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 $nto infty$.

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.

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