How to evaluate $\sum_{n=0}^{\infty}2^{-4n}\sum_{k=0}^{n}\frac{2^{-6k}(2k+2n)!}{k!^2(n+k)!(n-k)!}=\frac{\Gamma{(\frac{1}{4})}^2}{2\pi^{3/2}}$?

Question

Context:

While doing some investigation the following identity seems to be true: $$\sum_{n=0}^{\infty}2^{-4n}\sum_{k=0}^{n}\frac{2^{-6k}(2k+2n)!}{k!^2(n+k)!(n-k)!}=\frac{\Gamma{(\frac{1}{4})}^2}{2\pi^{3/2}}.\tag{1}$$ I computed both sides to at least 20 digits and seems to be correct. Sincerely I have no ideas to attack it but I think it could be proven with creative telescoping since the inner sum suggests it but I don't know if it's possible.

Question:

Can we prove $(1)$ with the WZ Method or any other approach? Thanks in advance for your efforts.

Answer 1

By changing the order of summation, we get $$ (*)=\sum_{k=0}^\infty A_k\quad\text{ where }A_k:=\sum_{n=k}^\infty \frac{(2k +2n)! 2^{-6 k -4 n}}{k !^{2} (n +k)!(n -k)!} $$ and \begin{align} A_k&=\frac{2^{-6k}}{k!^2 }\sum_{n=k}^\infty \frac{(2k +2n)! 2^{-4n}}{(n +k)!(n -k)!} =\frac{2^{-10k}}{k!^2 }\sum_{m=0}^\infty \frac{(4k+2m)! 2^{-4m}}{(2k+m)!m!} \\ & =\frac{2^{-10k}}{k!^2 }\times \frac{(4k)!}{(2k)!} \left(\frac{2}{\sqrt{3}}\right)^{4k+1} \end{align} using combinatorial identities and hypergeometric series.

Finally, $$ \sum_{k=0}^\infty A_k=\frac{\sqrt{\pi}}{\Gamma(3/4)^2} $$ from the expression for K(k), where $$ K(k)=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=\frac\pi{2}\sum_{\ell=0}^{\infty}\left(\frac{(2\ell)!}{2^{2\ell}\ell!^2}\right)^2k^{2\ell}. $$

Answer 2

Finally I was able to deal with it. The answer of Van der Wolf is opaque (I'm not sure of its validity) but gave me the idea to attack it. A more friendly form of the sum can be: $$S=\sum_{n=0}^{\infty}2^{-4n}\sum_{k=0}^{n}\frac{2^{-6k}(2k+2n)!}{k!^2(n+k)!(n-k)!}=\sum_{k=0}^{n}\frac{1}{2^{6k}k!^2}\sum_{n=0}^{\infty}\frac{2^{-4(n-k)}(2n)!}{n!(n-2k)!}\\=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{4n}n!}\sum_{k=0}^{n}\frac{1}{2^{2k}(n-2k)!k!^2}\tag{1}$$ We know that: $$\sum_{n=0}^{\infty}\frac{(2n)!^2}{2^{5n}n!^4}=\frac{\Gamma{(\frac{1}{4})}^2}{2\pi^{3/2}}\tag{2}.$$ Which follows from the series representations of the complete elliptic of the first kind: $$\frac{2}{\pi}K(k)=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}=\sum_{n=0}^{\infty}\frac{(2n)!^2k^{2n}}{2^{4n}n^4},\tag{3}$$ evaluated at $k=\frac{1}{\sqrt{2}}$. Comparing $(1)$ and $(2)$ suggests that: $$\sum_{k=0}^{n}\frac{1}{2^{2k}(n-2k)!k!^2}=\frac{(2n)!}{2^{n}n!^3}\tag{4}.$$ So the problem reduces to prove $(4)$. Without being much creative the remaining is an exercise of induction.

Remarks

Thanks to the comments of @user170231 there is a beatiful derivation of $(4)$ here: coin toss question

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In fact we can prove more: $$S(t)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{2^{-5k-3n}t^{2(k+n)}(2n+2k)!}{k!^2(n+k)!(n-k)!}=\frac{2}{\pi}K(t)\tag{5}.$$ $$S(t)=\sum_{k=0}^{\infty}\frac{2^{-5k}t^{2k}}{k!^2}\sum_{n=k}^{\infty}\frac{2^{-3n}t^{2n}(2n+2k)!}{(n+k)!(n-k)!}\overset{m=n+k}=\sum_{m=0}^{\infty}\frac{2^{-3m}t^{2m}(2m)!}{m!}\sum_{k=0}^{m}\frac{2^{-2k}}{k!^2(m-2k)!}\\\overset{(3),(4)}= \sum_{m=0}^{\infty}\frac{(2m)!^2t^{2m}}{2^{4m}m!^4}=\frac{2}{\pi}K(t).$$