prod_{i=2}^n\bigg(\frac{i}{i-1}\bigg)=n. $$ We can conclude using the Borel-Cantelli lemma.