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

Question

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

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.

Answer 1

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.