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Question Number 146886 by Khalmohmmad last updated on 16/Jul/21

Commented by EDWIN88 last updated on 16/Jul/21

$$!\mathrm{n}=\mathrm{n}!\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{!\mathrm{n}}{\mathrm{n}!}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\mathrm{n}!\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}}{\mathrm{n}!}$$$$$$

Answered by mathmax by abdo last updated on 16/Jul/21

$${\mathrm{A}}_{\mathrm{n}}=\frac{!\mathrm{n}}{\mathrm{n}!}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{1}{\mathrm{n}!}\times \mathrm{n}!\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=-\log 2$$

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