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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!

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closed as off-topic by Did, Mark Fantini, voldemort, Najib Idrissi, Care Bear Oct 4 '14 at 16:11

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Mark Fantini, voldemort, Najib Idrissi, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

What are your thoughts? – DanZimm Jul 25 '14 at 7:29
@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." – user61527 Jul 25 '14 at 7:58
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. – user61527 Jul 25 '14 at 8:24
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 8:28
up vote 4 down vote accepted

Notice that


and that


as $n\to\infty$. So



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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 $\frac{\pi}{2}$).

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i am sorry i forgot pi to put now i edited... – Salih Ucan Jul 25 '14 at 8:17
Can you please expand on why it is necessary for $\sin$ to be Lipshitz? – Mark Jul 25 '14 at 8:55
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 9:20


Evaluate your function for the sequence:

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

Then do the same for this sequence:

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

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@Dark Geek For $k\not = 0$, your $n_k$ is not an integer. But perhaps the question is with $n\in \mathbb{R}$.... – Kelenner Jul 25 '14 at 8:00
I cannot see how this argument can solve the problem. – Jack D'Aurizio Jul 25 '14 at 8:12

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 .

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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. – user61527 Jul 25 '14 at 7:50
If $n \rightarrow \infty$, then $\sqrt {n^2+n} \rightarrow \infty$ – Steven Van Geluwe Jul 25 '14 at 7:54
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. – user61527 Jul 25 '14 at 7:56
Yes. You're right. I'm confusing limits of sequences and limits of real functions. – Steven Van Geluwe Jul 25 '14 at 8:14

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