I like to find the following limit:
$\lim\limits_{n\to\infty}\sin^2(\pi\sqrt{ n^2+n})$.
Any ideas or insight would be greatly appreciated.
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1Hint: $\sqrt{n^2+n} \approx \sqrt{n^2}=n$. If this were equality then you would have $\sin(\pi n)=0$. Play with the error in this approximation. – Ian Mar 04 '16 at 17:56
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@Ian In fact a much better approximation is $;\sqrt{n^2+n}\cong n+\frac12;$ , and because of this the limit in fact is like $$;\sin^2\left(\pi\left(n+\frac12\right)\right)=1;$$ – DonAntonio Mar 04 '16 at 23:46
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@Joanpemo As it turns out, yes, but you can draw that conclusion by playing with the error in the leading order approximation (the $1/2$ being the first correction term). – Ian Mar 04 '16 at 23:55
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@Ian Thank you. Yes, that's true. – DonAntonio Mar 05 '16 at 08:57
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3Duplicate: $\lim\limits_{n \to \infty} \sin^{2} \left(\pi \sqrt{n^2 + n}\right)$. See also this same closed question. – Workaholic Mar 06 '16 at 18:17
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Hint: Note that $$\sqrt{n^2+n}=\sqrt{n^2+n}-n+n=\frac{n}{\sqrt{n^2+n}+n}+n=\frac{1}{\sqrt{1+1/n}+1}+n.$$
André Nicolas
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As $n$ tends to $+\infty$, you may write $$ \begin{align} \sin^2 \left( \pi \sqrt{n^2+n}\right) &=\sin^2 \left( \pi n \:\sqrt{1+\frac1n}\right)\\\\ &=\sin^2 \left( \pi n \:\left(1+\frac1{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\\\ &=\sin^2 \left( \pi n +\frac{\pi}2+\mathcal{O}\left(\frac{1}{n}\right)\right)\\\\ &=\cos^2 \left(\mathcal{O}\left(\frac{1}{n}\right)\right)\to 1. \end{align} $$
Olivier Oloa
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1@Dear Olivier Oloa, thank you very much, it was very nice and clear. – user62498 Mar 04 '16 at 18:10
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