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Question Number 134760 by rs4089 last updated on 07/Mar/21

Answered by mathmax by abdo last updated on 07/Mar/21

$$\mathrm{let}{\mathrm{U}}_{\mathrm{n}}={\left[\frac{(\mathrm{n}!)}{\mathrm{n}}\right]}^{\frac{1}{\mathrm{n}}}\Rightarrow {\mathrm{U}}_{\mathrm{n}}={\mathrm{e}}^{\frac{1}{\mathrm{n}}\log \left(\frac{\mathrm{n}!}{\mathrm{n}}\right)}$$$$\mathrm{we}\mathrm{have}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow \frac{\mathrm{n}!}{\mathrm{n}}\sim {\mathrm{n}}^{\mathrm{n}-1}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow$$$$\frac{1}{\mathrm{n}}\log \left(\frac{\mathrm{n}!}{\mathrm{n}}\right)\sim \frac{1}{\mathrm{n}}\{(\mathrm{n}-1)\log \left(\mathrm{n}\right)-\mathrm{n}+\frac{1}{2}\log \left(2\pi \mathrm{n}\right)\}$$$$=(1-\frac{1}{\mathrm{n}})\log \left(\mathrm{n}\right)-1+\frac{\log \left(2\pi \mathrm{n}\right)}{2\mathrm{n}}=\mathrm{logn}-1-\frac{\mathrm{logn}}{\mathrm{n}}+\frac{\log \left(2\pi \mathrm{n}\right)}{2\mathrm{n}}\to +\mathrm{\infty }\Rightarrow$$$${\mathrm{U}}_{\mathrm{n}}\to +\mathrm{\infty }$$$$$$

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