Finding lambda in a Stirling's formula problem

[james.farrow]

Help with Stirlings formula please!

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

 

[mathman]

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

 

[james.farrow]

Hi thanks for the reply, but can you explain/expand your answer 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

 

[mathman]

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.

 

[james.farrow]

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


Thanks for your help

James