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

$${\underset{n\to \mathrm{\infty }}{limsin}}^2\pi \sqrt{n^2+n}=?$$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

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

Commented by mathdanisur last updated on 05/Jul/21

$$coolthanksSer$$

Commented by mathdanisur last updated on 06/Jul/21

$$Ser,cantheanswer0?$$

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

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

Commented by mathdanisur last updated on 05/Jul/21

$$coolthanksSer$$

Commented by mathdanisur last updated on 06/Jul/21

$$Ser,cantheanswer0?$$

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

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

Commented by mathdanisur last updated on 05/Jul/21

$$coolthanksSer$$

Commented by mathdanisur last updated on 06/Jul/21

$$Ser,cantheanswer0?$$

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