Solved – Limit of \$t\$-distribution as \$n\$ goes to infinity

I found in my intro to stats textbook that $$t$$-distribution approaches the standard normal as $$n$$ goes to infinity. The textbook gives the density for $$t$$-distribution as follows, $$f(t)=frac{Gammaleft(frac{n+1}{2}right)}{sqrt{npi}Gammaleft(frac{n}{2}right)}left(1+frac{t^2}{n}right)^{-frac{n+1}{2}}$$

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

Contents

Stirling's approximation gives $$Gamma(z) = sqrt{frac{2pi}{z}},{left(frac{z}{e}right)}^z left(1 + Oleft(tfrac{1}{z}right)right)$$ so
$$frac{Gamma(frac{n+1}{2})}{Gamma(frac{n}{2})} = dfrac{sqrt{frac{2pi}{frac{n+1}{2}}},{left(frac{frac{n+1}{2}}{e}right)}^{frac{n+1}{2}}}{sqrt{frac{2pi}{frac{n}{2}}},{left(frac{frac{n}{2}}{e}right)}^{frac{n}{2}}}left(1 + Oleft(tfrac{1}{n}right)right)\= {sqrt{frac{frac{n+1}{2}}{e}}}left(1+frac1nright)^{frac{n}{2}}left(1 + Oleft(tfrac{1}{n}right)right) \= sqrt{frac{n}{2}} left(1 + Oleft(tfrac{1}{n}right)right)\ to sqrt{frac{n}{2}}$$ and you may have a slight typo in your question
In fact when considering limits as $$nto infty$$, you should not have $$n$$ in the solution; instead you can say the ratio tends to $$1$$ and it turns out here that the difference tends to $$0$$. Another point is that $$sqrt{frac{n}{2}-frac14}$$ is a better approximation, in that not only does the difference tend to $$0$$, but so too does the difference of the squares.