Limit of Gamma functions 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.
 A: As $p\to +\infty$ the Gamma function has the asymptotic approximation:
$$\Gamma\left(\frac{p+1}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$
and
$$\Gamma\left(\frac{p}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$
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.
