Method to obtain values of Gauss hypergeometric function.

Question

Context:

Reading this interesting paper: (https://arxiv.org/pdf/1607.04742.pdf) ("Special values of Gauss’s hypergeometric series derived from Appell’s series F1 with closed forms" by Akihito Ebisu). We find that the author obtains a lot of values of Gauss Hypergeometric function with a new approach. In particular we have: $$\sum_{n=0}^{\infty}\frac{(3n)!(2n)!(-1)^n}{n!^5}\left(\frac{1}{8640} \right)^n=\frac{9\cdot 5^{2/3}\cdot\Gamma(1/3)^6}{100\pi^4},\tag{1}$$ $$\sum_{n=0}^{\infty}\frac{(4n)!(-1)^n}{n!^4}\left(\frac{1}{6635520} \right)^n=\frac{3\cdot 5^{1/4}\cdot\Gamma(1/4)^4}{25 \pi^3}.\tag{2}$$

Guillera obtains values of Gauss Hypergeometric function with the WZ method: (https://arxiv.org/pdf/2001.08104.pdf), and then with Clausen's formula deduces some Ramanujan series for $1/\pi$.

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Updated Level 3:

From the comments of Tito Piezas III we are going to give the transformation that allows us to obtain (1) using Ebisu's values. We take this version of Clausen's formula: $$\left(\sum_{n=0}^{\infty}\frac{f(a)f(b)}{f(2b)n!}x^n\right)^2=(1-x)^{-a}\sum_{n=0}^{\infty}\frac{f(a)f(b)f(2b-a)}{f(b+1/2)f(2b)n!}\left(\frac{x^2}{4(x-1)} \right)^n\tag{3}$$ Where $f(x)=\frac{\Gamma{(x+n)}}{\Gamma{(x)}}=(x)_{n},$ is the Pochhammer symbol. Setting $a=1/3$ and $b=1/2$ in $(3)$ we have:

$$_2F_1\left(\frac{1}{3},\frac{1}{2};1;x\right)^2=\frac{1}{(1-x)^{1/3}}\sum_{n=0}^{\infty}\frac{(2n)!(3n)!}{n!^5}\left(\frac{x^2}{432(x-1)}\right)^{n}\tag{4}$$

Equivalently,

$$(1-x)^{1/3}\,_2F_1\left(\frac{1}{3},\frac{1}{2};1;x\right)^2= \,_3F_2\big(\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{2};1,1;108z\big) =\sum_{n=0}^{\infty}\frac{(2n)!(3n)!}{n!^5} z^n\tag{5}$$

where $z=\dfrac{x^2}{432(x-1)}$, or solving for $x$,

$$x=12\Big(18z\pm\sqrt{3z(108z-1)}\Big)$$

Thus a single $z$ yields two $x$. For example,

Let $z=-\frac1{8640}$ from $(1)$ yields $x = \color{blue}{-\frac14,\frac15}$ so,

$$\small\left(1+\frac14\right)^{1/3}\,_2F_1\left(\frac{1}{3},\frac{1}{2};1;-\frac14\right)^2 = \left(1-\frac15\right)^{1/3}\,_2F_1\left(\frac{1}{3},\frac{1}{2};1;\frac15\right)^2 = \frac{3}{5^{4/3}}\times\frac{3\,\Gamma(1/3)^6}{4\pi^4}$$

Let $z=-\frac1{326592}$ from Piezas's answer yields $x = \color{blue}{-\frac1{27},\frac1{28}}$ so,

$$\small\left(1+\frac1{27}\right)^{1/3}\,_2F_1\left(\frac{1}{3},\frac{1}{2};1;-\frac1{27}\right)^2 = \left(1-\frac1{28}\right)^{1/3}\,_2F_1\left(\frac{1}{3},\frac{1}{2};1;\frac1{28}\right)^2 = \frac{9}{7^{5/3}}\times\frac{3\,\Gamma(1/3)^6}{4\pi^4}$$

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Updated Level 2:

To obtain $(2)$, we again use $(3)$ and set $a=1/4$ and $b=1/2$ to get:

$$_2F_1\left(\frac{1}{4},\frac{1}{2};1;x\right)^2=\frac{1}{(1-x)^{1/4}}\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\left(\frac{x^2}{1024(x-1)}\right)^{n}\tag{6}$$

Equivalently,

$$(1-x)^{1/4} _2F_1\left(\frac{1}{4},\frac{1}{2};1;x\right)^2= \,_3F_2\big(\tfrac{1}{4},\tfrac{3}{4},\tfrac{1}{2};1,1;256z\big)=\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}z^{n}\tag{7}$$

where $z=\dfrac{x^2}{1024(x-1)}$, or solving for $x$,

$$x=32\Big(16z\pm\sqrt{z(256z-1)}\Big)$$

So again a single $z$ yields two $x$. For example,

Let $z=-\frac{1}{12288}$ yields $x = \color{blue}{-\frac13,\frac14}$ so,

$$\small\left(1+\frac13\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;-\frac13\right)^2 = \left(1-\frac14\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;\frac14\right)^2 = \frac{2^{-1/2}}{\,3^{5/4}}\times\frac{\Gamma(1/4)^4}{\pi^3}$$

