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

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.

Solution 1:

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.

Solution 2:

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.