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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.

Mark Viola
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Andresbm
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  • Note that $\frac {x-1}{x}\le \log(x) \le x-1$ for $0<x<1$ – Mark Viola Dec 17 '22 at 22:08
  • Your main question is only in your title. It should also be in the body. – coffeemath Dec 17 '22 at 22:08
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    We may assume $a_n<1$ since it is the case for $n$ large enough. Then $\prod (1-a_n)$ converges iff $\log\prod (1-a_n)$ converges iff $\sum \log(1-a_n)$ converges. Since $a_n\rightarrow 0$, $\log(1-a_n)\sim a_n$ and you're done ! – Tuvasbien Dec 17 '22 at 22:09

1 Answers1

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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.

Petronella
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