Question Number 89341 by Henri Boucatchou last updated on 17/Apr/20
$$\underset{{n}\rightarrow\infty} {{lim}}\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)}=? \\ $$
Commented by Henri Boucatchou last updated on 17/Apr/20
$$\boldsymbol{{Check}}\:\:\boldsymbol{{once}}\:\:\boldsymbol{{more}}\:\:\boldsymbol{{Sir}},\:\:\:\boldsymbol{{ln}}\underset{{i}} {\prod}\boldsymbol{{a}}_{\boldsymbol{{i}}} =\underset{\boldsymbol{{i}}} {\sum}\boldsymbol{{lna}}_{\boldsymbol{{i}}} … \\ $$
Commented by mathmax by abdo last updated on 17/Apr/20
$${you}\:{are}\:{right}…{thanks}.. \\ $$
Commented by mathmax by abdo last updated on 17/Apr/20
$${A}_{{n}} =\left(\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow{ln}\left({A}_{{n}} \right)\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {ln}\left(\:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right) \\ $$$$=\frac{\mathrm{2}}{\pi}×\frac{\pi}{\mathrm{2}{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left({sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right)\rightarrow\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)\:=−{ln}\left(\mathrm{2}\right)\Rightarrow{ln}\left({A}_{{n}} \right)\rightarrow{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\:\:{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$