"Convergence"/"Divergence" of $\prod_{n=1}^\infty (1 - \gamma_n)$

Question

While trying to understand a proof of a result in an article, I stumbled upon the product

$$\prod_{n=1}^\infty (1 - \gamma_n)$$

with $\gamma_n$ a real scalar belonging to $(0,1)$. I'm not really a mathematician, so I looked through some of the articles here about infinite products (like this, this and this), but got lost in the discussions about convergence and divergence to zero. What I would like to know is what are the conditions on $\gamma_n$ for the product to converge/diverge to zero as $n \rightarrow \infty$. Intuitively, I had thought that since $1 - \gamma_n < 1$, convergence was guaranteed, but this showed me it may not be that simple. All the other questions I cited are interested in when does the product converges to a nonzero limit, but I want to know when does it go to zero... Is it really obvious and that's why nobody asks this question (or maybe I didn't find the right one...)? If you could please also point me to a reference of any needed theorem or result that might be needed or useful, it would be really great.

Thanks to all in advance.

Answer 1

The condition is that if $$ \sum_{n=1}^\infty γ_n=\infty $$ then the product is zero. This can be relatively easy be seen by taking the logarithm or using $$ 1-x\le e^{-x}. $$

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On the other hand, it is easy to generalize Bernoulli's inequality to get $$ \prod_{n=N}^\infty (1-γ_n)\ge 1-\sum_{n=N}^\infty γ_n $$ so if the series converges one can find an $N$ such that the remainder is smaller than $1/2$ and thus $$ \prod_{n=1}^\infty (1-γ_n)\ge\frac12\prod_{n=1}^{N-1} (1-γ_n)>0. $$