Stirling's formula

Stirling showed that with the constant

k = e

the sequence

(x_{n})

with

x_{n} = n! \Large\frac{k^{n}}{n^{n+\frac 1 2}}

converges to

\sqrt{2\pi}

.

This means that for large

n

we have the approximation

n! \approx \sqrt {2\pi n} \Large (\frac{n}{e})\large ^{n}