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?