I need to prove: $$\lim_{p \to\infty}\frac{\Gamma \left ({\dfrac{p+1}{2}} \right )}{\Gamma(p/2)\sqrt{p\pi}}= \frac{1}{\sqrt{2\pi}}.$$

Where $\Gamma (x)$ represents the gamma function.

To provide some context, I need to find the limiting distribution of a student's t distribution , which should converge to the normal distribution. The key part of the proof translates to solving the above limit.

  • $\begingroup$ I edited the answer. It's about asymptotic series. $\endgroup$ – Von Neumann Feb 14 '18 at 8:33

As $p\to +\infty$ the Gamma function has the asymptotic approximation:

$$\Gamma\left(\frac{p+1}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$


$$\Gamma\left(\frac{p}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$

Hence substituting into the limit:

$$\frac{e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)}{e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)\cdot \sqrt{p\pi}} = \frac{\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)}{\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)\sqrt{p\pi}}$$

Which, as $p \to +\infty$ goes to

$$\frac{\sqrt{2\pi}}{2\pi} = \color{red}{\frac{1}{\sqrt{2\pi}}}$$

As wanted.

  • $\begingroup$ The gamma relations do not hold for p = 1 for the first relation and for p =2 for the second relation.... $\endgroup$ – Andy Feb 14 '18 at 6:48

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