Question Number 74041 by FCB last updated on 18/Nov/19
Commented by mathmax by abdo last updated on 18/Nov/19
$${let}\:{A}_{{n}} =\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right)^{\frac{\mathrm{1}}{{sin}\left(\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)}} \:\Rightarrow \\ $$$$\left.{ln}\left({A}_{{n}} \right)\right)\:=\frac{\mathrm{1}}{{sin}\left(\pi\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}\right)}{ln}\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right)\:\:{we}\:{have} \\ $$$${ln}\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right)\:\sim\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:\:\:\:{and}\:\pi\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}=\pi{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \\ $$$$\sim{n}\pi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)={n}\pi\:+\frac{\pi}{\mathrm{2}{n}}\:\Rightarrow{sin}\left(\pi\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}\right)\:\sim\left(−\mathrm{1}\right)^{{n}} \:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\:\Rightarrow \\ $$$${ln}\left({A}_{{n}} \right)\:\sim\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}×\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=\frac{\mathrm{1}}{{n}\:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\sim\frac{\mathrm{1}}{{n}×\frac{\pi}{\mathrm{2}{n}}}\:=\frac{\mathrm{2}}{\pi} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{lnA}_{{n}} =\frac{\mathrm{2}}{\pi}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:{A}_{{n}} ={e}^{\frac{\mathrm{2}}{\pi}} \\ $$$${any}\:{answer}\:{given}\:{is}\:{not}\:{correct}...! \\ $$
Commented by FCB last updated on 18/Nov/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by abdomathmax last updated on 18/Nov/19
$${you}\:{are}\:{welcome}. \\ $$