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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.

user
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    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. – Jack D'Aurizio Mar 20 '18 at 14:52
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    @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. – user Mar 20 '18 at 15:17
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    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 @?) – Did Mar 20 '18 at 17:33

2 Answers2

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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.

Did
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Gabriel Romon
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  • Thanks I’ll take a look! – user Mar 20 '18 at 11:41
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    @gimusi There's probably a simpler way (that Did seems to know)... – Gabriel Romon Mar 20 '18 at 11:43
  • You might want to answer this: https://math.stackexchange.com/questions/2692776/determining-if-a-series-is-absolutely-convergent-or-conditionally-convergent-wit, which is where the problem originates. – Martin R Mar 20 '18 at 12:33
  • @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 – user Mar 20 '18 at 12:51
  • @GabrielRomon we probabily will never know it – user Mar 20 '18 at 12:52
  • @GabrielRomon Sorry I can't see how the argument by contradiction works with Cesaro since we get from $\frac 1n \sum_{k=1}^n |\sin(k^3+1)|$ this $\sin(k^3+3k^2+3k+2)$, am I wrong? Thanks – user Mar 20 '18 at 12:55
  • @GabrielRomon How is the problem linked about https://math.stackexchange.com/q/1901485/505767 related to this one, could you please clarify? Thanks – user Mar 20 '18 at 12:57
  • @gimusi The only useful part for the problem at hand is: if $P(n)=a_d n^d+\cdots +a_0$ and $a_d$ is rationally independent from $\pi$, then $P(n)$ is equidistributed mod $2\pi$, thus $ \frac 1n \sum_{k=1}^n |P(k)|\to \frac{1}{2\pi}\int_0^{2\pi} |\sin(t)|dt=\frac{2}{\pi}$. – Gabriel Romon Mar 20 '18 at 14:17
  • @Gimusi Cesaro theorem states that if $u_n$ converges to $l$, then $\frac 1n \sum_{k=1}^n u_k$ converges to $l$. – Gabriel Romon Mar 20 '18 at 14:18
  • @GabrielRomon ah ok, you are relating the two results, sorry now it is clear! – user Mar 20 '18 at 14:21
  • @GabrielRomon Well your first solution is almost clear to me a part some minor details that I need to clarify better and as you noticed once we have that $\frac 1n \sum_{k=1}^n |P(k)|\to \frac{1}{2\pi}\int_0^{2\pi} |\sin(t)|dt=\frac{2}{\pi}$ by Cesaro we are done. I've also noticed that H.H.Rugh in his solution affirm that "One may note that the result (as well as the proof) goes through when replacing $n^2$ by any polynomial in $n$, $p(n)=a_d n^d+\cdots +a_0$ as long as the leading coefficient $a_d$ is rationally independent from $\pi$." – user Mar 20 '18 at 23:09
  • @GabrielRomon Also the way by the other approach is almost clear to me, a part some minor details that I need to digest better, even if I can't get how to generalize the result by discrete differentiation to the polynomial $(k^3+1)$ but I suppose that this is a some kind of well known result. I've also noticed that in this case fo the extension of the result it is said that "If $P$ is a polynomial with at least one coefficient not a rational multiple of $\pi$, then the sequence $(\sin(P(n)))_n$ does not converge to zero hence the series $\sum\limits_n\sin(P(n))$ diverges". – user Mar 20 '18 at 23:14
  • @GabrielRomon Why this difference for the generlization of the results, in one case is requested that "the leading coefficient is rationally independent from π" and in the other that "at least one coefficient not a rational multiple of π"? Thanks! – user Mar 20 '18 at 23:16
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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.

Jakobian
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