Show $\sum\limits_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} = -\Gamma\left(-\frac12\right) = 2\sqrt\pi$

Question

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.

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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

$$\sum_{k=1}^\infty \binom{\frac{8k-11}4}{\frac{8k-1}4} = 1$$

Answer 1

Not sure if this counts as "backtracking" since your answer in the first link already mentions this equation

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\sum\limits_{k=0}^{+\infty}(-1)^k\int\limits_0^1t^{k-3/4}(1-t)^{1/2}\,\mathrm dt$$

But it is easier to evaluate the sum by employing the Beta function and integrating the sum of the expression inside the integral. Swapping the order of the integral and summation before applying the geometric series leads to

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt$$

This resulting integral might seem more convoluted, however the simple substitution $u=\frac {1-t}{1+t}$ reduces it into another form of the beta function which can then be further simplified using Euler's Reflection formula.

\begin{align*} \int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt & =\sqrt2\int\limits_0^1u^{1/2}\left(1-u^2\right)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\int\limits_0^1u^{-1/4}(1-u)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)\\ & =\pi \end{align*}

This proves the result.

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}\color{blue}{=2\sqrt\pi}$$

Answer 2

Compute the partial sum $$S_p=\sum_{k=0}^p (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} $$ and using the Gaussian hypergeometric function $$S_p=\frac{8 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma \left(\frac{3}{4}\right)}+(-1)^p\frac{ \, _2F_1\left(1,p+\frac{5}{4};p+\frac{11}{4};-1\right) \Gamma \left(p+\frac{5}{4} \right)}{\Gamma \left(p+\frac{11}{4} \right)}$$ $$S_p=2 \sqrt{\pi }+(-1)^p \, _2\tilde{F}_1\left(1,p+\frac{5}{4};p+\frac{11}{4};-1\right) \Gamma \left(p+\frac{5}{4}\right)$$ For $p=10$, the second term is,numerically, $0.013661$. For $p=100$, it is $0.000493$.

Answer 3

Let $$S=\sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)}$$ Using beta-integral and gamma functions we can write $$S=\frac{2}{\Gamma(3/2)}\sum_{0}^{\infty} \int_{0}^{\pi/2} \sin^{2k-1/2}x~ \cos ^2 x~ dx.$$ Using IGP $$\implies S=\frac{2}{\Gamma(3/2)}\int_{0}^{\pi/2} \frac{\cos^2x}{\sqrt{\sin x} (1+\sin^2 x)} dx$$ Let $\sin x =t^2$, we get $$S=\frac{4}{\sqrt{\pi}}\int_{0}^{1}\frac{2\sqrt{1-t^4}}{1+t^4} dt$$ This integral has been evaluated in Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$, we get $$S=\frac{8}{\sqrt{\pi}} \frac{\pi}{4}=2 \sqrt{\pi}$$

Answer 4

Here is another evaluation using the hypergeometric function:

$$\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)}x^k=\frac{4\Gamma\left(\frac14\right)\,_2\text F_1\left(1,\frac14;\frac74;x\right)}{3\Gamma\left(\frac34\right)}$$

Now use this incomplete beta function identity:

$$\frac{4\Gamma\left(\frac14\right)\,_2\text F_1\left(1,\frac14;\frac74;-1\right)}{3\Gamma\left(\frac34\right)} =-\frac{(1+i)\Gamma^2\left(\frac14\right)\text B_{-1}\left(\frac34,-\frac12\right)}{\sqrt2\pi}$$

Now convert back to a hypergeometric function like so:

$$-\frac{(1+i)\Gamma^2\left(\frac14\right)\text B_{-1}\left(\frac34,-\frac12\right)}{\sqrt2\pi}=-\frac{4(1+i)(-1)^\frac34\Gamma^2\left(\frac14\right)}{3\sqrt2\pi}\,_2\text F_1\left(\frac32,\frac34; \frac74;-1\right) $$

to finish off use this identity:

$$-\frac{4(1+i)(-1)^\frac34\Gamma^2\left(\frac14\right)}{3\sqrt2\pi}\,_2\text F_1\left(\frac32,\frac34; \frac74;-1\right)=\frac{4\Gamma^2\left(\frac14\right)\Gamma\left(\frac34\right)\Gamma\left(\frac74\right)}{3\pi^\frac32}=2\sqrt\pi$$

No integration needed