Conjecture: ${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}{2n (2n-3+2x)}$

Question

Is the following conjecture for $0<x<1$ true and how do we prove it?

$${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}{2n (2n-3+2x)}$$

I encountered this function when answering a question a while back, where I have shown that:

$$S=\sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=\frac{\sqrt{\pi}}{2} \int_0^1 x ~{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right) dx$$

Where $G_k$ are so called Gregory coefficients.

Now I returned to this problem and considered the ratio:

$$\frac{{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n;1 \right)}{{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n+1;1 \right)}$$

For several cases for $n \geq 2$ Wolfram Alpha shown me the terms from the product above. This, together with an obvious identity:

$$\lim_{n \to \infty} {_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};n;1 \right)=1$$

Makes me believe that the infinite product converges.

However, I don't know how to prove the ratio. Gauss's contiguous relations connect the two functions and the derivative, but I'm not sure how to use them in this case.

The product is not telescoping (which is obvious, otherwise this hypergeometric function would be a rational function).

Another question: are there more infinite product identities like this one for hypergeometric functions?

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We can rewrite the product in a more convenient way:

$$\prod_{n=2}^\infty \frac{\left(1-\frac{2-x}{2n}\right)\left(1-\frac{1-x}{2n}\right)}{\left(1-\frac{3-2x}{2n}\right)}$$

Which looks a lot like a more complicated Gamma function.

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Claude Leibovici proposed a closed form for the product which means that the following holds:

$$\sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=\int_0^1 2^x \frac{x}{1+x} \frac{\Gamma \left(x+ \frac{1}{2} \right)}{\Gamma \left( x+1 \right)}~ dx$$

The series converges extremely slow, for example:

$$\sum_{k=1}^{500} \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1}=0.50667610440857923696$$

While the integral gives:

$$\int_0^1 2^x \frac{x}{1+x} \frac{\Gamma \left(x+ \frac{1}{2} \right)}{\Gamma \left( x+1 \right)}~ dx=0.50718128906684564110$$

Which is the correct numerical value.

Unfortunately, there's no in-built Gregory coefficient list in Mathematica, so it's hard to compute them for large $k$.

Answer 1

Too long for comments.

Considering $$f=\, _2F_1\left(\frac{1}{2}-\frac{x}{2},1-\frac{x}{2};2;1\right) \qquad \text{and} \qquad g=\frac{2^{x+1}\, \Gamma \left(x+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (x+2)}$$ I was naively thinking that the series expansions of their logarithms could be of interest.

The problem is that, for $\log(f)$, I have only be able to get numerical values of the coefficients (this required a lot of computing time and only the very first ones have been possible to obtain); however, for $25$ significant figures, they perfectly matched.

For the series expansions of $\log(g)$ built at $x=0$ and $x=1$, the following values are obtained (they look quite simple). $$\left( \begin{array}{ccc} k & \text{at } x=0 & \text{at } x=1 \\ 0 & \log (2) & 0 \\ 1 & -1-\log (2) & \frac{1}{2}-\log (2) \\ 2 & \frac{1}{2}+\frac{\pi ^2}{6} & -\frac{11}{8}+\frac{\pi ^2}{6} \\ 3 & -\frac{1}{3}-2 \zeta (3) & \frac{55}{24}-2 \zeta (3) \\ 4 & \frac{1}{4}+\frac{7 \pi ^4}{180} & -\frac{239}{64}+\frac{7 \pi ^4}{180} \\ 5 & -\frac{1}{5}-6 \zeta (5) & \frac{991}{160}-6 \zeta (5) \\ 6 & \frac{1}{6}+\frac{31 \pi ^6}{2835} & -\frac{4031}{384}+\frac{31 \pi ^6}{2835} \\ 7 & -\frac{1}{7}-18 \zeta (7) & \frac{16255}{896}-18 \zeta (7) \\ 8 & \frac{1}{8}+\frac{127 \pi ^8}{37800} & -\frac{65279}{2048}+\frac{127 \pi ^8}{37800} \\ 9 & -\frac{1}{9}-\frac{170 }{3}\zeta (9) & \frac{261631}{4608}-\frac{170 }{3} \zeta (9) \\ 10 & \frac{1}{10}+\frac{73 \pi ^{10}}{66825} & -\frac{1047551}{10240}+\frac{73 \pi ^{10}}{66825} \end{array} \right)$$

which reveal interesting patterns (notice in particular that the second piece of the coefficient is the same).

Answer 2

"Are there more infinite product identities like this one for hypergeometric functions?"

Well there is the obvious $$\frac{\sin x}{x}=\,_0F_1\left(;\tfrac32;-\tfrac{x^2}{4}\right)=\prod_{n\ge1}\left(1-\frac{x^2}{\pi^2n^2}\right)$$ and the corresponding product for the cosine.

Answer 3

The conjecture is indeed true for all $x\in\Bbb C$ when $\Re\, x>-1/2$.

If $\Re\, x>-1/2$ then DLMF $15.4.20$ tells us that $$ F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)=\frac{\Gamma(2)\Gamma(x+\frac{1}{2})}{\Gamma(\frac{x}{2}+\frac{3}{2})\Gamma(\frac{x}{2}+1)}. $$ Now notice that the sum of the arguments in the numerator equals the sum of arguments in the denominator, that is, $$ 2+x+\tfrac{1}{2}=\tfrac{x}{2}+\tfrac{3}{2}+\tfrac{x}{2}+1=x+\tfrac{5}{2}; $$ thus, according to DLMF $5.8.5$ we have the following infinite product expansion $$ F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)=\prod_{k=0}^\infty\frac{(k+\frac{x}{2}+\frac{3}{2})(k+\frac{x}{2}+1)}{(k+2)(k+x+\frac{1}{2})}, $$ which can be easily derived from the product expansion of the gamma function. Substituting $n=k+2$ and simplifying then gives $$ \begin{aligned} F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right) &=\prod_{n=2}^\infty\frac{(n-\frac{1}{2}+\frac{x}{2})(n-1+\frac{x}{2})}{n(n-\frac{3}{2}+x)}\\ &=\prod_{n=2}^\infty\frac{(2n-1+x)(2n-2+x)}{2n(2n-3+2x)}, \end{aligned} $$ which is the conjectured product expansion.