$\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\ \alpha_k\in(0,1)$ Proof verification I would like to ensure that my logic works here for proving the following:
$$\prod_{k=1}^{\infty}(1-\alpha_k)>0\iff\sum_{k=1}^{\infty}\alpha_k<\infty\;\;\;\;\;\;, \alpha_k\in(0,1)$$
So, for
$\Rightarrow $:
$$\sum_{k=1}^{\infty}\alpha_k=\infty \Rightarrow \alpha_k \not\to0
\Rightarrow (1-\alpha_k)\not\to1\Longrightarrow \\
1-\alpha_k\leq1 \land (1-\alpha_k<1 \;\text{almost always}) \Longrightarrow\\
\prod_{k=1}^{\infty}(1-\alpha_k)=0
$$
Where the before last statement expresses fact that this is infinitely many times strictly decreasing sequence, that is never increasing that is bounded by only by zero
And for $\Leftarrow$:
if $\sum_{k=1}^{\infty}\alpha_k<\infty$ then $\alpha_k\to0$. By using continuity, lets speak of
$$\exp(\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j))) 
$$
and it is enough to prove that
$$\lim_{k\to\infty}\log(\prod_{j=1}^{k}(1-\alpha_j)) \\
=\lim_{k\to\infty}\sum_{j=1}^{k}\log(1-\alpha_j)$$ exists and is not $\pm\infty$, because this means that the argument of the $\exp(*)$ is not $-\infty$ which means that the product is greater than $0$.
And this happens because $-\log(1-\alpha_j)>0$ and $\frac{-\log(1-\alpha_j)}{\alpha_j}\to_{\alpha_j\to0}1$ implies that our serie converges as we wished.
 A: The $\implies$ direction is wrong, the divergence of the series does not imply that $a_n \not\to 0$. You have the right idea in the $\impliedby$ direction, and that can be used for both directions.
We have the general inequality
$$
  \frac{x}{1+x} \le \log(1+x) \le x
$$
for $x > -1$, so that
$$
 - \sum_{k=1}^n \frac{\alpha_k}{1-\alpha_k} \le \log \prod_{k=1}^n (1-\alpha_k) \le -\sum_{k=1}^n \alpha_k \, .
$$
If $\sum_{k=1}^\infty \alpha_k = \infty$ then the right-hand side diverges to $-\infty$, so that $\prod_{k=1}^n (1-\alpha_k) \to 0$.
If $\sum_{k=1}^\infty \alpha_k < \infty$ then $\alpha_k < 1/2$ for all sufficiently large $k$, so that the left-hand side converges to a finite value $c$. It follows that $\prod_{k=1}^n (1-\alpha_k)$ is bounded below by the positive value $e^c$, and therefore $\prod_{k=1}^\infty (1-\alpha_k) \ge e^c > 0$.
Remark: Here we have used $\prod_{k=1}^{\infty}(1-\alpha_k) = \lim_{n\to \infty}\prod_{k=1}^{n}(1-\alpha_k)$ as definition of the infinite product, the limit exists because the partial products are positive and decreasing. Some sources do not allow zero as the limit of the partial product, or call such a product “diverging to zero.”
A: I just suggest a "quick" proof (I agree it's not an answer, but there is not enough place in the comment to write it as a comment).

If : You have that $\alpha _k\to 0$ when $k\to \infty $, and thus, $-\log(1-\alpha _k)\sim \alpha _k$ when $k\to \infty $. Therefore $\log\left(\prod_{k\geq 1}(1-\alpha _k)\right)>-\infty $, and thus $\prod_{k\geq 1}(1-\alpha _k)>0$.
Only if: You have that $-\sum_{k\geq 1}\log(1-\alpha _k)<\infty $ and thus $\alpha _k\to 0$ when $n\to \infty $. Therefore $\alpha _k\sim -\log(1-\alpha _k)$ when $k\to \infty $, and thus $\sum_{k\geq 1}\alpha _k<\infty $.
