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$$\sum_{n=0}^{\infty}(-1)^n\binom{2n}{n}^5\left(\frac{1+4n}{2^{10n}}\right)=\frac{\Gamma^4(1/4)}{2\pi^4}$$

Is there any way to prove this?

I don't even know where to start with this one.

The following Generating Function is known.

With $K$ as the Complete Elliptic Integral of the First Kind.

$$K^2\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)=\frac{\pi^2}{4}\sum_{n=0}^{\infty}\binom{2n}{n}^3\frac{x^{2n}}{2^{6n}}$$

But to go from here to the Fifth Power will be almost impossible and then we would have to differentiate it too.

So this approach is probably not the way to go.

I believe it might be like one of those Ramanujan Series for $1/\pi$ but this one has a Gamma Term so I am not sure.

Miracle Invoker
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    duplicated:https://math.stackexchange.com/questions/3746826/ – Setness Ramesory Feb 13 '24 at 11:37
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    @SetnessRamesory That is interesting, thanks for the link!

    Using the formulas mentioned there it also seems that,

    $$\int_0^1K^2dk=\frac{\pi^4}{32}\sum_{n=0}^{\infty}\binom{2n}{n}^6\left(\frac{1+4n}{2^{12n}}\right)$$

     – Miracle Invoker Feb 13 '24 at 12:00
  • @MiracleInvoker see this also by user @ Setness Ramesory https://math.stackexchange.com/questions/4397009/evaluate-int-01kx2-textdx-int-01-fracx-sqrt1-x2-2-x2?noredirect=1&lq=1 – User-Refolio Feb 27 '24 at 21:50
  • New bounty on this... @MiracleInvoker – User-Refolio Feb 27 '24 at 21:51
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    @User Wow! You put up bounties on two questions, I will work on that question I haven't seen it before. – Miracle Invoker Feb 28 '24 at 04:17
  • Hi. You had a post about certain Fourier series involving the elliptic integrals at some point, but it was probably deleted later. Do you mind sharing a few details about that post? – user97357329 Nov 24 '24 at 11:24
  • @user97357329 Yes I indeed had a post about Fourier Series of $K(\sin x)$ (not sure if it is the same one you have in thought), I can undelete it if you want, I just felt it was not rigorous enough. – Miracle Invoker Nov 25 '24 at 13:13
  • @MiracleInvoker I found all I needed in D. H. Bailey, J. M. Borwein, D. J. Broadhurst, and M. L. Glasser. “Elliptic integral evaluations of Bessel moments and applications.” J. Phys. A: Math.Theor., 41, 52035231, 2008. Thanks for your efforts. – user97357329 Nov 28 '24 at 21:09

1 Answers1

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Using directly generalized hypergeometric functions, $$f(x)=\sum_{n=0}^{\infty}(-1)^n\,\binom{2n}{n}^5 \,\frac{(1+4n)}{x^n}$$ $$f(x)=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac {1}{2};1,1,1,1;-\frac{2^{10}}{x}\right)-$$ $$\frac {2^7}x \, _5F_4\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac {3}{2};2,2,2,2;-\frac{2^{10}}{x}\right)$$

So, $$f\left(2^{10}\right)=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac {1}{2};1,1,1,1;-1\right)-$$ $$\frac{1}{8} \, _5F_4\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac {3}{2};2,2,2,2;-1\right)$$ which is $$f\left(2^{10}\right)=\frac{16 }{\pi ^3}\big[ K(-1)\big]^2$$ that is to say the rhs.

Claude Leibovici
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