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Question Number 167468 by LEKOUMA last updated on 17/Mar/22

Answered by MJS_new last updated on 17/Mar/22

$$\ln n!\approx \ln \left({\left(\frac{n}{\mathrm{e}}\right)}^n\sqrt{2\pi n}\right)=\frac{\ln 2\pi }{2}-n+(n+1/2)\ln n$$$$\ln n^n=n\ln n$$$$\frac{\ln n!}{\ln n^n}=1+\frac{\ln 2\pi }{2n\ln n}+\frac{1}{2n}-\frac{1}{\ln n}$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}1$$

Commented by LEKOUMA last updated on 17/Mar/22

$$Thanks$$

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