With Pochhammer symbol notation, we write
$$\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=\lim_{n\to\infty}\frac{n!\,n^z}{(z)_{n+1}}\frac{n!\,n^{z+1/2}}{(z+1/2)_{n+1}}. \tag{$\circ$}$$
Notice that
$$(z)_{n+1}(z+1/2)_{n+1}=z(z+1)\cdots(z+n)~\times~(z+1/2)(z+3/2)\cdots\left(z+\frac{2n+1}{2}\right) $$
$$=\frac{2z+0}{2}\frac{2z+2}{2}\cdots\frac{2z+2n}{2}~\times~\frac{2z+1}{2}\frac{2z+3}{2}\cdots\frac{2z+(2n+1)}{2}=\frac{(2z)_{2(n+1)}}{2^{2(n+1)}}$$
Therefore for $(\circ)$ we rewrite
$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{1}. \tag{$\bullet$}$$
We have an extra power of $2$, the Pochhammer's $\color{Red}2(n+1)$ is not compatible with the numerator's simple $(\color{Red}1\cdot n)!$ and $(\color{Red}1\cdot n)^{2z}$, and we are missing a $(1/2)_{n+1}$ down below. Now we notice that
$$\left(\frac{1}{2}\right)_{n+1}=\frac{1+0}{2}\cdots\frac{1+2n}{2}=\frac{1}{2^{n+1}}\frac{(2n+1)!}{(2\cdot1)\cdots(2\cdot n)}=\frac{(2n+1)!}{2^{2n+1}n!}.$$
We choose to further rewrite $(\bullet)$ as
$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{\color{Purple}{(1/2)_{n+1}}}\color{DarkBlue}{\frac{(2n+1)!}{2^{2n+1}n!}}=\color{LimeGreen}2\frac{(2n+1)!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{(1/2)_{n+1}}. \tag{$\triangle$}$$
There is but one thing left to do to make $(\triangle)$ into the form of $\Gamma(2z)\Gamma(1/2)$ (granted, the first $\Gamma$ will have a dummy variable $2n+1$ while the second simply has $n$). The $n^{2z}$ in the left numerator needs to be $(2n+1)^{2z}$, or something asymptotically $\sim$ to it, like - I don't know - $(\color{Orange}2n)^{2z}$?
Now can you tell why the $\color{Orange}2^{\color{Orange}{2z}\color{LimeGreen}{-1}}$ needs to be there for it to work out? ;-)
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.
$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}\color{blue}{=2\sqrt\pi}$$
Best Answer
We have (with all limits for $n\to\infty$): $$ \begin{aligned} \Gamma(z) &= \lim \frac{n!\; n^{z}}{z(z+1)(z+2)\dots(z+n)}\ , \\ \Gamma\left(z+\frac 12\right) &= \lim \frac{n!\; n^{z+\frac 12}} {\left(z+\frac 12\right)\left(z+\frac 32\right)\left(z+\frac 52\right)\dots\left(z+\frac {2n+1}2\right)}\ , \\[3mm] \Gamma(z)\Gamma\left(z+\frac 12\right) &= \lim \frac{n!^2\; n^{2z+\frac 12}}{z\left(z+\frac 12\right)(z+1)\left(z+\frac 32\right)(z+2)\left(z+\frac 52\right)\dots(z+n)\left(z+\frac {2n+1}2\right)}\ ,\\ &= \lim \frac{n!^2\; n^{2z+\frac 12}\; 2^{2n+2}}{(2z)(2z+1)(2z+2)\dots(2z+2n)(2z+2n+1)}\ , \\[3mm] \Gamma(2z) &= \lim \frac{n!\; n^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+n)}\ , \\ &= \lim \frac{(2n)!\; (2n)^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+2n)}\ ,\qquad\text{(subsequence)} \\[3mm] \frac {\Gamma(z)\Gamma\left(z+\frac 12\right)} {\Gamma(2z)} &= \lim\frac {n!^2\; n^{2z+\frac 12}\; 2^{2n+2}} {(2n)!\; (2n)^{2z}} \cdot \frac{1}{2z+2n+1}\qquad\text{ (now use Stirling)} \\ &= \lim\frac {\displaystyle {\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\; {\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\; {\color{green}{n^{2z}}}\cdot n^{\frac 12}\; \color{red}{2^{2n+\color{navy}2}} } {\displaystyle {\color{blue}{\left(\frac {{\color{red}{2}}n}e\right)^{2n}}}\sqrt {2\pi \;2n}\; \; 2^{2z}\; {\color{green}{n^{2z}}}} \cdot \frac{1}{2z+2n+1} \\ &= \lim\frac{\sqrt {2\pi}\cdot \sqrt {2\pi}}{\sqrt {2\pi}\cdot\sqrt 2} \cdot \frac{\sqrt n\cdot\sqrt n\cdot n^{\frac 12}}{\sqrt n(2z+2n+1)} \cdot \frac{\color{red}2^{\color{navy}2}}{2^{2z}} \\ &=\frac{\sqrt \pi}{2^{2z-1}}\ , \end{aligned} $$ hence the formula.