Ytukyg

Show that

$$lim _{x to infty} log left( frac{ sqrt{x} Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right) = frac{1}{2} log(2),$$ where $Gamma$ is the Gamma function.

I reduced this problem to calculate the limit:
$$lim_{x to infty} frac{Bleft(frac{x}{2},frac{x}{2}right)}{Bleft(frac{x+1}{2},frac{x+1}{2}right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.

Any help will be very appreciated, thanks!

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begin{align}
&bbox[10px,#ffd]{lim _{x to infty}
lnpars{root{x}Gammapars{x/2} over
Gammapars{bracks{x + 1}/2}}} =
{1 over 2},lnpars{2} + lim _{x to infty}
lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
\[5mm] = &
{1 over 2},lnpars{2} + lim _{x to infty}
lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
\[5mm] = &
{1 over 2},lnpars{2} + lim _{x to infty}
lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
expo{-pars{x - 1}} over
root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
\[5mm] = &
{1 over 2},lnpars{2} + lim _{x to infty}
lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
root{expo{}} over
x^{x}bracks{1 - pars{1/2}/x}^{x}}}
\[5mm] = &
{1 over 2},lnpars{2} +
underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
_{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
end{align}

Hint

Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.

$$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$

Now, using Stirling approximation, as Kemono Chen commented,
$$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
)right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$ Use this formula, simplify and continue with Taylor expansion to get
$$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.