Thread: Stirlings Formula

I need to use Stirlings Formula to determine a number ¬ such that

(4n)!/((2n)!)^2 ~ ¬ 2^4n/Route n as n ---> infinity

I have

By Stirling's Formula, and the Product and Quotient Rules for ~, we obtain

(4n)!/((2n)!)^2 ~ Route 8pi n(4n/e)^4n / (Route 4 pi n (2n/e)^2n)^2

= Route 8 pi n 2^4n / (route 4 pi n)^2

= ? not sure about this

¬ = ? or this

Can anyone help?

Originally Posted by Arron

I need to use Stirlings Formula to determine a number ¬ such that

(4n)!/((2n)!)^2 ~ ¬ 2^4n/Route n as n ---> infinity

I have

By Stirling's Formula, and the Product and Quotient Rules for ~, we obtain

(4n)!/((2n)!)^2 ~ Route 8pi n(4n/e)^4n / (Route 4 pi n (2n/e)^2n)^2

= Route 8 pi n 2^4n / (route 4 pi n)^2

= ? not sure about this

¬ = ? or this

Can anyone help?

A 'non very popular' version of the Stirling's formula is the following...

(1)

... so that from (1)...

(2)

... and...

(3)

Now from (2) and (3)...

(4)

... and the constant You are looking for is ...

Kind regards