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Question Number 122241 by TANMAY PANACEA last updated on 15/Nov/20

$$limit...$$

Commented by TANMAY PANACEA last updated on 15/Nov/20

Commented by Dwaipayan Shikari last updated on 15/Nov/20

$$\frac{\sqrt[n]{n!}}{n}.\frac{\left(\sum ^{\mathrm{\infty }}\frac{1}{n}\right)}{log\left(n\right)}$$$$\underset{n\to \mathrm{\infty }}{\lim }(\sqrt[n]{\frac{n!}{n^n}}=\sqrt[n]{\frac{n^n}{n^n}.\frac{1}{e^n}.\sqrt{2\pi n}}).\left(\frac{\sum ^{\mathrm{\infty }}\frac{1}{n}-logn+logn}{logn}\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{e}(\frac{\gamma }{logn}+1)(\gamma =\mathcal{E}ulerianconstant)$$$$=\frac{1}{e}$$

Commented by mnjuly1970 last updated on 15/Nov/20

$$merceymrdwaiypayan$$$$yourworkandeffort$$$$isveryadmirable.$$$$godkeep{you}^{″}young{sir}^{″}$$$$andgoodluck.$$

Commented by Dwaipayan Shikari last updated on 15/Nov/20

$$Withalwayspleasuresir!:)$$

Commented by TANMAY PANACEA last updated on 15/Nov/20

$$excellent$$

Commented by TANMAY PANACEA last updated on 15/Nov/20

Commented by TANMAY PANACEA last updated on 15/Nov/20

$$li\underset{n\to \mathrm{\infty }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}\times li\underset{n\to \mathrm{\infty }}{m}\frac{(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{n}-{log}_{e}n+{log}_{e}n)}{{log}_{e}n}}{m}$$$$Firstlimitli\underset{n\to \mathrm{\infty }}{m}{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}$$$$T_n=\frac{n!}{n^n}so{T}_{n+1}=\frac{(n+1)!}{{(n+1)}^{n+1}}$$$$li\underset{n\to \mathrm{\infty }}{m}\frac{T_{n+1}}{T_n}=\frac{(n+1)n!}{(n+1){(n+1)}^n}\times \frac{n^n}{n!}={\left(\frac{n}{n+1}\right)}^n={\left(\frac{1}{1+\frac{1}{n}}\right)}^n$$$$letP=\frac{1}{n}so\underset{y\to 0}{\lim }{\left(\frac{1}{1+y}\right)}^{\frac{1}{y}}=li\underset{y\to 0}{m}\frac{1}{{(1+y)}^{\frac{1}{y}}}$$$${log}_{e}p=-li\underset{y\to 0}{m}\frac{{log}_e(1+y)}{y}=-1$$$$p=e^{-1}$$$$secondlimit$$$$li\underset{n\to \mathrm{\infty }}{m}\left(\frac{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{n}-{log}_{e}n+{log}_{e}n}{{log}_{e}n}\right)$$$$(1+\frac{\gamma }{{log}_{e}n})\to (1+\frac{\gamma }{\mathrm{\infty }})=1$$$$soanswet=\frac{1}{e}\times 1=\frac{1}{e}$$

Commented by mnjuly1970 last updated on 15/Nov/20

$$excellent.thankyousomuch{}^{″}sir{tanmay}^{″}...$$

Commented by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{let}{\mathrm{U}}_{\mathrm{n}}=\frac{{(\mathrm{n}!)}^{\frac{1}{\mathrm{n}}}}{\mathrm{n}\log \left(\mathrm{n}\right)}(1+\frac{1}{2}+...+\frac{1}{\mathrm{n}})\Rightarrow {\mathrm{U}}_{\mathrm{n}}=\frac{{(\mathrm{n}!)}^{\frac{1}{\mathrm{n}}}}{\mathrm{nlog}\left(\mathrm{n}\right)}{\mathrm{H}}_{\mathrm{n}}$$$$\mathrm{we}\mathrm{have}{\mathrm{H}}_{\mathrm{n}}\sim \ln \left(\mathrm{n}\right)+\gamma \mathrm{and}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow$$$${\mathrm{U}}_{\mathrm{n}}\sim \frac{{\left({\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}}{\mathrm{nlog}\left(\mathrm{n}\right)}(\mathrm{logn}+\gamma )$$$$=\frac{\mathrm{n}{\mathrm{e}}^{-1}{\left(2\pi \mathrm{n}\right)}^{\frac{1}{2\mathrm{n}}}}{\mathrm{nlog}\left(\mathrm{n}\right)}(\gamma +\mathrm{logn})={\mathrm{e}}^{-1}{\left(2\pi \mathrm{n}\right)}^{\frac{1}{2\mathrm{n}}}(1+\frac{\gamma }{\mathrm{logn}})$$$$\mathrm{we}\mathrm{have}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\left(2\pi \mathrm{n}\right)}^{\frac{1}{2\mathrm{n}}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{e}}^{\frac{\log \left(2\pi \mathrm{n}\right)}{2\mathrm{n}}}={\mathrm{e}}^{\mathrm{o}}=1\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}=\frac{1}{\mathrm{e}}$$$$$$

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