Question Number 128124 by shaker last updated on 04/Jan/21
Answered by Dwaipayan Shikari last updated on 04/Jan/21
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{{n}}{{n}}\right)}={y} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+\frac{{k}}{{n}}\right)={logy} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}+{x}\right){dx}={logy}\Rightarrow\mathrm{2}{log}\left(\mathrm{2}\right)−\mathrm{1}={logy} \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}}{{e}} \\ $$