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How to show that $$\lim_{n \to \infty} \left ( n-1-2\left ( \frac{\Gamma \left ( \frac{n}{2} \right ) }{\Gamma \left ( \frac{n-1}{2} \right ) } \right )^2 \right ) = \frac{1}{2}$$ I find a relevant answer here:

How to show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$

But this get the answer 1, which is wrong. Wolfram confirmed the answer $\frac{1}{2}$ is right. I'm very confused how I can solve this limit, as the method in the link seemed not applicable to this problem.

Peter Chin
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  • https://dlmf.nist.gov/5.11.iii – Gary Dec 31 '24 at 09:45
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    $$A_n=\frac{\Gamma \left ( \frac{n}{2} \right ) }{\Gamma \left ( \frac{n-1}{2} \right ) }$$

    Take logarithms, apply twice Stirling approximation (three terms would be nice) and finish with Taylor. Exponentiate since $A_n=e^{\log(A_n)}$; Square it and finish. It will give you much more than the limit.

     – Claude Leibovici Dec 31 '24 at 10:13

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Strling's approximation: $$x!=\Gamma(x+1)\sim \sqrt{\frac{\pi}{72x}}(\tfrac xe)^x(12x+1)$$ Puttting this in OP's limit function, we have $$n-1-2\left(\sqrt{\tfrac{n-3}{n-2}} (\tfrac{n-2}{n-3})^{\frac {n-2}2}\sqrt{n-3}\frac1{\sqrt{2e}}\right)^2\left(\frac{6n-11}{6n-17}\right)^2$$ In the limit $n\to\infty$, with the help of WolframAlpha, $$n-1-\frac{(n-3)^2}{n-2}(1+\frac1{4n})^2(1+\frac 2n)=\frac12+O(\frac1n).$$ I must confess that the most work belongs to WA.

Bob Dobbs
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