How can we calculate the following ?: $$\lim_{r \to \infty}\,\sqrt{\,{r \over 2}\,}\,\ {\Gamma\left(\,\left(r - 1\right)/2\,\right) \over \Gamma\left(\,r/2\,\right)} = 1$$

• Did you try using Stirling's approximation? – Theorem Oct 21 '16 at 6:11
• @Stoke No... Can Stirling formula be applied to non-integers? But for advance, Thanks. If I use the formula, I can handle it. – kayak Oct 21 '16 at 6:15
• The approximation is also good for the Gamma function. You can use it by converting back to factorial by $\Gamma(n)=(n-1)!$ – Theorem Oct 21 '16 at 6:18

By Gautschi's inequality with $x+1=\frac{r}{2}$ and $s=\frac{1}{2}$,

$$\sqrt{x}\leq \frac{\Gamma(x+1)}{\Gamma(x+s)}\leq \sqrt{x+1} \tag{1}$$ the claim immediately follows by squeezing.

• I just saw this question, but I see that you've given the answer I was going to give. (+1) – robjohn Oct 24 '16 at 22:45

Hint

For this kind of problems, Stirling approximation is the key.

Consider $$A=\sqrt\frac{r}{2} \frac{\Gamma(\frac{r-1}{2})}{\Gamma(\frac{r}{2})}\implies \log(A)=\frac 12 \log(r)-\frac 12 \log(2)+\log\left(\Gamma(\frac{r-1}{2})\right)-\log\left(\Gamma(\frac{r}{2})\right)$$ Stirling approximation write $$\log\left(\Gamma(m)\right)=m (\log (m)-1)+\frac{1}{2} \left(-\log (m)+\log (2 \pi )\right)+O\left(\frac{1}{m}\right)$$ Apply to each factorial and simplify.

If you use it and continue with Taylor series for infinitely large values of $r$, you should find $$\log(A)=\frac{3}{4 r}+O\left(\frac{1}{r}\right)$$ and now, remembering that $A=e^{\log(A)}$ and Taylor again $$A=1+\frac{3}{4 r}+O\left(\frac{1}{r}\right)$$ For illustration purposes, using $r=100$, the exact value is $\approx 1.00758$ while the simple asymptotics gives $1.00750$.


We used the $\ds{\Gamma}$-Duplication Formula to get rid of $\ds{1/2}$-factors. In this way, the $\ds{\Gamma^{\, 2}\pars{r}}$ function in the denominator is quite convenient.