The given <Euler product for the gamma function> has the finite approximants
$$
\Gamma_m(z)=\frac{m!(m+1)^z}{z(z+1)\cdots(z+m)},\qquad\Gamma_m(z)\longrightarrow\Gamma(z)
$$
away from the poles. Split the denominator of $\Gamma_{2m+1}(2z)$ into even and odd factors. The resulting exact identity is
$$
\frac{\Gamma_{2m+1}(2z)}{2^{2z}\Gamma_m(z)\Gamma_m(z+\tfrac12)}
=\frac{(2m+1)!}{2^{2m+2}(m!)^2\sqrt{m+1}}.
$$
The right side does not depend on $z$, so its <limit> is a constant $C$. Evaluate the ratio at $z=1/2$: $\Gamma(1)=1$, and the <gamma reflection formula>, together with positivity on the positive real axis, gives $\Gamma(1/2)=\sqrt\pi$. Hence
$$
\boxed{C=\frac1{2\sqrt\pi},\qquad\Gamma(2z)=\frac{2^{2z-1}}{\sqrt\pi}\Gamma(z)\Gamma(z+\tfrac12).}
$$
This is the <gamma duplication formula>. The argument first applies where the factors are finite; the identity then holds as an identity of <meromorphic functions>, with values at poles interpreted accordingly.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 7E / Solution

{c}

For $z$ away from the poles, the <gamma function> satisfies
$$
\Gamma(z)=\frac1z\prod_{n=1}^\infty(1+1/n)^z(1+z/n)^{-1}.
$$
Its finite product is $m!(m+1)^z/[z(z+1)\cdots(z+m)]$. Splitting even and odd factors gives the <gamma duplication formula>.

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 Incoming links: Euler product for the gamma function