How can I prove the following conjectured identity?

$$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\stackrel?=\frac{\sqrt3}{2\,\pi}\left(2\sqrt{\frac8{\sqrt\alpha}-\alpha}-2\sqrt\alpha-3\right),$$

where

$$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}.$$

The conjecture is equivalent to saying that $\pi\,\mathcal{S}$ is the root of the polynomial

$$256 x^8-6912 x^6-814752 x^4-13364784 x^2+531441,$$

belonging to the interval $-1<x<0$.

The summand came as a solution to the recurrence relation

$$\begin{cases}a(1)=-\frac{81\sqrt3}{512\,\pi}\\\\a(n+1)=-\frac{9\,(2n+1)(4n+1)(4 n+3)}{32\,(n+1)(3n+2)(3n+4)}a(n)\end{cases}.$$

The conjectured closed form was found using computer based on results of numerical summation. The approximate numeric result is $\mathcal{S}=-0.06339748327393640606333225108136874…$ (click to see 1000 digits).

**Answer**

According to Mathematica, the sum is

$$ \frac{3}{\Gamma(\frac13)\Gamma(\frac23)}\left( -1 + {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; -1\right) \right). $$

This form is actually quite straightforward if you write out $(4n)!$ as

$$ 4^{4n}n!(1/4)_n (1/2)_n (3/4)_n $$

using rising powers (“Pochhammer symbols”) and then use the definition of a hypergeometric function.

The hypergeometric function there can be handled with equation 25 here: http://mathworld.wolfram.com/HypergeometricFunction.html:

$$ {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; y\right)=\frac{1}{1-x^k},$$

where $k=3$, $0\leq x\leq (1+k)^{-1/k}$ and

$$ y = \left(\frac{x(1-x^k)}{f_k}\right)^k, \qquad f_k = \frac{k}{(1+k)^{(1+1/k)}}. $$

Now setting $y=-1$, we get the polynomial equation in $x$

$$ \frac{256}{27} x^3 \left(1-x^3\right)^3 = -1,$$

which has two real roots, neither of them in the necessary interval $[0,(1+k)^{-1/k}=4^{-1/3}]$, since one is $-0.43\ldots$ and the other $1.124\ldots$. However, one of those roots, $x_1=-0.436250\ldots$ just happens to give the (numerically at least) right answer, so never mind that.

Also, note that

$$ \Gamma(1/3)\Gamma(2/3) = \frac{2\pi}{\sqrt{3}}. $$

The polynomial equation above is in terms of $x^3$, so we can simplify that too a little,

so the answer is that the sum equals

$$ \frac{3^{3/2}}{2\pi} \left(-1+(1-z_1)^{-1}\right), $$

where $z_1$ is a root of the polynomial equation

$$ 256z(1-z)^3+27=0, \qquad z_1=-0.0830249175076244\ldots $$

(The other real root is $\approx 1.42$.)

How did you find the conjectured closed form?

**Attribution***Source : Link , Question Author : Hanna K. , Answer Author : Kirill*