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:


$$P(t)=\prod_{k=0}^\infty 2 \left(1-\frac{\exp(t/2^{k+1})}{1+\exp(t/2^{k})} \right)$$

I tried series for the exponent, but didn't get telescoping either.


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?

  • $\begingroup$ $$P(x)=\frac23\frac{x^2+x+1}{x^2-1}\ln x$$ $\endgroup$ – Did Jun 17 '16 at 7:59
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    $\begingroup$ To show this, note that $$P(x^2)=2\frac{x^3+1}{(x^2+1)(x+1)}P(x)$$ hence $$P(x^2)\frac1{\ln(x^2)}\frac{(x^4-1)(x^2-1)}{x^6-1}=P(x)\frac1{\ln x}\frac{(x^2-1)(x-1)}{x^3-1}$$ and one can guess that $$P(x)\frac1{\ln x}\frac{(x^2-1)(x-1)}{x^3-1}$$ is constant. $\endgroup$ – Did Jun 17 '16 at 8:03
  • $\begingroup$ @Yuri S, This product has been discussed here, see the accepted answer. $\endgroup$ – Neves Jun 17 '16 at 14:16
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    $\begingroup$ Re the Edit, note that, for every $n$, $$\frac{1}{1-x^{1/2^n}}\prod_{k=0}^n \frac{1}{1+x^{1/2^k}}=\frac1{1-x^2}$$ hence, when $n\to\infty$, $$\prod_{k=0}^n \frac{2}{1+x^{1/2^k}}=\frac{2^{n+1}(1-x^{1/2^n})}{1-x^2}\longrightarrow{}{}{}{}{}\frac{2\log x}{x^2-1}.$$ $\endgroup$ – Did Jun 17 '16 at 14:18
  • $\begingroup$ @Did, thank you. This is completely clear, especially since the last limit is easily solved by the Taylor series for the exponential. Still don't know why you didn't post it as an answer though $\endgroup$ – Yuriy S Jun 17 '16 at 14:32

You could more or less infer the results from the infinite product you have posted. Denote


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


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.

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    $\begingroup$ I don't know the proof of the simple product either. I haven't seen it anywhere before, I just derived it numerically $\endgroup$ – Yuriy S Jun 17 '16 at 13:59
  • $\begingroup$ @YuriyS, I have added a brief proof, i.e., where the telescoping / cancelation happens. $\endgroup$ – Wiley Jun 17 '16 at 14:20

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