Sign of an infinite product

Question

Let $c\in (0,1)$ and $a=1-\frac{1}{\sqrt 2}$. How can I prove that the following product is strictly positive?

$$\prod_{i=0}^{\infty}\frac{1-ca^i}{1+ca^i}>0$$

Thank you very much.

Answer 1

hint

$$a=\frac{\sqrt{2}-1}{\sqrt{2}} \implies 0<a<1$$

So, for $ i>0$,

$$0<a^i<1$$ and $$0<ca^i<c<1$$

Answer 2

You could try to see if $\sum_{i=0}^{\infty} \ln(\frac{1-ca^i}{1+ca^i})$ converges.

Since $\ln(\frac{1-ca^i}{1+ca^i}) \sim -2ca^i$ the corresponding series are of the same nature, and since the later converges, so does the first one.
Hence $\sum_{i=0}^{\infty} \ln(\frac{1-ca^i}{1+ca^i}) \neq -\infty$, so your product ($\prod_{i=0}^{\infty}\frac{1-ca^i}{1+ca^i} =e^{\sum_{i=0}^{\infty} \ln(\frac{1-ca^i}{1+ca^i})}$) is strictly positive.

Answer 3

A nice result here:

Thm: If $0<b_n<1,$ then $\prod_{n=0}^\infty (1-b_n)>0$ iff $\sum b_n<\infty.$

The theorem follows by applying the logarithm to the product and using the inequality $-x/2>\ln(1-x)>-x,$ which holds for small $x>0.$ Assuming the theorem, note the factors in your product equal $1-b_n,$ where

$$b_n =\frac{2ca^n}{1+ca^n}.$$

The theorem then gives you the desired result.