5

If $x$ is a rational multiple of $\pi$, for a natural number $N$ big enough $\sin(n!\,x) = 0$ for all $n\geqslant N$ and then $\sin(n!\,x)\to 0$ as $n\to +\infty$.

However, I'm not so sure about the converse anymore: If it holds that $\sin(n!\,x)\to 0$ as $n\to +\infty$, does it necessarily follow that $x$ belongs to $\pi\mathbf{Q}$?

Fitzcarraldo
  • 358
  • 10
    What if $x=\pi e$? – Aphelli Aug 18 '21 at 21:12
  • 1
    @Mindlack Please do tell me! Does $\sin(n!, \pi e)\to 0$ when $n\to +\infty$? I got no idea. – Fitzcarraldo Aug 18 '21 at 21:36
  • 3
    When $x=2\pi e$ the limit is zero. https://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity/76098#76098 – Thomas Andrews Aug 18 '21 at 21:51
  • 5
    More generally, if $0\leq a_1<a_2<\dots<a_n<\dots$ are integers, then $2\pi\sum\frac{1}{a_n!}$ has this property. There are uncountably many such sequences, and thus uncountably many such values $2\pi \alpha$ which have this property. – Thomas Andrews Aug 18 '21 at 21:54
  • 2
    @Fitzcarraldo: well, $n!e -\sum_{k=0}^n{\frac{n!}{k!}}=\sum_{k=n+1}^{\infty}{\frac{1}{(n+1)\ldots k}} \leq \sum_{k=n+1}^{\infty}{\frac{1}{(n+1)^{k-n}}}$ so that there is an integer $N_n$ such that $n!e-N_n \in [0,1/n)$. Hence $|\sin{n\pi e}| \leq |\sin(\pi/n)| \leq \pi/n$. – Aphelli Aug 18 '21 at 22:02
  • 3
    @Mindlack Interestingly it is not known whether $\pi e$ is rational or irrational. – Infinity_hunter Aug 19 '21 at 04:24
  • @Thomas Andrews Very nice move of yours. From the cardinality of those $\alpha$ it follows that there must be uncountably many of them which are irrational! But by adapting word by word Fourier's proof of the irrationality of $e$, all the numbers of the form $\sum_{k=1}^{+\infty} 1/a_k!$ with $0\leqslant a_1 < a_2 <\dotsb$ are in fact irrational. – Fitzcarraldo Aug 19 '21 at 12:23
  • @Fitzcarraldo Technically, you need to prove each sequence determines a different number to prove there are uncountable many, but that isn’t hard to show. – Thomas Andrews Aug 19 '21 at 16:59

1 Answers1

4

Here we extend the idea discussed in the comments and discuss the condition for $x$ that makes the limit zero.

Let $x = \pi r$. Write the fractional part $\{r\} = r -\lfloor r \rfloor$ in the form $$ \{r\} = \sum_{j=2}^{\infty} \frac{a_j}{j!} \quad \text{for} \quad a_j \in \{\,0, 1,\ldots, j-1\,\}. $$

Then we note that $$ \begin{split} |\sin(n!\, \pi r)| = \Biggl| \sin\biggl(n!\, \pi\sum_{j=2}^{\infty} \frac{a_j}{j!} \biggr) \Biggr| = \Biggl| \sin\biggl(\pi \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots(n+k)} \biggr) \Biggr| \end{split} $$ and $$ \begin{split} \frac{a_{n+1}}{n+1} &\leq \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots (n+k)} \\ &\leq \frac{a_{n+1}}{n+1} +\sum_{k=2}^{\infty} \frac{n+k-1}{(n+1)(n+2)\cdots (n+k)} \\ &= \frac{a_{n+1} + 1}{n+1}. \end{split} $$

From this, it is not hard to check that $\sin(n!\, \pi r) \to 0$ if and only if the sequence $(a_{n+1}/(n+1))$ has only limit points in $\{\,0, 1\,\}$, or equivalently, $\operatorname{dist}(a_{n+1}/(n+1), \mathbb{Z})\to 0$. Thomas's example corresponds to the case where all $a_j$ are either $0$ or $1$.

Finally, since the set of all such $r$ is uncountable, there is an irrational number $r$ such that $\sin(n!\, \pi r) \to 0$.

Fitzcarraldo
  • 358
Sangchul Lee
  • 181,930
  • I do dare to ask, how do you know that ${r}$, which I assume to be not necessarily rational, got such an explicit series expansion? – Fitzcarraldo Aug 20 '21 at 18:58
  • 1
    @Fitzcarraldo, The proof is quite similar to how you would show that every real number admits a decimal (or in general $n$-ary) expansion. For instance, you can define $a_n$ as $$a_n=\lfloor n! r\rfloor \text{ mod } n=\lfloor n! r\rfloor - n \lfloor (n-1)! r\rfloor . $$ It is not hard to check that this $a_n$'s satisfy all the desired properties. – Sangchul Lee Aug 20 '21 at 19:02
  • 1
    @Fitzcarraldo: The following is a general result in arithmetic: > Theorem: Let $(a_n:n\in\mathbb{N})$ be a sequence of positive integer numbers larger than $1$. Then, any real number $\alpha\in R$ can be uniquely expressed in the form $$\begin{align} \alpha=c_0+\sum^\infty_{n=1}\frac{c_n}{a_1\cdot\ldots\cdot a_n}\tag{0}\label{a-basis} \end{align}$$ where $(c_n:n\in\mathbb{Z}_+)\subset\mathbb{Z}$, satisfy
    (i). $c_0\in\mathbb{Z}$, $0\leq c_n<a_n$ for $n\geq1$, and
    (ii). $#{n\in\mathbb{N}: c_n<a_n-1}=\infty$.     – Mittens Aug 21 '21 at 04:37
  • @SangchulLee Very sophisticated argument! So, any bounded non-negative integer sequence $(a_k){k\geqslant 2}$ would define a number $r = \sum{k=2}^{+\infty} a_k/k! < 1$ such that ${n!, r}\to 0$ when $n\to +\infty$, while if $(a_k)_{k\geqslant 2}$ is unbounded and such that that for every $\epsilon > 0$ small enough from some point onward every $a_k\in \mathopen{]}(1 -\epsilon)k, k\mathclose{[}$, then ${n!, r}\to 1$ when $n\to +\infty$. Actually, one could say that a sequence of the first kind is dominated by a multiple of the corresponding constant sequence of $e - 2$. – Fitzcarraldo Aug 21 '21 at 22:17
  • @SangchulLee A general integer sequence $(a_k)_{k\geqslant 2}$ with $0\leqslant a_k\leqslant k-1$ would contain subsequences of those two kinds, so what would look like such a sequence but just with ${n!, r}\to l \neq 0, 1$ when $n\to +\infty$? – Fitzcarraldo Aug 21 '21 at 22:19