I want to estimate the order of $\Gamma(n/2)$.
We have from Stirling interpolation , for sufficiently large value of $n$,
\begin{align}
\Gamma(n/2)&\approx \sqrt{\frac{4\pi}{n}}\left(\frac{n}{2e}\right)^{n/2}\\
&=\sqrt{4\pi}\exp\left\{-\frac12\log n+\frac n2\log(n/2)-\frac n2\right\}
\end{align}
Now, $-\frac12\log n+\frac n2\log(n/2)-\frac n2\le n\log n-\left(\frac{\log n+n}{2}\right)\le n\log n-\sqrt{n\log n}\le n\log n.$
Thus we get, $\displaystyle \Gamma(n/2)=O(\exp(n\log n))$.
From here how can I say about the order of $\Gamma(n/2)$ ?
Best Answer
(A weak form of) Stirling's approximation gives that
$$ \log\Gamma(n) = n\log n-n-\frac{1}{2}\log n+\log\sqrt{2\pi}+o(1) $$ hence by replacing $n$ with $\frac{n}{2}$ we simply get
$$ \log\Gamma(n/2) = \frac{n}{2}\log n-\frac{1+\log 2}{2}\,n-\frac{1}{2}\log n+\log(2\sqrt{\pi})+o(1) $$ as expected from Legendre duplication formula.