Evaluating $\lim_{x\rightarrow\infty} \frac{1}{x}\left(\frac{\Gamma(2x-1)}{\Gamma(x-1)}\right)^{1/x}$

Question

Evaluate the limit: $$\lim_{x\rightarrow\infty} \frac{1}{x}\left(\frac{\Gamma(2x-1)}{\Gamma(x-1)}\right)^{1/x}$$

Any help will be appreciated.

I don't know how to deal with the factorial.

Answer 1

Via duplication formula for Gamma function we can write $$\Gamma(2x-1)=(2x-2)2^{2x-3}\Gamma (x-1)\Gamma (x-1/2)\pi^{-1/2}$$ and therefore the desired limit is equal to $$4\lim_{x\to\infty}\frac{1}{x}\Gamma^{1/x}(x-1/2)$$ which is same as $$4\exp\left(\lim_{x\to\infty}\frac{\log\Gamma(x-1/2)}{x}-\log x\right)$$ Stirling approximation now easily gives the limit as $$4\exp\left(\lim_{x\to\infty}\log(x-1/2)- 1-\log x\right) $$ which is same as $4/e$.

If the limit is for positive integer $x$ then we can use Cesaro-Stolz and get the same answer easily without the use of Stirling. We just need to consider $$f(x)=\frac{\Gamma (2x-1)} {x^{x}\Gamma (x-1)} $$ and consider the ratio $$\frac{f(x+1)}{f(x)}=\frac{2x(2x-1)}{x^{2}-1}\cdot\left(\frac{x}{x+1}\right)^{x}\to \frac{4}{e}$$ and hence $\sqrt[x] {f(x)} \to 4/e$.

I later checked that Cesaro-Stolz works for continuous variable $x$ also provided the function involved is bounded on every finite interval (continuity ensures this). So the above proof is valid and the result is achieved without Stirling.

Answer 2

Hint

$$A=\frac 1x\left(\frac{\Gamma (2 x-1)}{\Gamma (x-1)}\right)^{\frac{1}{x}}$$ Take logarithms $$\log(A)=-\log(x)+\frac 1x \log\left({\Gamma (2 x-1)}\right)-\frac 1x \log\left({\Gamma ( x-1)}\right)$$ and use Stirling approximation $$\log(\Gamma(p))=p (\log (p)-1)+\frac{1}{2} \left(-\log \left({p}\right)+\log (2 \pi )\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$

Edit

Since you got the result from multiple answers, doing what I suggested would lead to $$\log(A)=\log \left(\frac{4}{e}\right)-\frac{3\log (2)}{2 x}+O\left(\frac{1}{x^2}\right)$$ and continuing with Taylor $$A=e^{\log(A)}=\frac{4}{e}\left(1-\frac{3\log (2)}{2 x}\right)+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.

Answer 3

An alternative way for bringing Paramanand Singh's approach to a solution.
Once the problem has been reduced to the computation of $$ \lim_{x\to +\infty}\left(\tfrac{1}{x}\log\Gamma\left(x-\tfrac{1}{2}\right)-\log x\right)\stackrel{\text{d.H.}}{=}-1+\lim_{x\to +\infty}\left(\psi\left(x-\tfrac{1}{2}\right)-\log x\right)$$ by invoking Frullani's integral (also giving the integral representation for $\psi$) it is enough to compute

$$ \lim_{x\to +\infty}\int_{0}^{+\infty}\color{red}{\left(\frac{1}{t}-\frac{e^{t/2} }{1-e^{-t}}\right)}e^{-xt}\,dt$$ which is simply zero, since the red term is regular in a right neighbourhood of the origin and its absolute value does not grow faster that $e^{t/2}$. It follows that the wanted limit equals $\color{red}{\frac{4}{e}}$.