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?

 A: 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.
