Evaluate $\lim_{n\rightarrow \infty} \Gamma(n+\frac{1}{2})/ \left( \sqrt{2n\pi} \Gamma(n) \right)$ 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.
 A: 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
$$
  \lim_{n\rightarrow \infty} \frac{\Gamma\left(n + \frac{1}{2}\right)}{\Gamma\left(n\right)\sqrt{2n\pi}}
  = \frac{1}{\sqrt{2\pi}}\,.
 $$
A: 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 
$$
\lim_{n\to\infty} \frac{g(n)}{f(n)}?
$$
