Why $\lim_{n\to\infty} \frac{\Gamma\left(n - \frac{1}{2}\right)}{\Gamma\left(n\right)} = e^{-\frac{1}{2}}$

Question

Why is it true that $$\lim_{n\to\infty} \frac{\Gamma\left(n - \frac{1}{2}\right)}{\Gamma\left(n\right)} = e^{-\frac{1}{2}}$$

I only know the integral definition of gamma function. My notes writes $$\lim_{n\to\infty} \frac{\Gamma\left(n - \frac{1}{2}\right)}{\Gamma\left(n\right)} = \lim_n\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right) = e^{-1/2}$$

I don't know why the first equality holds, nor why the second equality holds...

Answer 1

This is not true. If it were true, we'd have

\begin{eqnarray*} \lim_{n\to\infty}\frac{\Gamma\left(n-1\right)}{\Gamma(n)} &=& \lim_{n\to\infty}\left(\frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\cdot\frac{\Gamma(n-1)}{\Gamma\left(n-\frac12\right)}\right) \\ &=& \lim_{n\to\infty}\frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\cdot\lim_{n\to\infty}\frac{\Gamma(n-1)}{\Gamma\left(n-\frac12\right)} \\ &=& \mathrm e^{-\frac12}\cdot\mathrm e^{-\frac12} \\ &=& \mathrm e^{-1}\;, \end{eqnarray*}

whereas this limit is in fact $0$ (since the quotient is $\frac1{n-1}$).

The correct statement that this might be intended to state might be

$$ \frac{\Gamma\left(n-\frac12\right)}{\Gamma(n)}\sim n^{-\frac12}\;. $$

Answer 2

The first equality is not true reads the comments.

for the second equality cannot be true

note that by telescopic sum we have

$$ \ln\left(\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right)\right) = \sum_{k=1}^{n-1}\ln\left(2k - 1\right)-\ln\left(2k \right)= \sum_{j=2}^{2n-2}\ln\left(j- 1\right)-\ln\left(j \right)\\ = \ln\left(2- 1\right)-\ln\left(2n-2 \right)\to-\infty $$

Whence

$$\prod_{k=1}^{n-1}\left(1 - \frac{1/2}{k}\right) \to0.$$

Answer 3

Considering the general case$$f_a=\frac{\Gamma (n-a)}{\Gamma (n)}\implies \log(f_a)=\log (\Gamma (n-a))-\log (\Gamma (n))$$ and use Stirling approximation to get $$\log(f_a)=-a \log \left({n}\right)+\frac{a(a+1)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor series $$f_a=e^{\log(f_a)}=n^{-a} \left(1+\frac{a(a+1)}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$