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I was reading about this question just now Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) and I immediately got intrigued for I would have never came up with such a trick to evaluate the limit.

So I thought about: how could we prove that limit exists, without using such a trick?

For example, I studied about the Heine method, for limits with successions. This for example rapidly can prove the limit $\lim_{n\to +\infty} \sin(n)$ does not exist. Indeed I can choose a sequence $a_n = \pi n$ such that $a_n \to +\infty$ as $n\to +\infty$ or another sequence $b_n = \frac{2n+1}{2} \pi$, and if the limits exists then it does not depend upon the chosen sequence. In this case, I would get $\sin(a_n) = 0$ whereas $\sin(b_n) = \pm 1$, so the limit does not exist.

Can we use a similar argument to prove the existence of the other limit (the one in the linked question)?

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

Thank you so much.

Arctic Char
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Heidegger
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  • Which other limit? – Kurt G. Jan 07 '23 at 19:52
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    @Numb3rs there is no need to be rude. I didn't understand your question either. How could you use a method designed to prove non-existence to show that a particular limit does, in fact, exist? It's not much use to find a whole lot of subsequences which converge to the same thing, after all...that might be suggestive but unless you can show more, it probably won't lead to a proof. – lulu Jan 07 '23 at 20:02
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    The limit does not exist. Please do not be rude with people who are giving up their time to help you. – copper.hat Jan 07 '23 at 20:03
  • $\sqrt{n^2+1}$ is irrational. Thus $\pi \sqrt{n^2+1}$ is never an integer multiple of $\pi.$ Therefore $\sin\left(\pi\sqrt{n^2+1}\right)\neq 0$ for all $n\in \mathbb{N}.$ The function values actually jump around and cannot converge. – Marius S.L. Jan 07 '23 at 20:03
  • @lulu I wrote down a proof for a limit that does not exist, unsing the Heine method for limits. What I am asking is, since that method can be used wither to prove also that certain limits exist, if we could use it, or another way instad of what the other guy did – Heidegger Jan 07 '23 at 20:04
  • @copper.hat I apologise if I was rude. In the other answer though, they say it exists... – Heidegger Jan 07 '23 at 20:04
  • @Numb3rs My turn to apologise. I forgot to account for $n$ being natural. – copper.hat Jan 07 '23 at 20:07
  • @copper.hat What about Marius SL comment? Well I'm getting confused o.O – Heidegger Jan 07 '23 at 20:08
  • @MariusS.L. It certainly approaches $0$, because $\pi \sqrt{n^2+1}$ gets indefinitely closer to $n\pi$. – Vishu Jan 07 '23 at 20:10
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    @Numb3rs You need to look at $\sqrt{n^2+1} \pmod 1$ because of the nature of $\sin$. – copper.hat Jan 07 '23 at 20:12
  • @copper.hat So it suffices to say that as $n\to +\infty$, with $n\in\mathbb{N}$, $\sqrt{n^2+1}\pi$ goes like $n\pi$ and then the sine is zero? (Which I couldn't do if $n \in \mathbb{R}$) – Heidegger Jan 07 '23 at 20:23
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    @Numb3rs it reduces to showing $\lim_n (\sqrt{n^2+1}-n)=0$. – copper.hat Jan 07 '23 at 20:25
  • @copper.hat Yes, with using the method the other guy used. But assuming we do not use that... – Heidegger Jan 07 '23 at 20:26
  • @Vishu But it is $n\pi + \text{sth.}$ and this something doesn't ever vanish. – Marius S.L. Jan 07 '23 at 20:28
  • @MariusS.L. This something gets smaller and smaller – Vishu Jan 07 '23 at 20:29
  • @Vishu Oops! I (mistakenly) thought it started right at the first digit. – Marius S.L. Jan 07 '23 at 20:33
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    Numb3rs Your method to prove that $\lim_{n\to +\infty} \sin(n)$ does not exist is flawed because your $a_n$'s and $b_n$'s are not integers. – Anne Bauval Jan 07 '23 at 20:47
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    @AnneBauval Sorry that was my mistake: for that limit I meant $n$ as a real number! – Heidegger Jan 07 '23 at 21:30
  • You can also take a look at this one: https://math.stackexchange.com/a/4458695/1061348 – Henry Jan 08 '23 at 07:31

2 Answers2

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We have $$ \sqrt{n^2+1}=n\sqrt{1+\frac{1}{n^2}}=n\left(1+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=n+o(1) $$ therefore $\sin(\pi\sqrt{n^2+1})=\sin(n\pi+o(1))$. If you mean that $n$ takes integer values, then the limit exists and is $0$. If you want $n$ to be real (as mentioned in the comments) then the limit does not exist, taking $a_n=n$ and $b_n=2n+1/2$ gives you two different limits, namely $0$ and $1$ respectively.

Tuvasbien
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Assuming that $n$ is an integer and using $b_n=\sqrt{n^2+1}-n$

$$a_n= \sin \left(\pi \sqrt{n^2+1}\right)=\sin (n\pi +b_n\pi )=\color{red}{(-1)^n}\,\sin(b_n\pi)+\cos(b_n\pi)$$

Now, by Taylor $$b_n=\sqrt{n^2+1}-n=\frac{1}{2 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$

$$\sin(b_n\pi)=\frac{\pi }{2 n}-\frac{\pi \left(6+\pi ^2\right)}{48 n^3}+O\left(\frac{1}{n^5}\right)$$ $$\cos(b_n\pi)=1-\frac{\pi ^2}{8 n^2}+\frac{\pi ^2 \left(24+\pi ^2\right)}{384 n^4}+O\left(\frac{1}{n^6}\right)$$

So, $$a_{2n}=\frac{\pi }{4 n}-\frac{\pi \left(6+\pi ^2\right)}{384 n^3}+O\left(\frac{1}{n^5}\right)\quad \to ~0^+$$ $$a_{2n+1}=-\frac{\pi }{4 n}+\frac{\pi }{8 n^2}+\frac{\pi \left(\pi ^2-18\right)}{384 n^3}-\frac{\pi \left(\pi ^2-2\right)}{256 n^4}+O\left(\frac{1}{n^5}\right)\quad \to ~0^-$$

Claude Leibovici
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