Upperbound of tail event 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?

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.
