prove-that-lim-n-e-n-n-n-n-n-2pi-

Question Number 195259 by mathlove last updated on 28/Jul/23

p r o v e t h a t

\lim_{n \to \infty} \frac{e^{n} \cdot (n!)}{n^{n} \sqrt{n}} = \sqrt{2 π}

Commented by Frix last updated on 28/Jul/23

Easy because for n \to \infty : n! \to \frac{n^{n}}{e^{n}} \sqrt{2 π n}

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