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Question Number 206845 by hardmath last updated on 27/Apr/24

$$\mathrm{Find}:\frac{\mathrm{\infty }!}{{\mathrm{\infty }}^{\mathrm{\infty }}}=?$$

Commented by A5T last updated on 27/Apr/24

$$Doyouwishtofindthis:\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{n^n}?$$

Commented by hardmath last updated on 27/Apr/24

$$\mathrm{Yes}\mathrm{Ser}$$

Commented by A5T last updated on 27/Apr/24

$$0$$

Answered by Frix last updated on 27/Apr/24

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n!}{n^n}\underset{\mathrm{Formula}]}{\stackrel{[{\mathrm{Sterling}}^{\prime }\mathrm{s}}{=}}\underset{n\to \mathrm{\infty }}{\lim }\frac{\frac{n^n}{{\mathrm{e}}^n}\sqrt{2\pi n}}{n^n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{\sqrt{2\pi n}}{{\mathrm{e}}^n}=0$$

Commented by hardmath last updated on 28/Apr/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}$$

Commented by MM42 last updated on 30/Apr/24

$$\mathrm{\forall }n>2\to 0<\frac{n!}{n^n}<\frac{n!}{(n+1)!}=\frac{1}{n+1}$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}\frac{n!}{n^n}=0✓$$

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