Limit of $\sin (a^{n}\theta\pi)$ as $ n \to \infty$ where $a$ is an integer greater than $2$

Question

In Hardy's Pure Mathematics, Hardy discusses the limit $$\lim_{n\to\infty}\sin (2^{n}\theta\pi)$$ and says that if this limit exists it must be zero and then $\theta$ must be a rational number whose denominator is a power of $2$. Then he asks the reader to consider the limit $$\lim_{n\to\infty}\sin (a^{n}\theta\pi)$$ where $a$ is an integer greater than $2$.

It is also mentioned that when $a > 2$ then the limit can be non-zero and as an example Hardy states that if $a = 9$ and $\theta = 1/2$ then the above limit is $1$. In this case I reasoned (based on Hardy's technique for the case $a = 2$) that $\lim_{n\to\infty}\sin((9^{n}\pi)/2)$ will be $1$ only when we can write $$\frac{9^{n}\pi}{2} = 2b_{n}\pi + \frac{\pi}{2} + c_{n}$$ for all sufficiently large values of $n$, where $\{b_{n}\}$ is a sequence which takes only integer values and $\{c_{n}\}$ is a sequence which tends to zero as $n \to \infty$. Thus we get

$\displaystyle \begin{aligned}9^{n}\pi &= 4b_{n}\pi + \pi + 2c_{n}\\ \Rightarrow (9^{n} - 1)\pi &= 4b_{n}\pi + 2c_{n}\end{aligned}$

Since $9^{n} - 1$ is divisible by $9 - 1 = 8$ and hence also by $4$, it follows that we can take $b_{n} = (9^{n} - 1)/4$ and $c_{n} = 0$. The same logic could be applied when $a$ is any integer of the form $a = 4m + 1$ and $\theta = 1/2$.

But, and this is my question, how does one handle the case for general integer $a > 2$ and any real $\theta$ rational or irrational?

EDIT: I should also provide more details so that readers get the full context. For the specific case when $a = 2$ Hardy reasons that if the limit exists and, say is equal to $l$, then we must have $|l| \leq 1$ so that we have a constant $\alpha = \sin^{-1} l$ lying in interval $[-\pi/2, \pi/2]$. Since the solution of $\sin x = \sin \alpha$ is given by $x = m\pi + (-1)^{m}\alpha$ for all integers $m$, we can see that the existence of limit $l$ entails that $$2^{n}\theta\pi = b_{n}\pi + (-1)^{b_{n}}\alpha + c_{n}$$ where $b_{n}$ takes integer values and $c_{n} \to 0$ as $n\to\infty$. We can now see that $$2^{n}\theta = b_{n} + (-1)^{b_{n}}\beta + d_{n}$$ where $\beta = \alpha/\pi$ so that $\beta \in [-1/2, 1/2]$ and $d_{n} = c_{n}/\pi \to 0$ as $n \to\infty$. Hardy somehow assumes that $b_{n}$ will always be even and then proceeds as follows:

$\displaystyle 2^{n}\theta = b_{n} + \beta + d_{n}$ so that multiplication by $2$ gives

$\displaystyle 2^{n + 1}\theta = 2b_{n} + 2\beta + 2d_{n}$ and also we have

$\displaystyle 2^{n + 1}\theta = b_{n + 1} + \beta + d_{n + 1}$ and therefore we get

$\displaystyle (b_{n + 1} - 2b_{n}) - \beta + (d_{n + 1} - 2d_{n}) = 0$

After this Hardy uses simple arguments to show that $\beta = 0$ and $d_{n}$ should be identically zero from a certain value of $n = n_{0}$ so that $2^{n_{0}}\theta = b_{n_{0}}$ and thus $\theta$ is rational with denominator a power of $2$.

In the above I don't understand why he assumes $b_{n}$ as even. Also I think the argument can be continued without assuming parity of $b_{n}$ but I am not sure. I wonder how this can be carried forward for higher values of $a$.

Answer 1

I came across this old question recently, and since I have been thinking about this problem through a different source, I decided to add some thoughts here.

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First, I readdress the case $a=2$. I came with a similar method as described by Hardy, however, I think that one can get results without having to worry about the parity of the integer ($b_n$ in your notation).

If $A:=\lim_{n\rightarrow\infty}\sin(2^n\theta)$ exists, then there is $|\psi|\leq\frac{\pi}{2}$ and integers $k_n$ such that $$ 2^n\theta = \pi k_n+\psi +\varepsilon_n$$ where $\varepsilon_n\rightarrow0$. The limit $A$ being either $\sin(\psi)$ or $\sin(\pi-\psi)$. It follows that $$0 =\pi(k_{n+1}-2k_n)-\psi+(\varepsilon_{n+1}-2\varepsilon_n)$$ Being $k_{n+1}-2k_n$ a convergent integer valued sequence, it follows that there is $N\in\mathbb{N}$ such that $$k_{n+1}-2k_n=\frac{\psi}{\pi},\qquad n\geq N$$ From this, one concluded that $\psi=0$ for $\frac{\psi}{\pi}\in\mathbb{Z}$ and $\Big|\frac{\psi}{\pi}\Big|\leq\frac{1}{2}$. In turn, this implies that $k_n=2^n2^{-N}k_N$ for all $n\geq N$. Putting things together, one concludes that $$\theta=2^{-N}k_N\pi+2^{-n}\varepsilon_n\xrightarrow{n\rightarrow\infty}2^{-N}k_N\pi$$ Conversely, if $\theta/\pi$ is a dyadic number, clearly the limit $A$ exists and is $0$.

