Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$ This question is the last part of a problem leading to proof of Stirling's approximation. I've already proved that $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists and that $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$.
Hence, the question asks to assume $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists, and then asks to use $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$ to show $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$.
My attempt goes like this:
Since $\sqrt{x}$ is continuous, we can use $\sqrt{\underset{n\to \infty }{\text{lim}}f(n)}=\underset{n\to \
\infty }{\text{lim}}\sqrt{f(n)}$ to get $\underset{n\to \infty }{\text{lim}}\frac{2^{2 n} (n!)^2}{(2 n)! \
\sqrt{2 n+1}}=\sqrt{\frac{\pi }{2}}$.
Then we can eliminate $2^{2 n}n!$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{\sqrt{1+2 n} \left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots  \
\frac{3}{2}\frac{1}{2}}$$
Then factor out $n^n$ and adjust $\sqrt{1+2n}$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{n^{n+\frac{1}{2}} \sqrt{2+\frac{1}{n}} \left(1-\frac{1}{2n}\right) \left(1-\frac{3}{2n}\right) \cdots  \
\frac{3}{2n}\frac{1}{2n}}$$
The $\sqrt{2+\frac{1}{n}}$ factor will give a $\sqrt{2}$, so the remaining $\prod _k^n \left(1-\frac{2 k-1}{2 n}\right)$ must somehow relate to $\sqrt{2} e^{-n}$.
However, I'm not sure how to do this.
 A: Elaborating on Daniel Schepler's comment: if $L = \lim_{n \to \infty} \frac{n!}{n^{n+1/2} e^{-n}}$ then
$$\sqrt{\pi/2}
= \lim_{n\to \infty} \frac{2^{2n} (n!)^2}{(2n)!\sqrt{2n+1}}
= \lim_{n \to \infty} \frac{2^{2n}n^{2n+1} e^{-2n} L^2}{(2n)^{2n+1/2} e^{-2n} L \sqrt{2n+1}}
= L\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{2}\sqrt{2n+1}}
= \frac{L}{2}.$$

For what it's worth, I tried to continue your approach by approximating $\log \left(\prod_{k=1}^n (1 - \frac{2k-1}{2n})\right)$ with the integral $n \int_0^{1-1/(2n)} \log(1-x) \, dx$ but it was terribly messy and didn't seem tight enough to get the exact equivalence with $\sqrt{2} e^{-n}$. But I think the above approach is the intended approach.
A: I didn't use the given limit, but Stirling's approximation can solve this in one shot. You can rewrite the given limit as:
$$L = \lim_{n \to \infty}\frac{n!}{\left( \frac ne\right)^n \sqrt n}$$
Then, by stirling's approximation $n! = \sqrt{2\pi n}\ \left( \frac ne \right)^n + O(\frac{1}{n})$, the limit becomes
$$L = \frac{\sqrt{2\pi} \cdot n!}{n!} = \sqrt{2\pi}$$
A: Might this relate to the fact that we can take that product out as a limit and notice that
${\underset{n\to \infty }{\text{lim}} (1-\frac 1{n})^n}=e^{-1}$. Then if we can separate the limits and the product will be $e^{-n}$ Then maybe there is some relation to the multiples of two in the denominator and the $\sqrt {2}$? I am not sure if there are some binomial tricks (considering we can make $n$ as small as we like) to bound this changing numerator $2k+1$ when $n \to \infty$.
