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can you help me with the following:

$\lim_{n \rightarrow \infty} \sin^{2} \pi \sqrt{n^2 + n}$

Thanks a lot!

Salih Ucan
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  • What are your thoughts? – DanZimm Jul 25 '14 at 07:29
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    @JanHasenbichler Your comment doesn't make any sense; whether or not $\sin$ is defined "at infinity" is irrelevant, and just because the sequence of inputs tends to infinity doesn't mean that there "is no real answer." –  Jul 25 '14 at 07:58
  • Editing the question to include a factor of $\pi$ substantially changes the techniques involved (and simplifies the question a lot). Since your original question was already answered, you should ask the corrected version as a new question, rather than editing so substantially. –  Jul 25 '14 at 08:24
  • It was just a matter of seconds to modify my answer in order to cover this easier case, but I totally agree with @T.Bongers, you should avoid modifying questions in such a radical way. – Jack D'Aurizio Jul 25 '14 at 08:28

4 Answers4

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For the limit $\lim_{n\to+\infty}\sin^2\sqrt{n^2+n}$, since $\{e^{in}\}_{n\in\mathbb{N}}$ is dense in the unit circle (it is a consequence of the irrationality of $\pi$ and the Dirichlet box principle), the same holds for the sequence $\{e^{i(n+1/2)}\}_{n\in\mathbb{N}}$ (a translation of the unit circle preserves density).

By taking imaginary parts (the projection preserves density too), we have that the sequence $\{\sin(n+1/2)\}_{n\in\mathbb{N}}$ is dense in the $[-1,1]$ interval.

Since the sine is a Lipschitz function and the distance between $\sqrt{n^2+n}$ and $n+1/2$ is bounded by $\frac{1}{4n}$, the sequence $\{\sin\sqrt{n^2+n}\}_{n\in\mathbb{N}}$ is dense in the interval $[-1,1]$, so your limit does not exist.

For the modified limit, $$\lim_{n\to +\infty}\sin^2 \pi\sqrt{n^2+n},$$ the fact that the distance between $\sqrt{n^2+n}$ and $n+\frac{1}{2}$ is bounded by $\frac{1}{4n}$, together with the fact that the sine is a Lipschitz function, ensures that the limit is just $1$ (the sine function equals $\pm 1$ in the odd multiples of $\frac{\pi}{2}$).

Jack D'Aurizio
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  • i am sorry i forgot pi to put now i edited... – Salih Ucan Jul 25 '14 at 08:17
  • Can you please expand on why it is necessary for $\sin$ to be Lipshitz? – Mark Jul 25 '14 at 08:55
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    It is not necessary, but the Lipschitz condition grants that $\lim_{n\to +\infty}\sin^2 \pi\sqrt{n^2+n}=\lim_{n\to +\infty}\sin^2\pi(n+1/2)$, since the difference of the two main terms is $O(1/n)$. – Jack D'Aurizio Jul 25 '14 at 09:20
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Notice that

$$(\sin(\pi(\sqrt{n^{2}+n}-n)))^{2}=((-1)^{n}\sin(\pi\sqrt{n^{2}+n}))^{2}=(\sin(\pi\sqrt{n^{2}+n}))^{2}$$

and that

$$\sqrt{n^{2}+n}-n=\frac{n}{\sqrt{n^{2}+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\to\frac{1}{2}$$

as $n\to\infty$. So

$$\lim_{n\to\infty}(\sin(\pi\sqrt{n^{2}+n}))^{2}=\lim_{n\to\infty}(\sin(\pi(\sqrt{n^{2}+n}-n)))^{2}=\lim_{n\to\infty}\bigg(\sin\bigg(\pi\frac{1}{\sqrt{1+\frac{1}{n}}+1}\bigg)\bigg)^{2}$$

$$=(\sin(\frac{\pi}{2}))^{2}=1$$

user71352
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Hint:

Evaluate your function for the sequence:

$$n_k = \frac{\sqrt{1+4k^2\pi^2}-1}{2}$$

Then do the same for this sequence:

$$n_k = \frac{\sqrt{1+4\pi^2(k^2+k+1/4)}-1}{2}$$

Darth Geek
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The sine function doesn't have a limit if $x \rightarrow \infty$. The sine function goes up and down between -1 and 1 infinitely many times .

Steven Van Geluwe
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  • Although this is true, it doesn't prove that the limit in this case doesn't exist. There are a not of sequences $a_k$ tending to infinity for which $\sin a_k$ has a limit. –  Jul 25 '14 at 07:50
  • If $n \rightarrow \infty$, then $\sqrt {n^2+n} \rightarrow \infty$ – Steven Van Geluwe Jul 25 '14 at 07:54
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    Yes, and as $n \to \infty$, then $n \pi \to \infty$. But $\sin n\pi \to 0$. I'm just saying that knowing that the sequence of inputs tends to $\infty$ is not enough to conclude that the sequence diverges. –  Jul 25 '14 at 07:56
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    Yes. You're right. I'm confusing limits of sequences and limits of real functions. – Steven Van Geluwe Jul 25 '14 at 08:14