Question

How do you find the nth root of n!?

Answer 1

`(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 `n!` using the recursive definition:

`{ (0! = 1), (n! = n*(n-1)!) :}`

then take the `n`th root.

However, note that a good approximation to `n!` is given by the Stirling series:

`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 `10` significant digits.

See also https://socratic.org/s/aAV5WyiQ for an alternative formula for `n!`