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Let, $$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}$$ In this post, it was asked if, $$\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}=\frac{\sqrt3}{2\,\pi}\left(2\sqrt{\frac8{\sqrt\alpha}-\alpha}-2\sqrt\alpha-3\right)$$ I was curious about this result, and a little experimentation showed that, $$\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\color{blue}{\frac13}+n\right)\,n!^2\,(-256)^n}=\frac{\sqrt3}{2\,\pi}\small\left(\frac16\sqrt{\frac{16}{\sqrt{2\alpha+8}}-\frac{\alpha-8}{2}}+\frac{1}{12}\sqrt{2\alpha+8}-1\right)$$ More generally, defining, $$x=\frac{3}{4}+\frac{\pi}{2\sqrt3}\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256/c)^n}\tag1$$ $$y=3+2\sqrt3\,\pi\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\color{blue}{\frac13}+n\right)\,n!^2\,(-256/c)^n}\tag2$$ how do we show that $x,y$ in fact are algebraic numbers, being roots of the simple quartics, $$cx^4+4x=3\tag3$$ $$c\Big(\tfrac{y}{y+1}\Big)^4+4\Big(\tfrac{y}{y+1}\Big)=3\tag4$$

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Tito Piezas III
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  • $(3)$ is the consequence of Ramanujan's solution of trinomial equation in terms of power series in $c$. From $(1)$ it follows that $y/4=x+3cx'$, where $x'=\frac{dx}{dc}$. Differentiating $(2)$ gives $4(cx^3+1)x'+x^4=0$. Combining these equations one can easily show that $y=\frac{x}{1-x}$. – Martin Nicholson Nov 13 '16 at 08:17

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