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Question Number 53466 by maxmathsup by imad last updated on 22/Jan/19

$$let{U}_n=\frac{1}{n}{\left\{\prod _{k=1}^n(n+k)\right\}}^{\frac{1}{n}}$$$$find{lim}_{n\to +\mathrm{\infty }}{U}_n$$

Answered by Prithwish sen last updated on 22/Jan/19

$$={\lim }_{\mathrm{n}\to \mathrm{\infty }}{[(1+\frac{1}{\mathrm{n}})(\mathrm{l}+\frac{2}{\mathrm{n}}).......(1+\frac{\mathrm{n}}{\mathrm{n}})]}^{\frac{1}{\mathrm{n}}}$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}\log {\mathrm{u}}_{\mathrm{n}}={\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\log (1+\frac{\mathrm{r}}{\mathrm{n}})$$$$={\int }_0^1\log (1+\mathrm{x})\mathrm{dx}$$$$={[(1+\mathrm{x})\log (1+\mathrm{x})-\mathrm{x}]}_0^1$$$$=2\log 2-1$$$$=\log 4-1$$$$\therefore {\lim }_{\mathrm{n}\to \mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\log 4-1}=\frac{4}{\mathrm{e}}$$

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