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I was looking at this site on section [97], Ramanujan gave the infinite sum

$$\sum_{k=0}^{\infty}(-1)^k(4k+1)\left[\frac{(2k-1)!!}{(2k)!!}\right]^5=\frac{2}{\Gamma^4\left(\frac{3}{4}\right)}\tag1$$

We conjectured a similar to $(1)$,

$$\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k-1}\left[\frac{(2k-1)!!}{(2k)!!}\right]^2=-\frac{\Gamma^2\left(\frac{1}{4}\right)+4\Gamma^2\left(\frac{3}{4}\right)}{2\pi\sqrt{2\pi}}\tag2$$

$$\sum_{k=0}^{\infty}(-1)^k\frac{k}{2k-1}\left[\frac{(2k-1)!!}{(2k)!!}\right]^2=-\sqrt{5(\pi-3)}\cdot\frac{\Gamma^2 \left(\frac{1}{4}\right)-4\Gamma^2\left(\frac{3}{4}\right)}{4\pi\sqrt{2\pi}}\tag3$$

How can we prove $(1)$ and $(2)$?

Endgame
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  • The link is broken – Frank W Jan 11 '19 at 16:34
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    This post is very minimalistic, and could be improved by adding additional context. When you consulted the sources provided in MathWorld, what questions did you have? The sources listed for 97 are Hardy 1923; Hardy 1924; Whipple 1926; Watson 1931; Bailey 1935; Hardy 1999, p. 7. For (2), please explain the reasoning that led to the conjecture. You can edit the post to add additional context. – Carl Mummert Jan 11 '19 at 16:41
  • You may just consider the Fourier-Legendre expansions of $\left[x(1-x)\right]^\nu$ with $4\nu\in\mathbb{Z}$. Suitable inner products convert hypergeometric values into values of the Beta function. $(3)$ does not look to be correct. – Jack D'Aurizio Jan 11 '19 at 17:30

1 Answers1

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After converting the double factorials to Gamma functions, Maple evaluates your first sum as $$ -2\,{\frac {{\it EllipticE} \left( i \right) }{\pi}}$$ and your second as $$ - \frac{\Gamma(3/4)^2}{\sqrt{2} \pi^{3/2}}$$

The first appears to agree numerically with your conjecture, but the second doesn't. Yours is off by about $7.69 \times 10^{-6}$.

Robert Israel
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