# How will the limit $\lim \sin(A^{n})$ behave for $|A| > 1$?

While answering to the following question

Problem. Find the range of $x$ for which $\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \sin(x^{-n})$ converges.

Unlike some careless answerers who merely claimed that this converges only for $|x| > 1$, I faced a hard question:

Question. Is there any $|A| > 1$ such that $\sin (A^{n})$ converges? (Of course, I put $A = x^{-1}$ here.)

I literally have no idea on it since the behavior of the sequence depends on $n \mapsto A^{n} \text{ mod } 2\pi$ and I have no idea how it behaves for values of $A$.

Actually I suspect that there is a number $A > 1$ for which $\sin (A^{n})$ converges. But this is just a groundless guess, anyway.

Do you have any idea regarding this one?

• Do you have to answer your question to solve the problem? or you already have solution to the problem, but just curious about the question? – Sungjin Kim Feb 18 '14 at 7:08
• @i707107 As you see, my question is logically just a necessary condition for the problem. I failed to manage both of them. In fact, this is a problem I saw in some QnA site, and none gave a satisfactory answer. I'm sure that even the one who made this question was not aware of this delicacy. – Sangchul Lee Feb 18 '14 at 7:47

It can be shown that $$\mathcal{D}=\{|A|>1 : \sin(A^n) \textrm{ diverges.}\}$$ is dense. (I conjecture that $\mathcal{C}=\{|A|>1: \sin(A^n) \textrm{ converges.}\}$ has measure zero, do not know how to deal with complex exponentials ---> now I do know.)

This follows from a result on equidistribution:

For real numbers $\theta>1$, is is shown in Koksma's paper that $\theta^n$ is equidistributed modulo $1$ for almost all $\theta>1$. Equidistribution modulo $2\pi$ is just a minor modification of Satz 2 in the paper.

Then the density of $\mathcal{D}$ follows from considering $A=re^{i\pi\frac{k}{m}}$ for $k,m$ fixed integers. In fact $A^{2mn}=(r^{2m})^n$, and for almost all $r^{2m}>1$, $(r^{2m})^n$ is equidistributed modulo $2\pi$.

A stronger result of the above(in terms of error term) is shown in Leveque's paper.

Edit: Proof of my conjecture

Let $A=re^{i \theta}$, so $A^n=r^ne^{in\theta}$. Suppose that $\theta\neq k\pi$ for any integer $k$. Then there is a subsequence $n_j$ for $j=1,2,\cdots$ such that $$\sin n_j\theta >\frac 12 \ \ \textrm{ for all }j.$$

Now, $$\sin A^n = \frac{e^{-r^n\sin n\theta}e^{i r^n\cos n\theta}-e^{r^n\sin n\theta}e^{-i r^n\cos n\theta}}{2i}$$

Since $|v-w|\geq ||v|-|w||$ for any complex $v, w$,

$$|\sin A^{n_j}|\geq \frac{1}{2} \left|e^{r^{n_j}\sin n_j\theta}-e^{-r^{n_j}\sin n_j\theta}\right|$$

Since $r>1$, the above $\rightarrow\infty$ as $j\rightarrow\infty$.

Hence, if $A=re^{i\theta}$, with $\theta \neq\pi \textrm{ mod }2\pi$, then the sequence $\sin A^n$ diverges. This proves my conjecture. $$\mathcal{C}\textrm{ has measure zero.}$$

Further, we see that $\mathcal{C}\subset \mathbb{R}$, and has measure zero for the Lebesgue measure on $\mathbb{R}$.

Edit2 I finally found references that partially settles your problem.

We have (1) from Pisot, (2) from Kawapisz.

(1) If a real number $\lambda$ is transcendental, then the series $$\sum_{n=0}^{\infty} \sin^2 (\pi\lambda x^n)$$ is divergent for all $x\geq 1$.

(2) If $x$ is algebraic, and there exists $\lambda\neq 0$ such that $$\sin (\pi\lambda x^n)$$ converges to $0$, then $\lambda \in \mathbb{Q}(x)$.

With $\lambda=1/\pi$, there does not exist any real number $x\geq 1$ such that $$\sum_{n=0}^{\infty} \sin^2 x^n$$ converges.

Therefore, the a partial conclusion to your problem is:

The series $$\sum_{n=0}^{\infty} (-1)^{n} \sin(x^{-n})$$ absolutely converges if and only if $|x|>1$. If $|x|<1$ and $x$ is algebraic, then the series does not converge.

Still, whether or not there exists a real number $x>1$ such that $\sin x^n$ converges, is open. What we know so far is: if $x>1$ algebraic then $\sin x^n$ does not converge to $0$.

One observation.

Let, w.l.o.g. $1<A<2\pi$. Then, let, $n=\min_{k}\{k\in \mathbb{N}:A^{k}<2\pi,\ A^{k+1}\ge2\pi\}$. So the $n+1$ st term of the sequence $A^n\mod{2\pi}$ is $A^{n+1}-2\pi m$ for some $m\ge 1$. Let, $A_1:=A^{n+1}-2\pi m$. Now, $A_1\le 2\pi A-2\pi$. So, if $2\pi A-2\pi \le A\Rightarrow A\le \frac{2\pi}{2\pi -1}$, then, $A_1\le A$. now, again define $n_1:=\min_{k}\{k\in \mathbb{N},k\ge n+1: A_1^{k}<2\pi, A_1^{k+1}\ge 2\pi\}$ and again set $A_2:=A_1^{n_1+1}-2\pi m_1$ and now, we can see that $A_2\le A_1$. In this way one can find a decreasing sub-sequence ${A_n}$ in the orbit of $A^n \mod{2\pi}$.

• $\sin((A+2\pi)^2) \ne \sin(A^2)$. – aschepler Feb 22 '14 at 1:30
• I am not using this anywhere, all I'm saying is that once you have a term $A^n$ which is the first term to cross $2\pi$, then call it $A_1$ and similarly if you define $A_2,A_3,\cdots$ etc. then they should form a non-increasing sequence. – Samrat Mukhopadhyay Feb 24 '14 at 9:32