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Question Number 145531 by mathdanisur last updated on 05/Jul/21

lim_(n→∞) sin^2 π (√(n^2 +n)) = ?

$$\underset{{n}\rightarrow\infty} {{lim}sin}^{\mathrm{2}} \pi\:\sqrt{{n}^{\mathrm{2}} +{n}}\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

sin^2 π(√(n^2 +n)) = sin^2 πn(√(1+(1/n))) sin^2 π(√(n^2 +n)) ∼_∞ sin^2 πn(1+(1/(2n))) sin^2 π(√(n^2 +n)) ∼_∞ sin^2 (πn+(π/2)) sin^2 π(√(n^2 +n)) ∼_∞ cos^2 (πn) = ((−1)^n )^2 = 1

$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:=\:\mathrm{sin}^{\mathrm{2}} \pi{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{sin}^{\mathrm{2}} \pi{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{sin}^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \pi\sqrt{{n}^{\mathrm{2}} +{n}}\:\underset{\infty} {\sim}\:\mathrm{cos}^{\mathrm{2}} \left(\pi{n}\right)\:=\:\left(\left(−\mathrm{1}\right)^{{n}} \right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$

Commented by mathdanisur last updated on 05/Jul/21

cool thanks Ser

$${cool}\:{thanks}\:{Ser} \\ $$

Commented by mathdanisur last updated on 06/Jul/21

Ser, can the answer 0?

$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$

Answered by mathmax by abdo last updated on 05/Jul/21

we have π(√(n^2 +n))=nπ(√(1+(1/n)))=nπ(1+(1/n))^(1/2) ∼nπ{1+(1/(2n))+((((1/2))((1/2)−1))/(2n^2 ))+o((1/n^3 ))} ∼nπ +(π/2)−(π/(8n)) +o((1/n^2 )) ⇒ sin(π(√(n^2 +n)))∼sin(nπ+(π/2)−(π/(8n))+o((1/n^2 )))∼(−1)^n ⇒ sin^2 (π(√(n^2 +n)))∼1 ⇒lim_(n→+∞) sin^2 π(√(n^2 +n))=1

$$\mathrm{we}\:\mathrm{have}\:\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}=\mathrm{n}\pi\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}=\mathrm{n}\pi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sim\mathrm{n}\pi\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right)\right\} \\ $$$$\sim\mathrm{n}\pi\:+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8n}}\:+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$$\mathrm{sin}\left(\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}\right)\sim\mathrm{sin}\left(\mathrm{n}\pi+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right)\sim\left(−\mathrm{1}\right)^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\pi\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}\right)\sim\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{sin}^{\mathrm{2}} \pi\sqrt{\mathrm{n}^{\mathrm{2}} \:+\mathrm{n}}=\mathrm{1} \\ $$

Commented by mathdanisur last updated on 05/Jul/21

cool thanks Ser

$${cool}\:{thanks}\:{Ser} \\ $$

Commented by mathdanisur last updated on 06/Jul/21

Ser, can the answer 0?

$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$

Answered by Dwaipayan Shikari last updated on 05/Jul/21

sin^2 (π(√((n+(1/2))^2 −(1/4)))) sin^2 (π(n+(1/2))(√(1−(1/((2n+1)^2 ))))) lim_(n→∞) sin^2 ((π/2)(2n+1)(1−(1/(2(2n+1)^2 )))) =sin^2 (πn+(π/2)−(1/(2(2n+1)^2 )))≈sin^2 (πn+(π/2)) =(±1)^2 =1

$${sin}^{\mathrm{2}} \left(\pi\sqrt{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$$${sin}^{\mathrm{2}} \left(\pi\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\right) \\ $$$$=\mathrm{sin}\:^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\approx\mathrm{sin}\:^{\mathrm{2}} \left(\pi{n}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\left(\pm\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Commented by mathdanisur last updated on 05/Jul/21

cool thanks Ser

$${cool}\:{thanks}\:{Ser} \\ $$

Commented by mathdanisur last updated on 06/Jul/21

Ser, can the answer 0?

$${Ser},\:{can}\:{the}\:{answer}\:\mathrm{0}? \\ $$

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