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I think $\lim_{n\to \infty} n\sin n$ does not exist, for $\sin x$ is a wave function and when $n\to \infty$, then so does factor $n$. But I am confused that how to give a strict proof for the divergence.

I would appreciate it if someone could give some suggestions and comments.

Sahiba Arora
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Jacob.Lee
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3 Answers3

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Every convergent sequence is bounded.

Edit: Let $m \in \mathbb{Z}.$ Then $$\sin x\geq \sin\left(\frac12\right)>0\,\, \forall x \in \left[\frac12+2m\pi,\frac32+2m\pi\right].$$ Moreover there exists $n_m \in \left[\frac12+2m\pi,\frac32+2m\pi\right] \cap \mathbb{N}.$ Therefore $$n_m\sin(n_m)\geq \left(\frac12+2m\pi\right)\sin\left(\frac12\right).$$It follows that $(n \sin n)_n$ is unbounded.

Sahiba Arora
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Given that $$\lim_{n \to \infty} \frac{1 }{n} = 0$$ And $\lim_{n \to \infty} \sin n$ doesn't exist, it's easy to prove that. The former can be proved by the definition while the latter with the proof by contradiction. Suppose $$ \lim _{n \to \infty}n\sin n = L$$ So we have $$ \lim _{n \to \infty}n\sin n \times \frac{1}{n}= \lim _{n \to \infty} \sin n = 0$$ And this is a contradiction.

Edit: If we want to prove that $\lim_{n \to \infty} \sin n$ doesn't exist, according to the Jonas Teuwen's answer in the mentioned page, suppose $\lim_{n \to \infty} \sin n = l$. It's known that a sequence converges iff every subsequence converges. So $\lim_{n \to \infty} \sin 2n = \lim_{n \to \infty} \sin 2(n+1) = l$ and $\lim_{n \to \infty} \cos2n = \lim_{n \to \infty} 1-2\sin^2 n = 1 - 2l^2$ and also $\lim_{n \to \infty} \cos2(n+1) = 1 - 2l^2$. Then $$\sin(2) = \sin(2(n+1) - 2n) = \sin(2(n+1))\cos(2n) - \sin(2n)\cos(2(n+1))$$Taking limit yields $$\lim_{n \to \infty} \sin(2) = l\times (1-2l^2) - l\times (1-2l^2) = 0$$So this implies $$\sin(2) = 0$$ Which is clearly a contradiction.

S.H.W
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  • $\lim_{n\to\infty} \frac{\sin n}{n} \neq 1$. – Clement C. Jun 12 '20 at 01:50
  • @ClementC. Sorry for that. Is it correct now? – S.H.W Jun 12 '20 at 01:54
  • Looks good to me, under the assumption that $\lim \cos n$ doesn't exist is known/usable. (Also, you can use $L'=0$ at the end.) – Clement C. Jun 12 '20 at 01:57
  • @ClementC. Thanks for mentioning my silly mistake. – S.H.W Jun 12 '20 at 02:04
  • (+1) All too easy. – Mark Viola Jun 12 '20 at 03:07
  • @MarkViola My pleasure. – S.H.W Jun 12 '20 at 08:43
  • Very quick and nice argument. The last line should be $\lim cos(2n)=1$. – Sungjin Kim Jun 12 '20 at 20:09
  • @SungjinKim That's very kind of you, you're right. Actually I changed $\frac{\sin}{n}$ to $\frac{1}{n}$ which clearly is a better option for reaching to a contradiction. – S.H.W Jun 12 '20 at 20:41
  • We return to the question of whether $n$ goes to infinity through integers only, or through real numbers. If through real numbers, sure, it is obvious that the limit does not exist. But if $n$ only runs through integers, at least some small result about the nature of $\pi$ (at least its size) is necessary, as in another answer. – paul garrett Jun 12 '20 at 20:44
  • @paulgarrett I think my answer shows the limit doesn't exist if $n \in \mathbb{N}$. If there is a mistake in that please say so that I improve that or remove if is necessary. – S.H.W Jun 12 '20 at 20:50
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    @S.H.W, true, but needs to be confirmed, that $\sin n$ for $n\in\mathbb Z$ does not go to $0$. They don't, but why? It needs some property of $\pi$. Otherwise, with a function $f$ instead of $\sin$, periodic of period $1$ (rather than $2\pi$), vanishing at $0$, then it would vanish at all integers. It's a small thing, and your conclusion is correct (of course), but there is this small potential hazard. Again, $\sin n$ for $n$ integer does not assume all the values of $\sin$ (the range $[-1,1]$), but only a discrete subset... and some details about that subset must be clarified... – paul garrett Jun 12 '20 at 20:55
  • @paulgarrett Actually the link which I've provided shows that $\lim_{n \to \infty} \sin n$ doesn't exist and it is proved by contradiction. – S.H.W Jun 12 '20 at 20:58
  • Mmm, I'd not want to have to depend on any argument written in that style... Better to redo it "from scratch", etc. It's not completely obvious, nor do I see why it should be "by contradiction". In such simple things, there is no reason to be vague or obscure, I think. – paul garrett Jun 12 '20 at 21:19
  • @paulgarrett Thanks, I added the proof. – S.H.W Jun 12 '20 at 21:42
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We have that $a_n = n \sin{(n)}$. Rewrite it during that $\sin{(x)}\in\left[-1,1\right]$: $$-n\leq n \sin{(n)} \leq n$$ Than, using the lim: $$-lim_{n\rightarrow\infty}n \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq lim_{n\rightarrow\infty}n$$ $$-\infty \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq \infty$$ This shows, that $a_n$ is unbounded and thats because limit does not exists: $$\nexists lim_{n\rightarrow\infty} n \sin{(n)}$$

Gosrabios
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  • Also, you can see that $a_n$ is unbounded if you plot the graph. – Gosrabios Jun 12 '20 at 10:32
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    $-n \leq \frac1n \leq n \implies -\lim n \leq \lim \frac1n \leq \lim n \implies -\infty\leq \lim \frac1n \leq \infty \implies \frac1n$ is unbounded? – Sahiba Arora Jun 12 '20 at 11:07
  • @SahibaArora It's unbounded on infinity. Maybe I can't fully understand word "unbounded" because English is not my native language. – Gosrabios Jun 12 '20 at 11:12
  • Your proof does not show $(n \sin n)$ is unbounded. A sequence $(a_n)$ is said to be unbounded if for each $M >0,$ there exists $ n \in \mathbb{N}$ such that $a_n>M.$ Your proof suggests that if $-n \leq a_n \leq n,$ then $(a_n)$ is unbounded. This is not true, for example, if $a_n=\frac1n.$ – Sahiba Arora Jun 12 '20 at 11:20