Prove $\, _6F_5\left(\{\frac12\}_5,\frac{5}{4};\frac{1}{4},\{1\}_4;-1\right)=\frac{2}{\Gamma \left(\frac{3}{4}\right)^4}$ and another

Question

I found $2$ intersting hypergeometric identities on this site, which ultimately reduces to $$\small \ _4F_3\left(-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,2;1\right)-\frac{1}{8} \ _4F_3\left(\frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,3;1\right)=\frac{8}{\pi ^2}$$ $$\scriptsize \ _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1\right)-\frac{1}{8} \ _5F_4\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1\right)=\frac{2}{\Gamma \left(\frac{3}{4}\right)^4}$$ How to prove these identities? Any help will be appreciated.

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Update: I found another proof for the second result. Due to certain corollary of Dougall formula (see Thm $3.4.6$ in Special functions, Andrews&Askey&Roy), i.e.

$$\, _6F_5\left(a,\frac{a}{2}+1,b,c,d,e;\frac{a}{2},a-b+1,a-c+1,a-d+1,a-e+1;-1\right)=\frac{\Gamma (a-d+1) \Gamma (a-e+1)}{\Gamma (a+1) \Gamma (a-d-e+1)} \ _3F_2(a-b-c+1,d,e;a-b+1,a-c+1;1)$$

We may set all $5$ parameters to be $\frac 12$ then recall from Clausen formula that $\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;z\right)$ $=\frac{4 K\left(\frac{1}{2} \left(1-\sqrt{1-z}\right)\right)^2}{\pi ^2}$ and special value of $K\left(\frac{1}{2}\right)$ to arrive at $$\, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{5}{4};\frac{1}{4},1,1,1,1;-1\right)=\frac{2}{\Gamma \left(\frac{3}{4}\right)^4}$$ Furthermore the very well-poised parameter pair $\frac{5}{4};\frac{1}{4}$ allows us to decompose the series, completing the proof.

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Update $2$: Using Jack's method and FL expansion given here one may prove an important result (also obtainable via Dougall $_5F_4$):

$$\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{5}{4},1-s,1-t;\frac{1}{4},s+\frac{1}{2},t+\frac{1}{2},1;1\right)=\frac{B(s+t-1,s+t-1)}{B(s,s) B(t,t)}$$

Provided that $s+t>1$. Letting $s\to\frac32, t\to \frac12$ and eliminating the very first term yields $$\, _6F_5\left(\frac{1}{2},1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{9}{4};\frac{5}{4},2,2,2,3;1\right)=\frac{32}{5} \left(1-\frac{8}{\pi ^2}\right)$$ Which is equivalent to the first result after simplifications. In one word, both $2$ identities are not-so-trivial corollaries of Dougall formula.

Answer 1

The first identity can be written as

$$ -\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4\frac{1}{(n+1)(2n-1)} +2\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4\frac{(2n+1)^3}{(n+1)^3(n+2)}=\frac{8}{\pi^2}$$

and this identity can be proved by reindexing and by considering the FL-expansions of $\left[x(1-x)\right]^\mu$ for $\mu\in\frac{1}{4}\mathbb{Z}$, as stated in the intro of this forthcoming article.

^{I hope this will speed up the review process, more than a year has passed since the submission of this article, which might bring something useful to the table, i.e. the fact that fractional operators can be used together with standard operators in stressing the interplay between hypergeometric functions and Euler sums.}

Actually the first identity is equivalent to

$$ 1 = \int_{0}^{1}\sqrt{x(1-x)}\frac{dx}{\sqrt{x(1-x)}} = \sum_{n\geq 0}\frac{c_{2n} d_{2n}}{4n+1} $$ where $$ \sqrt{x(1-x)}\stackrel{L^2(0,1)}{=}\sum_{n\geq 0}c_{2n}P_{2n}(2x-1),\qquad \frac{1}{\sqrt{x(1-x)}}\stackrel{\mathcal{D}}{=}\sum_{n\geq 0}d_{2n}P_{2n}(2x-1).$$

The second identity, involving $\left[\frac{1}{4^n}\binom{2n}{n}\right]^5$, is a consequence of Brafman's formula for $s=\frac{1}{2}$ and the evaluation of FL-expansions at $x=\frac{1}{2}$, together with the special value $K\left(\frac{1}{2}\right)=\frac{1}{4\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2$. Indeed

$$ K(x)K(1-x)\stackrel{L^2(0,1)}{=}\frac{\pi^3}{8}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4(4n+1)P_{2n}(2x-1) $$ and by evaluating both sides at $x=\frac{1}{2}$ $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^5(4m+1)(-1)^m = \frac{\Gamma\left(\frac{1}{4}\right)^4}{2\pi^4}.$$

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Besides this, a closed form for the simple-looking $\phantom{}_4 F_3$

$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4 $$

still eludes me.