Calculate $\lim_{s\rightarrow \infty}\frac{\sqrt{s}\Gamma(2s+1)}{(2^s\Gamma(s+1))^2}$

Question

I have to calculate $$\lim_{s\rightarrow \infty}\frac{\sqrt{s}\Gamma(2s+1)}{(2^s\Gamma(s+1))^2}$$

I came to this limit $$\lim_{s\rightarrow \infty}\frac{1}{2^{2s-1}s^{\frac{1}{2}}B(s,s)}$$

Any help, how to continue?

Answer 1

Too long for a comment

The Stirling's approximation is, probably, the shortcut to the answer. But we can also play with the asymptotics of Beta-function. $$B(s;s)=\int_0^1(1-x)^{s-1}x^{s-1}dx=\int_0^1e^{(s-1)(\ln(1-x)+\ln x)}dx$$ Using the Laplace's method to find the asymptotics for $s\to\infty$ $$f(x)=\ln(1-x)+\ln x\,\,\Rightarrow\,\, f'(x)=\frac{1}{x}-\frac{1}{1-x}=\frac{1-2x}{x(1-x)}$$ $f'(x)=0$ at $\,x-\frac{1}{2};\,\, f\big(\frac{1}{2}\big)=2\ln\frac{1}{2}; \,\,f''\big(\frac{1}{2}\big)=-8$

$$f(x)=f\big(\frac{1}{2}\big)+f'\big(\frac{1}{2}\big)\big(x-\frac{1}{2}\big)+\frac{1}{2}f''\big(\frac{1}{2}\big)\big(x-\frac{1}{2}\big)^2+...$$ $$B(s;s)\sim e^{(s-1)2\ln\frac{1}{2}}\int_0^1e^{-4(s-1)(x-\frac{1}{2})^2}dx$$

Expanding the integration to $\pm\infty$, the main asymptotics term $$B(s;s)\sim\frac{1}{2^{2(s-1)}}\sqrt\frac{\pi}{4(s-1)}\sim\frac{\sqrt\pi}{2^{2s-1}\sqrt s}$$