# Limit of tt-distribution as nn goes to infinity

I found in my intro to stats textbook that $$tt$$-distribution approaches the standard normal as $$nn$$ goes to infinity. The textbook gives the density for $$tt$$-distribution as follows, $$f(t)=Γ(n+12)√nπΓ(n2)(1+t2n)−n+12f(t)=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\sqrt{n\pi}\Gamma\left(\frac{n}{2}\right)}\left(1+\frac{t^2}{n}\right)^{-\frac{n+1}{2}}$$

I think it might be possible to show that this density converges (uniformly) to the density of normal as $$nn$$ goes to infinity. Given $$lim\lim_{n\to \infty}\left(1+\frac{t^2}{n}\right)^{-\frac{n+1}{2}}=e^{-\frac{t^2}{2}}$$, it would be great if we can show
$$\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\to \sqrt{\frac{n}{2}}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\to \sqrt{\frac{n}{2}}$$ as $$n\to \inftyn\to \infty$$, yet I am stuck here. Can someone point out how to proceed or an alternative way to show that $$tt$$-distribution converges to normal as $$n\to \inftyn\to \infty$$.

Stirling’s approximation gives $$\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right)\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right)$$ so
$$\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{e}\right)}^{\frac{n}{2}}}\left(1 + O\left(\tfrac{1}{n}\right)\right)\\= {\sqrt{\frac{\frac{n+1}{2}}{e}}}\left(1+\frac1n\right)^{\frac{n}{2}}\left(1 + O\left(\tfrac{1}{n}\right)\right) \\= \sqrt{\frac{n}{2}} \left(1 + O\left(\tfrac{1}{n}\right)\right)\\ \to \sqrt{\frac{n}{2}}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{e}\right)}^{\frac{n}{2}}}\left(1 + O\left(\tfrac{1}{n}\right)\right)\\= {\sqrt{\frac{\frac{n+1}{2}}{e}}}\left(1+\frac1n\right)^{\frac{n}{2}}\left(1 + O\left(\tfrac{1}{n}\right)\right) \\= \sqrt{\frac{n}{2}} \left(1 + O\left(\tfrac{1}{n}\right)\right)\\ \to \sqrt{\frac{n}{2}}$$ and you may have a slight typo in your question
In fact when considering limits as $$n\to \inftyn\to \infty$$, you should not have $$nn$$ in the solution; instead you can say the ratio tends to $$11$$ and it turns out here that the difference tends to $$00$$. Another point is that $$\sqrt{\frac{n}{2}-\frac14}\sqrt{\frac{n}{2}-\frac14}$$ is a better approximation, in that not only does the difference tend to $$00$$, but so too does the difference of the squares.