A few days ago, I posted a partial solution to an old question re. showing
$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4} \, dx = \frac\pi4$$
by converting $\frac1{1+x^4}$ to a Taylor series and I ended up recovering a sum involving a ratio of Gamma functions. The missing piece is to show
$$\sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} = 2\sqrt\pi$$
Similar resolved questions seem to suggest the duplication or multiplication formulae would be of some help and/or the identities showcased here, but I'm not sure how to align the sum I have to match the forms given in the identity.
I'm most interested in a hint - what manipulations should I consider to get a form that more closely resembles the known results above? Any other methods are acceptable, but I'm specifically trying to compute this sum in order to evaluate the aforementioned integral.
By converting Gammas to factorials to binomial coefficients, I got
$$\begin{align*} \sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} &= \left(-\frac32\right)! \sum_{k=0}^\infty (-1)^k \frac{\left(k-\frac34\right)!}{\left(k+\frac34\right)!\left(-\frac32\right)!} \\[1ex] &= \Gamma\left(-\frac12\right) \sum_{k=0}^\infty (-1)^k \binom{k-\frac34}{k+\frac34} \\[1ex] &= 2\sqrt\pi \sum_{k=0}^\infty \binom{\frac{8k-11}4}{\frac{8k-1}4} \end{align*}$$
where in the last equality, I applied Pascal's rule $\binom nk=\binom{n-1}k+\binom{n-1}{k-1}$ (though I'm not 100% sure it holds for $n,k\in\Bbb Q$) to eliminate the positive terms in the series. So the new missing piece is
$$\sum_{k=1}^\infty \binom{\frac{8k-11}4}{\frac{8k-1}4} = 1$$