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