1. Stirling's Formula

I need to use Stirling's Formula to find a value λ so that

(4n)!/ ((2n)!)^2 ~ (λ/√n)(16)^n as n → ∞

So far I have calculated

(√(8πn)(4n/e)^4n)/ (√(4πn)(2n/e)^2n)^2

= (√8πn)/4πn

At which point I get stuck.
Can anyone tell me whether I'm on the right track or not and where to go from here. Much appreciated.

2. n!~$\displaystyle \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$

hence we can write your expression in the following way

$\displaystyle \frac{(4n)!}{((2n)!)^{2}}$~$\displaystyle \frac{\sqrt{2\pi(4n)}\left(\frac{4n}{e}\right)^{4n }}{\left[\sqrt{2\pi(2n)}\left(\frac{2n}{e}\right)^{2n}\righ t]^{2}}$

$\displaystyle =\frac{2\sqrt{2\pi n}2^{4n}\left(\frac{2n}{e}\right)^{4n}}{2(2\pi n)\left(\frac{2n}{e}\right)^{4n}}$

$\displaystyle =\frac{16^{n}}{\sqrt{2\pi n}}$

now, if we let $\displaystyle n\rightarrow n$ then the $\displaystyle 2\pi$ tern doesnt realy make any contributions, hence we can conclude that

$\displaystyle \frac{(4n)!}{((2n)!)^{2}}$~$\displaystyle \frac{16^{n}}{\sqrt{n}}$

or we can find the particular $\displaystyle \lambda$ which will then just be $\displaystyle \frac{1}{\sqrt{2\pi}}$

3. I totally understand how you've calculated ((2n)!) but I'm still having difficulty with the top part of the calculation.

Why have you used x and 3n in the first step?
Also, how does (4n/e)^4n become 2^4n(2n/e)^4n?

4. I made a typo, fixed it though

$\displaystyle \left(\frac{2\cdot2n}{e}\right)^{4n}=2^{4n}\left(\ frac{2n}{e}\right)^{4n}=\left(2^{4}\right)^{n}\lef t(\frac{2n}{e}\right)^{4n}=\left(16\right)^{n}\lef t(\frac{2n}{e}\right)^{4n}$

5. Got it!