Proof for approximation $\Gamma(x+\frac{1}{2})\approx\sqrt{x}\,\Gamma(x)$

Question

I am working on finding that $\Gamma(x+\frac{1}{2})$ can be approximated with $\sqrt{x}\Gamma(x)$. I have tried many approximations, e.g., Stirling, Taylor series, etc, but I cannot find an appropriate solution. Can anyone give me any guidelines? Thanks!

Update: I forgot to mention that they almost match in Matlab simulations. e.g., plot([1:100],sqrt([1:100]),'r');hold on;plot([1:100],gamma([1:100]+1/2)./gamma([1:100]),'o')

Answer 1

A particular case of Gautschi's inequality gives $$\forall x>0,\qquad \sqrt{x}<\frac{\Gamma(x+1)}{\Gamma\left(x+\frac{1}{2}\right)}<\sqrt{x+1} \tag{1}$$ hence $$\frac{x}{\sqrt{x+1}}<\frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x)}<\sqrt{x}. \tag{2}$$

Gautschi's inequality is a quite straightforward consequence of the log-convexity of the $\Gamma$ function.

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A more accurate approximation comes from Stirling's inequality $$ \log\Gamma(x) = \left(x-\frac{1}{2}\right)\log x-x+\frac{1}{2}\log(2\pi)+\frac{O(1)}{x}\tag{3} $$ leading to: $$ \log\frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma\left(x\right)} -\left(\frac{1}{2}\log x-\frac{1}{8x}\right) = \frac{O(1)}{x^3}\tag{4}$$

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Essentially the same accuracy can also be recovered from the fact that

If $\{a_n\}_{n\geq 1}$ is defined through $$ a_n=\binom{2n}{n}\frac{\sqrt{\pi\left(n+\frac{1}{4}\right)}}{4^n} $$ $\{a_n\}_{n\geq 1}$ is an increasing sequence converging towards $1$.

This fact can be proved through Weierstrass products and creative telescoping, too.
The Legendre duplication formula then gives $$ \frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x)}\sim \frac{x}{\sqrt{x+\frac{1}{4}}}. \tag{5}$$

Answer 2

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Stirling Asymptotic Expansion '$+$' Duplication Formula:

\begin{align} \ln\pars{\Gamma\pars{z + {1 \over 2}}} & \sim z\ln\pars{z} - z + {1 \over 2}\,\ln\pars{2\pi} - {1 \over 24z} + {7 \over 2880z^{3}} - {31 \over 40320z^{5}} + \mrm{O}\pars{1 \over z^{7}} \\[5mm] \ln\pars{\Gamma\pars{z}} & \sim \pars{z - \color{#f00}{1 \over 2}}\ln z - z + {1 \over 2}\ln\pars{2\pi} + \frac{1}{12z} - \frac{1}{360z^{3}} + {1 \over 1260z^{5}} + \mrm{O}\pars{1 \over z^{7}} \end{align}

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$$ \ln\pars{\Gamma\pars{z + 1/2} \over \root{z}\Gamma\pars{z}} \sim -\,{1 \over 8z} + {1 \over 192z^{3}} + \mrm{O}\pars{1 \over z^{5}}\quad \mbox{as}\ \verts{z} \to \infty $$