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

$$provethat$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{e^n\cdot (n!)}{n^n\sqrt{n}}=\sqrt{2\pi }$$

Commented by Frix last updated on 28/Jul/23

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

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