If, $0<a_n< 1, \forall n$, show that $\sum_{n=1}^{+\infty} a_n$ converges iff $\prod_{n=1}^{+\infty} (1 - a_n)$ converges to a non-zero number.
As $1-x \leq e^{-x}$, we have that $\prod_{n=1}^N (1 - a_n) \leq \prod_{n=1}^N e^{-a_n} = e^{-\sum_{n=1}^N a_n}$. Then, by the Monotone Convergence Theorem, if $\sum_{n=1}^{+\infty}a_n$ converges, then $\prod_{n=1}^{+\infty} (1 - a_n)$ converges.
I'm having some trouble to conclude the other direction.
Thank you very much!
Use that $\displaystyle e^x< {1\over 1-x}$ (Bernoulli's inequality) when $|x|<1$. Then WLOG we may assume $|a_n|<1$. As $\prod_{i=1}^\infty (1-a_n)$ converges to a nonzero number iff $\prod_{i=1}^\infty (1-a_n)^{-1}$ does, we see that
$$e^{\sum a_n}<\prod_{i=1}^\infty {1\over 1-a_n}$$
proving the other direction.
As Clement C. has noted, your conclusion does not show the product converges to something non-zero. For that, you can use that as the sum $\sum a_n<\infty$ converges, since then we just note that
$$\lim_{n\to\infty} {\log (1-a_n)\over a_n}=-1$$
so $\sum \log(1-a_n)$ converges, which is only possible if the product does not converge to $0$.
hint
use the equivalence test $$-\ln (1-a_n)\sim a_n \;\;(n\to+\infty) $$ so the series $$\sum a_n \;\;\text {and}\;\; \sum -\ln (1-a_n) $$ are both convergent or both divergent.
if $\prod (1-a_n)$ converges to zero, it means that $\sum -\ln (1-a_n) $ is divergent and so is $\sum a_n $.
