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How can we prove that for $k\in \mathbb{Z}\quad k\to \infty$

$$\sin(1 + k^3) \not \to 0$$

Lately I've encoutered it a couple of times in some OP posted here about series and by the discussion I had, also with expert users, it seems there is not a simple solution.

One possible strategy I had in mind is to show that for

$$\sin(1+k^3) \approx0$$

then

$$\sin(1+(k+1)^3)=\sin(1+k^3+3k^2+3k+1)=\sin(1+k^3)\cos(3k^2+3k+1)+\sin(3k^2+3k+1)\cos(1+k^3)\approx \pm \sin(3k^2+3k+1)$$

and show that $\sin(3k^2+3k+1)$ is "far" from zero.

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    $\begingroup$ I believe the revised version of the question if fully compliant about the rules for asking a good question. I have removed the previous comments but I still have something to suggest: 1) no one is perfect. If a question has some issues, please point them out, in a polite and respectful way; 2) to further stress the previous point, please avoid escalating and make a wise use of the flags, if necessary. $\endgroup$ – Jack D'Aurizio Mar 20 '18 at 14:52
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    $\begingroup$ @JackD'Aurizio I'm sorry and I apologize for what happened, I will take in great attention your suggestions for the future in order to avoid this kind of public escalations into personal disputes. Thanks for your time, Regards. $\endgroup$ – gimusi Mar 20 '18 at 15:17
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    $\begingroup$ Nothing personal from my side (if ever this is the nature of the implicit allegation here), but simply a bland, banal, factual reminder of the rules of the site, that we should all try to follow, right? That such reminders seem to send some into a whirl of insults is something I deplore, and that I have to endure, but for which I cannot feel responsible. (@JackD'Aurizio When posting comments obviously referring to some users and obviously pretending to address their behaviour, why forget to use the @?) $\endgroup$ – Did Mar 20 '18 at 17:33
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See the answer here: $\displaystyle \frac 1n \sum_{k=1}^n |\sin(k^3+1)|$ converges to $\displaystyle\frac 2\pi$ as $n\to \infty$.

If $\sin(k^3+1)$ converged to $0$, so would $|\sin(k^3+1)|$ and Cesaro theorem would imply that $\displaystyle \frac 1n \sum_{k=1}^n |\sin(k^3+1)|\to 0$, a contradiction.


Equidistribution is overkill for the problem at hand. There is a more elementary way, as explained in this answer by Did.

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  • $\begingroup$ Thanks I’ll take a look! $\endgroup$ – gimusi Mar 20 '18 at 11:41
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    $\begingroup$ @gimusi There's probably a simpler way (that Did seems to know)... $\endgroup$ – Gabriel Romon Mar 20 '18 at 11:43
  • $\begingroup$ You might want to answer this: math.stackexchange.com/questions/2692776/…, which is where the problem originates. $\endgroup$ – Martin R Mar 20 '18 at 12:33
  • $\begingroup$ @MartinR Yes it is strictly related but this is a specific part of that question, since I can't get an answer I thought it was useful to ask a question about this $\endgroup$ – gimusi Mar 20 '18 at 12:51
  • $\begingroup$ @GabrielRomon we probabily will never know it $\endgroup$ – gimusi Mar 20 '18 at 12:52
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We will show that under certain conditions, sequence $\sin(P(n)),\,n=1, 2, ...$ where $P(n)$ is a "good enough" polynomial, is dense on the interval $[-1, 1]$.

We will use that, for any polynomial $P$,
$$P(x) = a_mx^m+a_{m-1}x^{m-1}+...+a_0 $$ If at least one of the coefficients $a_j$, $j>0$ is irrational, then $P(n)$ is uniformly distributed modulo $1$. In fact, we will only need the fact that the sequence $\{P(n)\}$ is dense in $[0, 1)$.

Let's choose any number from $[-1, 1]$, and let's write it in the form $\sin(a)$. $$|\sin(a)-\sin(P(n))|=2|\sin(\frac{a-P(n)}{2})\cos(\frac{a+P(n)}{2})| \leq2|\sin(\frac{a-P(n)}{2})| \\= 2|\sin(\frac{a-P(n)}{2}\,\text{mod}\,2\pi)| $$

$$\frac{a-P(n)}{2}\,\text{mod}\,2\pi = 2\pi(\frac{a-P(n)}{4\pi}\,\text{mod}\,1) $$ Recall what we said before, if at least one of $\frac{a_j}{\pi}$ is irrational, for $m\geq j\geq 1$, then $\frac{a-P(n)}{4\pi}\,\text{mod}\,1$ is dense in $[0, 1)$. Then for any $\varepsilon>0$ we can choose $n$, so that: $$\frac{a-P(n)}{4\pi}\,\text{mod}\,1 < \frac{\varepsilon}{4\pi}$$

Then: $$2|\sin(\frac{a-P(n)}{2}\,\text{mod}\,2\pi)|<\varepsilon $$ From which follows that: $$|\sin(a)-\sin(P(n))|<\varepsilon $$ So the sequence is dense on the interval $[-1, 1]$.

You can see that this is a much stronger result. Since $\frac{1}{\pi}$ is irrational, $\sin(1+k^3)$ is dense in $[-1, 1]$, hence doesn't converge to any real number.

I came up with this proof while reading the book I am learning from right now, it's called "Uniform Distribution of Sequences" by L. Kuipers and H. Niederreiter. You can find the proof that sequences $P(n)$ of that form are uniformly distributed there. It's Theorem 3.2 on page 27.

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