calculus - Evaluate $lim_{nrightarrow infty} Gamma(n+frac{1}{2})/ left( sqrt{2npi} Gamma(n) right)$ using Stirling's formula.

I am working on the limit

$$\displaystyle\lim_{n\rightarrow \infty} \frac{\Gamma(n+\frac{1}{2})}{ \sqrt{2n\pi}\, \Gamma(n)}\,.$$

I am thinking I may be able to use Stirling's formula, but they are slightly different, and I am having trouble relating the two. Any help is appreciated.

Stirling's formula says that the limit is 1 as $n$ approaches infinity of the following:

$$\Gamma(n) / ( \sqrt{2\pi} n^{n - \frac{1}{2}}e^{-n})$$

In particular, how do I relate $\Gamma(n)$ to $n^{n}$ and $e^{-n}$? Not sure how do deal with those two terms.

Answer

You know that

$$\lim_{n\to\infty} \frac{\Gamma(n)}{\sqrt{2\pi} n^{n - 1/2}e^{-n}} = 1. \tag1$$

It seems to me that this is how $\Gamma(n)$ relates to
$n^n$ and $e^{-n}$.
What you need to know, though, is whether this will help you
relate $\Gamma\left(n+\frac12\right)$ to $\Gamma(n)\sqrt{2n\pi},$
and if it does, how does it?

Notice what happens if we replace $n$ by $n+\frac12$ in Equation $(1).$
We get

$$\lim_{n\to\infty} \frac{\Gamma\left(n+\frac12\right)} {\sqrt{2\pi} \left(n+\frac12\right)^n e^{-(n+1/2)}} = 1. \tag2$$

Let $f(n)$ be the left-hand side of $(1)$
and $g(n)$ be the left-hand side of $(2).$
What can you say about

$$\lim_{n\to\infty} \frac{g(n)}{f(n)}?$$