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? ;-)
Best Answer
Use $\sqrt{y}=y^{\frac32-1}$.
Pull the constant multiple out, and you've calculated $\Gamma(\frac32)$. Now, apply the functional equation $\Gamma(\frac32)=\frac12\Gamma(\frac12)$.
But... be careful with that constant multiple. It looks like you lost a factor of $2$ doing the substitution.