Evaluate $\lim_{n\rightarrow \infty} \Gamma(n+\frac{1}{2})/ \left( \sqrt{2n\pi} \Gamma(n) \right)$ using Stirling's formula.

Question

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 1

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

Answer 2

By approaching this in a more general way, first we establish the following lemma.

Lemma: Let $p,q\in\mathbb{R}^{+}$. Then, $$ \lim_{p\rightarrow \infty} \frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}} = \frac{1}{2^{q/2}}\,. $$

Proof: It is easy to verify using standard calculus that the following claim holds.

Claim: For $a,b,c\in\mathbb{R}$,

$$ \lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^{bx+c} = e^{ab}\,. $$ Now, knowing that with Stirling's formula $ \Gamma(n) \approx \sqrt{2\pi}(n-1)^{n-\frac{1}{2}}e^{-n}$, it follows that \begin{align*} \frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}} &\approx \frac{\sqrt{2\pi} \left(\frac{p+q}{2}-1\right)^{\frac{p+q}{2}-\frac{1}{2}}e^{-\frac{p+q}{2}}} {\sqrt{2\pi} \left(\frac{p}{2}-1\right)^{\frac{p}{2}-\frac{1}{2}}e^{-\frac{p}{2}}\,p^{q/2}} \\ &= \frac{2^{-\frac{p+q}{2}+\frac{1}{2}}\, p^{\frac{p+q}{2}-\frac{1}{2}} \left(1+\frac{q-2}{p}\right)^{\frac{p+q}{2}-\frac{1}{2}}e^{-\frac{p+q}{2}}} {2^{-\frac{p}{2}+\frac{1}{2}}\, p^{\frac{p}{2}-\frac{1}{2}} \left(1+\frac{-2}{p}\right)^{\frac{p}{2}-\frac{1}{2}}e^{-\frac{p}{2}}\,p^{q/2}}\\ &= \frac{1}{2^{q/2}}\cdot \frac{\left(1+\frac{q-2}{p}\right)^{\frac{p+q}{2}-\frac{1}{2}}}{\left(1+\frac{-2}{p}\right)^{\frac{p}{2}-\frac{1}{2}}}\cdot e^{-\frac{q}{2}}\,. \end{align*} Taking the limit both sides knowing that both converge to the same value, it easily follows with our claim that \begin{align*} \lim_{p\rightarrow \infty} \frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}} =& \lim_{p\rightarrow \infty} \frac{1}{2^{q/2}}\cdot \frac{\left(1+\frac{q-2}{p}\right)^{\frac{1}{2}p + \frac{q}{2}-\frac{1}{2}}}{\left(1+\frac{-2}{p}\right)^{\frac{1}{2}p-\frac{1}{2}}}\cdot e^{-\frac{q}{2}} \\ =&\, \frac{1}{2^{q/2}}\cdot \frac{e^{\frac{1}{2}(q-2)}}{e^{\frac{1}{2}(-2)}}\cdot e^{-\frac{q}{2}} \\ =&\, \frac{1}{2^{q/2}} \end{align*} This proves our lemma.

THE MOMENT YOU'VE BEEN WAITING FOR:

As a special case, when $p=2n$ and $q=1$ $$ \lim_{n\rightarrow \infty} \frac{\Gamma\left(n + \frac{1}{2}\right)}{\Gamma\left(n\right)\sqrt{2n}} = \frac{1}{\sqrt{2}}\,. $$ Therefore, it immediately follows from the limit of a constant times a function that $$ \lim_{n\rightarrow \infty} \frac{\Gamma\left(n + \frac{1}{2}\right)}{\Gamma\left(n\right)\sqrt{2n\pi}} = \frac{1}{\sqrt{2\pi}}\,. $$