1. Sep 2, 2009

james.farrow

Show that

4n is equivalent to lambda/n^1/2(16)^n as n-> infinity
2n

Now using

p is eqquivalent to p!/q!(p-q)!
q

and n = (2*pi*n)^1/2(n/e)^n

so 4n = (8*pi*n)^1/2(4n/e)^4n

and 2n = (4*pi*n)^1/2(2n/e)^2n

now try as I might I can't get it to work out - thus being able to find lambda??
perhaps my algebra is not what is used to be!!

Can anyone give me some gentle prods please......

Many thanks

James

2. Sep 2, 2009

mathman

Re: Help with Stirlings formula please!!

4n! ~ (8*pi*n)1/2(4n/e)4n
(2n!)2 ~ (4*pi*n)(2n/e))4n

4n!/(2n!)2 ~ [(2*pi*n)-1/2]24n = [(2*pi*n)-1/2]16n

3. Sep 3, 2009

james.farrow

Re: Help with Stirlings formula please!!

I can see 4n! ~ (8*pi*n)1/2(4n/e)4n

but where does (2n!)^2 ~ (4*pi*n)(2n/e))4n come from?

This maybe a really silly question but here goes.....

Why is (2n!)^2?

Many Thanks

James

4. Sep 3, 2009

mathman

Re: Help with Stirlings formula please!!

Your question is for [4n,2n] (the combination symbol) which is 4n!/{(2n!)(2n!)}

(2n!)^2 is just (2n!)(2n!), where the second 2n is simply 4n-2n.

5. Sep 4, 2009

james.farrow

Re: Help with Stirlings formula please!!

Thanks Mathman, after a bit of chewing on last night I 'got it'