An estimator for the shape parameter for the Weibull distribution is derived from the relation:
$\displaystyle{\frac{\sigma^2}{\mu^2}} = \displaystyle{\frac{\Gamma\left(1+\frac{2}{k}\right)}{\Gamma\left(1+\frac{1}{k}\right)}} - 1$
Can the ratio
$\displaystyle{\frac{\Gamma(1+2x)}{\Gamma(1+x)}}$
be further simplified, or is that as compact as it can get?
Note that $\Gamma(z+1)=z\Gamma(z)$, and also is known the duplication formula defined by
$$\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt\pi\Gamma(2z),\quad\forall z\notin\left\{-\frac{n}2:n\in\Bbb N_{\ge 0}\right\}$$
Hence
$$\frac{\Gamma(1+2z)}{\Gamma(1+z)}=2\frac{\Gamma(2z)}{\Gamma(z)}=\frac{4^z}{\sqrt\pi}\Gamma(z+1/2)$$
what seems an expression more easy to handle when $z$ is not of the form $-n/2$ for $n\in\Bbb N_{\ge 0}$.
