I am stuck in one step of an application of Borel Cantelli Lemma.
I am trying to show the following : if $a_n$ is a sequence such that $0 < a_n < 1$ for all $n$ then $\prod_{n=1}^\infty 1 - a_n = 0 \Rightarrow \sum_{n=1}^\infty a_n = +\infty$. By using the log I can show that $\prod_{n=1}^\infty 1 - a_n = 0 \Leftrightarrow \sum_{n=1}^\infty \log(1-a_n) = - \infty $ (since $1-a_n > 0, \forall n$) but I cannot conclude about the divergence of $\sum_{n=1}^\infty a_n$. Any hint ?
Hint:
If $\sum_{n=1}^\infty a_n < \infty$, given $\epsilon \in (0,1)$ there exists $N$ such that $\sum_{n=N}^\infty a_n < \epsilon$, and for all $m > N$
$$\prod_{n= N}^m(1-a_n) > 1 - \sum_{n=N}^ma_n > 1 - \epsilon > 0$$