Let $z=-\frac{1}{6635520}$ yields $x = \color{blue}{-\frac1{80},\frac1{81}}$ so,

$$\small\left(1+\frac1{80}\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;-\frac1{80}\right)^2 = \left(1-\frac1{81}\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;\frac1{81}\right)^2 = \frac{3}{5^{7/4}}\times\frac{\Gamma(1/4)^4}{\pi^3}$$

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Updated Level 2': Also:

Let $z=-\frac{1}{63^2}$ yields $x = \color{blue}{-\frac{32}{49},\frac{32}{81}}$,

$$A=\left(1+\frac{32}{49}\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;-\frac{32}{49}\right)^2 \tag{8}$$ $$B=\left(1-\frac{32}{81}\right)^{1/4}\,_2F_1\left(\frac{1}{4},\frac{1}{2};1;\,\frac{32}{81}\right)^2 \tag{9}$$

and both are equal to the $_3F_2$ series: $$A\,=\,B\,=\,\sum_{n=0}^{\infty}\frac{(-1)^n(4n)!}{63^{2n}n!^{4}}=\frac{3\cdot\Gamma{(1/7)}^2\Gamma{(2/7)}^2\Gamma{(4/7)}^2}{32\pi^{4}}\tag{10}.$$ With some effort one can show Ramanujan's series: $$\sum_{n=0}^{\infty}\frac{(-1)^n(65n+8)(4n)!}{63^{2n}n!^{4}}=\frac{9\sqrt{7}}{\pi}.\tag{11}$$

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Question: So Akihito's method is less opaque than WZ method. Do you think that we can use his method and prove all the rational series known for $1/\pi$ ?

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Reflections: This question can be misunderstood as to what the WZ method refers. I think this method is revolutionary and brilliant because it's "trivial" except for the fact when Carlson's theorem is needed. Also as Tito Piezas III has pointed, Jesús Guillera has proved series for $1/\pi^2$ using it and no other proofs are known till today. This method goes far beyond and deals with any hypergeometric series so you can see how magnificent it is.

Answer 1

The proof for the Ramanujan-Sato series $1/\pi$ for levels 1,2,3,4 has already been found by the Borwein brothers and later authors such as Chan and Cooper (who also proved the higher levels).

But it is the general proof for Ramanujan-type series for $1/\pi^2, 1/\pi^3,$ etc. that has proved elusive, and Guillera used the WZ method to prove some of the $1/\pi^2.$ The unique $1/\pi^3$ and $1/\pi^4$ are still unproven.

However, there seems to be some missing details in the OP's post. The sums above involve the generalized hypergeometric function $_pF_q$,

\begin{align} \sum_{n=0}^{\infty}\frac{(3n)!(2n)!}{n!^5}\,x^n &=\,_3F_2\Big(\tfrac13,\tfrac23,\tfrac12;1,1;108x\Big)\\ \sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\,x^n &=\,_3F_2\Big(\tfrac14,\tfrac34,\tfrac12;1,1;256x\Big)\end{align}

not the usual Gauss hypergeometric function $_2F_1$ or Appell's function $\text{F}_1$ both discussed in Akihito Ebisu's paper. For example, we have the OP's,

$$\,_3F_2\Big(\tfrac13,\tfrac23,\tfrac12;1,1;\color{blue}{\tfrac{-1\;}{80}}\Big) = \frac{9\cdot 5^{2/3}\cdot\Gamma(1/3)^6}{100\cdot\pi^{4}}$$

since $108x = \frac{-1\,}{80}.$ But in the paper we have,

$$\,_2F_1\Big(\tfrac13,\tfrac23;\tfrac76;\color{blue}{\tfrac5{32}}\Big) = \frac{\sqrt3\cdot\Gamma(1/3)^6}{20\cdot\pi^{3}}$$

and similar identities, and the paper does not seem to specify how to transform one blue number to the other blue number, nor say the identities are related to series for $1/\pi$.

So unless the transformation is clarified, then the appearance of $\Gamma(1/3)$ may only be coincidence, and Ebisu's paper was not intended to prove any series for $1/\pi$.

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Addendum:

Just in case there is indeed a general transformation, then there are several $_3F_2$ waiting to be expressed as a $_2F_1$, namely,

Level 2.

$$\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\left(\frac{-1}{3\cdot64^2} \right)^n\,=\,\frac{2^{-1/2}}{\,3^{5/4}}\times\frac{\Gamma(1/4)^4}{\pi^3}$$

$$\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\left(\frac{-1}{5\cdot1152^2} \right)^n=\frac{3}{5^{7/4}}\times\frac{\Gamma(1/4)^4}{\pi^3}$$

Level 3.

$$\sum_{n=0}^{\infty}\frac{(3n)!(2n)!\,}{n!^5}\left(\frac{-1}{3\cdot4^3} \right)^n=\frac{1}{3^{3/3}}\times\frac{3\,\Gamma(1/3)^6}{4\pi^4}$$

$$\sum_{n=0}^{\infty}\frac{(3n)!(2n)!}{n!^5}\left(\frac{-1}{5\cdot12^3} \right)^n=\frac{3}{5^{4/3}}\times\frac{3\,\Gamma(1/3)^6}{4\pi^4}$$

$$\sum_{n=0}^{\infty}\frac{(3n)!(2n)!}{n!^5}\left(\frac{-1}{7\cdot36^3} \right)^n=\frac{9}{7^{5/3}}\times\frac{3\,\Gamma(1/3)^6}{4\pi^4}$$

How to transform the negative numbers into the blue numbers of a $_2F_1$ is the big question.