How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$

How can I prove the following conjectured identity?
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).


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

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

Leave a Comment