Closed form of infinite product $\prod\limits_{k=0}^\infty 2 \left(1-\frac{x^{1/2^{k+1}}}{1+x^{1/2^{k}}} \right)$ I encountered this infinite product while solving another problem:
$$P(x)=\prod_{k=0}^\infty 2 \left(1-\frac{x^{1/2^{k+1}}}{1+x^{1/2^{k}}} \right)$$
$$P(x)=P \left( \frac{1}{x} \right)$$
I strongly believe it has a closed form in general, because it has a closed form for all the values I tried (checked numerically by Wolfram Alpha with high precision):
$$P(2)=\frac{14}{9} \ln 2$$
$$P(3)=\frac{13}{12} \ln 3$$
$$P(4)=\frac{14}{15} \ln 4$$
$$P(5)=\frac{31}{36} \ln 5$$
So the general closed form should be:
$$P(x)=R(x) \ln x$$

What is $R(x)$? And how to prove the closed form?

The product looks like it telescopes, but I couldn't find an appropriate form.
Another thought was to make a substitution:
$$x=e^t$$
$$P(t)=\prod_{k=0}^\infty 2 \left(1-\frac{\exp(t/2^{k+1})}{1+\exp(t/2^{k})} \right)$$
I tried series for the exponent, but didn't get telescoping either.

Edit
Turns out, there is a related product which has more simple form (I derived it numerically, I don't know how to prove it either, except in the way @Did did).

$$\prod_{k=0}^\infty \frac{2}{1+x^{1/2^k}}=\frac{2}{x^2-1} \ln x$$



So far @Did's proof looks like magic. Is there any way to derive this product by using the definition and properties of the natural logarithm?


You could more or less infer the results from the infinite product you have posted.  Denote
$$f_k(t)=\frac{2}{1+t^{1/2^k}},$$
we have
$$\frac{f_k(x)~f_k(x^{1/2})}{f_k(x^{3/2})}=\frac{2}{1+x^{1/2^k}}\frac{1+\left(x^{1/2^{k+1}}\right)^3}{1+x^{1/2^{k+1}}}\\
=\frac{2}{1+x^{1/2^k}}\left(1+x^{1/2^{k+1}}+x^{1/2^k}\right)=2\left(1-\frac{x^{1/2^{k+1}}}{1+x^{1/2^k}}\right),$$
which is the term in your product.  Therefore
$$P(x)=\frac{\prod_{k=0}^\infty f_k(x)~\prod_{k=0}^\infty f_k(x^{1/2})}{\prod_{k=0}^\infty f_k(x^{3/2})}=\frac{\frac{2\ln x}{x^2-1}\frac{\ln x}{x-1}}{\frac{3\ln x}{x^3-1}}=\frac{2(x^2+x+1)\ln x}{3(x^2-1)}.$$

To make it a bit more rigorous, consider the following equality
$$x^2-1=2^{n+1}(x^{1/2^n}-1)\prod_{k=0}^{n}\frac{x^{1/2^k}+1}{2}$$
by applying $x^2-y^2=(x-y)(x+y)$ repeatedly, so the equality you posted is equivalent to
$$\lim_{n\to\infty}2^n\left(x^{1/2^n}-1\right)=\lim_{y\to0}\frac{x^y-1}{y}=\lim_{y\to0}~x^y\ln x=\ln x,$$
where $y=1/2^n$ and we have used L'Hôpital's rule.  You can get similar expressions for $x^{1/2}$ and $x^{3/2}$.  Multiply them together as what has been done above, then take the limit, you would get the same answer.
