Question Number 53466 by maxmathsup by imad last updated on 22/Jan/19
$${let}\:{U}_{{n}} =\frac{\mathrm{1}}{{n}}\left\{\prod_{{k}=\mathrm{1}} ^{{n}} \left({n}+{k}\right)\right\}^{\frac{\mathrm{1}}{{n}}} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$
Answered by Prithwish sen last updated on 22/Jan/19
![=lim_(n→∞) [(1+(1/n))(l+(2/n)).......(1+(n/n))]^(1/n) lim_(n→∞) log u_n = lim_(n→∞) (1/n) Σ_(r=1) ^n log(1+(r/n)) = ∫_0 ^1 log(1+x)dx = [(1+x)log(1+x)−x]_0 ^1 = 2log2 − 1 = log4 −1 ∴ lim_(n→∞) u_n = e^(log4−1) = (4/e)](https://www.tinkutara.com/question/Q53481.png)
$$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)\left(\mathrm{l}+\frac{\mathrm{2}}{\mathrm{n}}\right)…….\left(\mathrm{1}+\frac{\mathrm{n}}{\mathrm{n}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \mathrm{log}\:\mathrm{u}_{\mathrm{n}} =\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{r}}{\mathrm{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\left[\left(\mathrm{1}+\mathrm{x}\right)\mathrm{log}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2log2}\:−\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{log4}\:−\mathrm{1} \\ $$$$\:\:\:\:\:\therefore\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \:\mathrm{u}_{\mathrm{n}} \:=\:\mathrm{e}^{\mathrm{log4}−\mathrm{1}} \:=\:\frac{\mathrm{4}}{\mathrm{e}} \\ $$