Question Number 146886 by Khalmohmmad last updated on 16/Jul/21
Commented by EDWIN88 last updated on 16/Jul/21
$$!\mathrm{n}\:=\mathrm{n}!\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}!} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{!\mathrm{n}}{\mathrm{n}!}\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{n}!\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}!}}{\mathrm{n}!} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 16/Jul/21
$$\mathrm{A}_{\mathrm{n}} =\frac{!\mathrm{n}}{\mathrm{n}!}\:\Rightarrow\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}!}×\mathrm{n}!\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:} \frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} =−\mathrm{log2} \\ $$