Question Number 145531 by mathdanisur last updated on 05/Jul/21
$$\underset{{n}\rightarrow\infty} {{lim}sin}^{\mathrm{2}} \pi\:\sqrt{{n}^{\mathrm{2}} +{n}}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 05/Jul/21
$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:=\:\mathrm{sin}^{\mathrm{2}} \pi{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{sin}^{\mathrm{2}} \pi{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{sin}^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{cos}^{\mathrm{2}} \left(\pi{n}\right)\:=\:\left(\left(−\mathrm{1}\right)^{{n}} \right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$
Commented by mathdanisur last updated on 05/Jul/21
$${cool}\:{thanks}\:{Ser} \\ $$
Commented by mathdanisur last updated on 06/Jul/21
$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$
Answered by mathmax by abdo last updated on 05/Jul/21
$$\mathrm{we}\:\mathrm{have}\:\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}=\mathrm{n}\pi\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}=\mathrm{n}\pi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sim\mathrm{n}\pi\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right)\right\} \\ $$$$\sim\mathrm{n}\pi\:+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8n}}\:+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$$\mathrm{sin}\left(\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}\right)\sim\mathrm{sin}\left(\mathrm{n}\pi+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right)\sim\left(−\mathrm{1}\right)^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}\right)\sim\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{sin}^{\mathrm{2}} \pi\sqrt{\mathrm{n}^{\mathrm{2}} \:+\mathrm{n}}=\mathrm{1} \\ $$
Commented by mathdanisur last updated on 05/Jul/21
$${cool}\:{thanks}\:{Ser} \\ $$
Commented by mathdanisur last updated on 06/Jul/21
$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$
Answered by Dwaipayan Shikari last updated on 05/Jul/21
$${sin}^{\mathrm{2}} \left(\pi\sqrt{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$$${sin}^{\mathrm{2}} \left(\pi\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\right) \\ $$$$=\mathrm{sin}\:^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\approx\mathrm{sin}\:^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\left(\pm\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by mathdanisur last updated on 05/Jul/21
$${cool}\:{thanks}\:{Ser} \\ $$
Commented by mathdanisur last updated on 06/Jul/21
$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$