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Now let's try to push the argument to $a>2$ and see how far we (or rather I) can get. Suppose $A=\lim_n\sin(a^n\theta)$ exists. Then, as before, there is $|\psi|\leq\frac{\pi}{2}$ and integers $k_n$ such that $$ a^n\theta =\pi k_n +\psi +\varepsilon_n$$ with $\varepsilon_n\xrightarrow{n\rightarrow\infty}0$. Then $$0=\pi(k_{n+1}-ak_n)-(a-1)\psi +(\varepsilon_{n+1}-a\varepsilon_n)$$ The same argument used for $a=2$ applies here and so, there is $N\in\mathbb{Z}_+$ such that $$k_{n+1}-ak_n=\frac{\psi}{\pi}(a-1),\qquad n\geq N$$ It follows that $\widetilde{\psi}=\frac{\psi}{\pi}(a-1)\in\mathbb{Z}$ and $|\widetilde{\psi}|\leq \frac{a-1}{2}$, That give us $2\big[\tfrac{a-1}{2}\big]+1$ possible values for $\psi$. By induction, we obtain that $$ k_{m+N}=a^mk_N + \widetilde{\psi}\frac{a^m-1}{a-1}=a^mk_N + \frac{\psi}{\pi}(a^m-1)=a^m\big(k_N+\frac{\psi}{\pi}\big)-\frac{\psi}{\pi}\in\mathbb{Q} $$ Putting things together, we have that $$\theta=a^{-N}\big(k_N+\frac{\psi}{\pi}\big)\pi +a^{-n}\varepsilon_n\xrightarrow{n\rightarrow\infty}a^{-N}\big(k_N+\frac{\psi}{\pi}\big)\pi$$

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To be cautious, I will also state that this is a partial solution, in that we have necessary conditions on $\theta$ when $\lim_{n\rightarrow\infty}\sin(a^n\theta)$ exists.

• As a simple application of this finding one can see that the sequence $\sin(4^n)$ ($\theta=1$) is divergent; otherwise, for some $\psi$ with $\frac{\psi}{\pi}\in\mathbb{Q}$, $k\in\mathbb{Z}$ and $N\in\mathbb{Z}_+$ we would have $$1=4^{-N}\big(k+\frac{\psi}{\pi}\big)\pi\in\mathbb{Q}$$ which is not possible.

• Similarly, for any $a\in\mathbb{N}$, $a>2$, and $\theta\in\mathbb{Q}\setminus\{0\}$, we can rule out the convergence of $\sin(a^n\theta)$.

• If $\frac{p}{q}\in\mathbb{Q}$, $(p,q)=1$ and $q\big| a$, then the sequence $\sin(a^n\theta)=0$ for $\theta=a^{-N}\big(k+\frac{p}{q}\big)\pi$ and $n\geq N$.

• In this final example we show a case with a non zero limit. If $a=2m+1$, $r\in\mathbb{Z}$, $(p,q)=1$ and $q\big|\,m$, then with $\theta=a^{-N}\big(2r+\frac{p}{q}\big)\pi$, the series $\sin(a^n\theta)=\sin\big(\frac{p}{q}\pi\big)$ for $n\geq N$.

Answer 2

Here is a very partial answer, but at least it contains some ideas

If $\theta=m/a^p$ for integers $m,p$ then $\sin a^n\theta\pi=\sin a^{n-p}m\pi=0$ as soon as $n\geq p$.

The article cited by Norbert shows that this is the only case where the limit is zero.

Note that $\sin a^n\pi/2=\sin (a+4k)^n\pi/2$ for any integer $k$. In particular, $\sin 9^n\pi/2=\sin\pi/2=1$ for any $n$.

The same idea occurs with other denominators. For example $$ \sin 7^n\pi/3=\sin(6+1)^n\pi/3=\sin\pi/3=\frac{\sqrt3}2\ \ \mbox{ for all } n. $$

More generally, if we consider $\sin a^n\,m\pi/q$, we can write $a=2kq+r$ with $k$ integer and $0\leq r<q$, and in that case $$ \sin \frac{a^n\,m\pi}q=\sin \frac{r^n\,\pi}q. $$ This last expression will usually oscillate when $r\ne0,1$.