A proof about convergence of $\prod_{n=1}^{\infty} (1-a_n)$, given $\sum_{n=1}^{\infty} a_n$ convergence.

Question

Hypothesis

1. $a_n > 0$, for all $n\ge 1$.
2. $\sum_{n=1}^{\infty} a_n$ converges.

Why I am asking

I'm reading Continued Fractions by A. Ya. Khinchin. This statement helps to prove the convergence of an infinite continued fraction.

Comments

I actually found the same question searching here, and tried to understand them. But I don't understand the proofs given. An example I have seen is a proof by equivalence test. If it is possible, I need a more basic prove. Otherwise, any other is welcome :) PD: Sorry for my english xd. I hope it is understandable.

Answer 1

Convergence of $\sum a_n$ implies $a_n \rightarrow 0$. Then there is an $N$ such that for all $n \geq N$ we have $a_n <1$.

If we put $$ b_n = \prod_{i=N}^{N+n} \left( 1-a_i \right) $$ then the sequence $\left\{b_n\right\}$ is decreasing (since $0<1-a_n<1$ for $n \geq N$) and bounded (since $0<b_n<1$ for all $n$).

Hence $b=\lim_{n \rightarrow \infty} b_n$ exists and is finite so that $$ \prod_{n=1}^\infty \left( 1-a_n \right) = \left[\prod_{n=1}^{N-1} \left( 1-a_n \right)\right] \cdot \left[\prod_{n=N}^\infty \left( 1-a_n \right)\right] = \left[\prod_{n=1}^{N-1} \left( 1-a_n \right)\right] \cdot b $$ and the result follows.