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Stirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n. So n!/n^n ~ sqrt(2 pi n)e^-n.
by OscarCunningham 7y ago
Stirling's approximation says that n! ~ sqrt(2 pi n) (n/e)^n.
So n!/n^n ~ sqrt(2 pi n)e^-n.
- hyperpallium 7y agoCool! So the 1/e^n factor will bring it to 0, as n goes to infinity, I think?
- OscarCunningham 7y agoYeah, e^n grows much faster than sqrt(2 pi n), so sqrt(2 pi n)/e^n will tend to 0.