Edit: The Mathexchange gods were not in my favor and at no point of writing this post was the question: How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$? suggested to me, which answers my question thoroughly. I have voted to close this question as a duplicate.
This is from Khinchin's Continued Fractions, English edition pages 63-64. Here $\varphi$ is a is a positive function and it is assumed that $\sum_{n=1}^\infty \frac{1}{\varphi(n)}$ diverges. Then I agree that for any $m$ the series $\sum_{l=2}^\infty\frac{1}{3(1 + \varphi(m+l))}$ diverges as well. Khinchin then writes that
From the theory of infinite products, it follows that the product $$\prod_{l=2}^n (1-\frac{1}{3(1 + \varphi(m+l))}) = 0$$ approaches zero as $n\to\infty$.
It has been some time since I have worked with infinite products, but what I recall is the basic result that if $a_n > 0$ for all $n$, then whether the product $\prod_{n=1}^\infty a_n$ converges to a non-zero value is equivalent to whether the series $\sum_{n=1}^\infty \log(a_n)$ converges/diverges. But I cannot recall any result that would immediately show that the series over the terms $\log\left(1-\frac{1}{3(1 + \varphi(m+l))}\right)$ w.r.t. $l\geq 2$ would converge. I also know and know how to prove that if $0\leq a_n < 1$ for all $n$, then $$\sum_{n=1}^\infty a_n < \infty \implies \prod_{n=1}^\infty (1 - a_n) > 0$$
but it seems to me that Khinchin's claim requires a sorta reverse implication of this. That is,
$$\sum_{n=1}^\infty a_n = \infty \implies \prod_{n=1}^\infty (1 - a_n) = 0$$
and as of writing I am just not seeing how this can be done.
