Proving the duplication formula I'm trying to prove the famous Legendre duplication formula:
$$ \Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}}   $$
I have to prove this using these two identities/theorems:
$$ \Gamma(z)= \lim_{n \rightarrow \infty}  \frac{n! n^{z}}{z(z+1)...(z+n)}$$
and
$$ (2n)! \sim \frac{2^{2n} (n!)^2}{\sqrt{\pi n}}   $$ as $n \rightarrow \infty $
I put in $ z = 2z $ in the first one and obtained:
$$ \Gamma(2z) = \lim_{n \rightarrow \infty}  \frac{n! n^{2z}}{2z(2z+1)...(2z+n)} $$.
I then tried multiplying both the numerator and denominator by various factors to try to incorporate the other identity, but it was of no use. Can anyone tell me what direction I'm supposed to go from here?
 A: 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.
A: Hint:
now do the same for the product $\Gamma(z) \Gamma(z+1/2)$ and compare
A: This is not an answer since not using all the identities/theorems.
Showing that $$\Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}}$$ is the same as showing that
$$\log (\Gamma (2 z))-\log \left(\frac{2^{2 z-1} \Gamma (z) \Gamma \left(z+\frac{1}{2}\right)}{\sqrt{\pi  }}\right)=0$$ Expand the rhs in terms of logarithms; use Stirling approximation (up to any order) three times and continue with Taylor expansions for large values of $z$
