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Question Number 22871 by keshavnimgade85@gmail.com last updated on 23/Oct/17

$$\mathrm{If}f\left(n\right)=\frac{1}{n}{[(n+1)(n+2)(n+3)...(n+n)]}^{1/n},$$$$\mathrm{then}\underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)=$$

Answered by ajfour last updated on 23/Oct/17

$$f\left(n\right)={[(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n})]}^{1/n}$$$$\underset{n\to \mathrm{\infty }}{\lim }\left(\ln \left[f\left(n\right)\right]\right)={\int }_0^1\ln (1+x)dx$$$$=x\ln (1+x){\mid }_0^1-{\int }_0^1\frac{x}{1+x}dx$$$$=\ln 2-[x-\ln (1+x)]{\mid }_0^1$$$$=\ln 2-1+\ln 2=2\ln 2-1$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)=e^{\ln \left(4/e\right)}=\frac{4}{\mathit{e}}.$$

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