How do you find the nth root of n!?
1 Answer
Dec 31, 2016
#(n!)^(1/n) ~~ (2pin)^(1/(2n))(n/e)(1+1/(12n)+1/(288n^2)-139/(51840n^3))^(1/n)#
Explanation:
The precise value requires you to calculate
#{ (0! = 1), (n! = n*(n-1)!) :}#
then take the
However, note that a good approximation to
#n! ~~ sqrt(2pin)(n/e)^n(1+1/(12n)+1/(288n^2)-139/(51840n^3)+O(n^(-4)))#
Then:
#(n!)^(1/n) ~~ (2pin)^(1/(2n))(n/e)(1+1/(12n)+1/(288n^2)-139/(51840n^3))^(1/n)#
This will be good for about
See also https://socratic.org/s/aAV5WyiQ for an alternative formula for