I was working through the proof of convergence of the infinite product on the wikipedia website (see: https://en.wikipedia.org/wiki/Infinite_product): $$ \prod_{k=1}^{\infty}a_k\text{ converges to } c>0 \iff \sum_{k=1}^{\infty} \log(a_k)\text { converges, where } a_k>0. $$ The proof is split up into two parts. In the first the part the above statement is shown if $a_n\geq 1$ and in the second part the case $0<a_k\leq 1 $ is handled. The proof seems to me rather convoluted, so I am not sure if I have understood the line of thought correctly. I will try to reproduce the second part
"The same proof also shows that if $a_{k}=1-q_{k}$ for some $0\leq q_{k}<1$, then $\prod\limits_{k=1}^{\infty }(1-q_{k})$ converges to a non-zero number if and only if $\sum\limits_{k=1}^{\infty }q_{k}$ converges."
in my own words.
Basically we use the series, $\sum\limits_{k=1}^n(-q_k)$, to establish a connection between $\prod\limits_{k=1}^{\infty}a_k$ and $\sum\limits_{k=1}^{\infty} \log(a_k)$.
For an index $n\in\mathbb{N}$ we have $$ 1-\sum\limits_{k=1}^nq_k\leq\prod\limits_{k=1}^n(1-q_k)\leq \exp\left(\sum\limits_{k=1}^n(-q_k)\right),\text{ because } \log(1-q_k)\leq -q_k \text{ for each } k. $$ (To be rigourous we would have to show inductively that $1-q_1-q_2\cdots -q_n\leq 1-q_1-q_2+q_1q_2\cdots=(1-q_1)(1-q_2)\cdots(1-q_n)$ is satisfied for all $n$; but I will skip that.)
If $\prod\limits_{k=1}^n(1-q_k)$ converges to a non zero number $c>0$, then $\exp\left(\sum\limits_{k=1}^n(-q_k)\right)$ converges by monotone convergence theorem. So $\sum\limits_{k=1}^n(-q_k)$ must converge and $\lim\limits_{k\to\infty}(-q_k)=0$. Conversely, if $\sum\limits_{k=1}^n-q_k$ converges, then $\prod\limits_{k=1}^n(1-q_k)$ converges because it's bounded below and monotonously decreasing. We observe: $$ \lim\limits_{k\to\infty}\frac{\log(1-q_k)}{-q_k}=\lim\limits_{q_k\to0}\frac{\log(1-q_k)}{-q_k}~\underset{\text{L'Hospital}}{=}~\lim\limits_{q_k\to0}\frac{-\frac{1}{1-q_k}}{-1}= 1. $$ By limit comparison test $\sum\limits_{k=1}^n\log(1-q_k)$ and $\sum\limits_{k=1}^n-q_k$ both converge or both diverge.
So we can conclude that $$ \prod_{k=1}^{n}a_k\text{ converges to } c>0 \iff \sum\limits_{k=1}^n(-q_k)\text{ converges}\iff \sum_{k=1}^{\infty} \log(a_k)\text { converges}, $$ which shows the original statement in the case of $0<a_k\leq1$.
Is this correct? I am a bit unsure because if $\sum\limits_{k=1}^n-q_k$ converges to $1$ then it could be that $\prod\limits_{k=1}^n (1-q_k)$ converges to $0$.