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Question Number 194466 by horsebrand11 last updated on 08/Jul/23

$$⋐$$

Answered by cortano12 last updated on 08/Jul/23

$$\underbrace{⋐}$$

Answered by witcher3 last updated on 08/Jul/23

$${\mathrm{e}}^{\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\ln (1+\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}}))}$$$$\ln (1+\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}}))=\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}})-\frac{1}{{2\mathrm{n}}^2}{(1+\frac{\mathrm{k}}{\mathrm{n}})}^2+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{{2\mathrm{n}}^2}{(1+\frac{\mathrm{k}}{\mathrm{n}})}^2⩽\frac{1}{2\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}}{(1+\frac{\mathrm{k}}{\mathrm{n}})}^2<\frac{1}{2\mathrm{n}}{\int }_0^1{(1+\mathrm{x})}^2\mathrm{dx}$$$$=\frac{1}{\mathrm{n}}(\frac{8}{3}-\frac{1}{3})=\mathrm{O}\left(\frac{1}{\mathrm{n}}\right)$$$$\mathrm{\Sigma }\ln (1+\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}}))=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}})+\mathrm{O}\left(\frac{1}{\mathrm{n}}\right)$$$$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\ln (1+\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}}))=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}}(1+\frac{\mathrm{k}}{\mathrm{n}})$$$$={\int }_0^1(1+\mathrm{x})\mathrm{dx}=\frac{3}{2}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}(1+\frac{1}{\mathrm{n}}+\frac{\mathrm{k}}{{\mathrm{n}}^2})={\mathrm{e}}^{\frac{3}{2}}$$

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