Question Number 134760 by rs4089 last updated on 07/Mar/21
Answered by mathmax by abdo last updated on 07/Mar/21
![let U_n =[(((n!))/n)]^(1/n) ⇒U_n =e^((1/n)log(((n!)/n))) we have n!∼n^n e^(−n) (√(2πn))⇒((n!)/n)∼n^(n−1) e^(−n) (√(2πn)) ⇒ (1/n)log(((n!)/n)) ∼(1/n){(n−1)log(n)−n +(1/2)log(2πn)} =(1−(1/n))log(n)−1 +((log(2πn))/(2n)) =logn−1−((logn)/n)+((log(2πn))/(2n))→+∞ ⇒ U_n →+∞](Q134824.png)
$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left[\frac{\left(\mathrm{n}!\right)}{\mathrm{n}}\right]^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\frac{\mathrm{n}!}{\mathrm{n}}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\Rightarrow\frac{\mathrm{n}!}{\mathrm{n}}\sim\mathrm{n}^{\mathrm{n}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\frac{\mathrm{n}!}{\mathrm{n}}\right)\:\sim\frac{\mathrm{1}}{\mathrm{n}}\left\{\left(\mathrm{n}−\mathrm{1}\right)\mathrm{log}\left(\mathrm{n}\right)−\mathrm{n}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\pi\mathrm{n}\right)\right\} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}\right)\mathrm{log}\left(\mathrm{n}\right)−\mathrm{1}\:+\frac{\mathrm{log}\left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{2n}}\:=\mathrm{logn}−\mathrm{1}−\frac{\mathrm{logn}}{\mathrm{n}}+\frac{\mathrm{log}\left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{2n}}\rightarrow+\infty\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} \rightarrow+\infty \\ $$$$ \\ $$