I'm trying to show that the series: $\sin(m) + \sin(\sin(m)) + \sin(\sin(\sin(m)) + \cdots$ converges for all real numbers $m$. To be specific, the series is defined as follows:

$\sum_1^\infty{a_k}$ where $a_{k+1}=\sin(a_k)$ for $k=1, 2, \ldots$ and $a_1=\sin(m)$.

I've concluded that the terms in the series tend to $0$ as $k\rightarrow\infty$. I also tried applying the ratio test:

$\dfrac{|a_k+1|}{|a_k|}=\dfrac{|\sin(\sin\cdots(\sin(m))|}{|\sin(\sin(\cdots(m)|} =\dfrac{|\sin(a_k)|}{|a_k|}\rightarrow1$ as $k\rightarrow\infty$, but since the limit is $1$ the test is inconclusive. I don't know an awful lot about series, but I do know about the the root test, the integral test and about doing comparisons. I can't see how I can apply any of those methods here.

I would be really grateful for any help with this problem!

  • $\begingroup$ I would not expect this series to converge. After getting below, say, $1/10$ quickly, the individual terms approach zero really, really slowly. I used the sequence of terms for finding fractional iterates of $\sin x;$ that's why i have an opinion... $\endgroup$ – Will Jagy Sep 23 '14 at 19:58
  • $\begingroup$ So did you use $\sin x < x$ and then iterate, to show the sequence tends to $0$? How did you know the ratio tended to $1$? This is just for my own curiosity $\endgroup$ – snulty Sep 23 '14 at 20:03
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    $\begingroup$ (This has been asked before on the site.) Except if $m$ is such that some $a_k$ is zero (and then the series converges), one can show that $a_k\sim\alpha\sqrt{3/k}$ when $k\to\infty$ where $\alpha=\pm1$ depends on $a_1$, hence the series diverges. $\endgroup$ – Did Sep 23 '14 at 20:14
  • $\begingroup$ @snulty No, I didn't really make a formal proof of the fact that the terms tend to zero, but I convinced myself drawing the unit circle. No matter what value m is, $\sin{m}$ will always be a number between -1 and 1. Let's say that the value is positive. Inserting this value (which is now an angle measured in radians) into $sin$ will give you something even smaller - I concluded this by drawing a unit circle and exploring the angles... I know that the ratio tends to 1 because of the $\frac{\sin{x}}{x}\rightarrow1$ when $x\rightarrow0$ and $a_k\rightarrow0$ when $k\rightarrow\infty$. $\endgroup$ – Jarvi79 Sep 23 '14 at 20:17
  • $\begingroup$ @Jarvi79 oh so it's like the sequential version of a limit, that's cool. $\endgroup$ – snulty Sep 23 '14 at 20:20

Heuristics: Say $a_n$ is near zero but not zero, then $\sin(x)=x-\frac16x^3+o(x^3)$ when $x\to0$ hence $$a_{n+1}=a_n-\frac16a_n^3+o(a_n^3).$$ If $a_n\sim c/n^b$, this imposes that $c^2=3$ and $b=\frac12$, hence the idea that $a_n\sim\alpha\sqrt{\alpha/n}$ with $\alpha=\pm1$ depending on $a_1$.

Full proof: Change variables and consider $$b_n=\frac1{a_n^2},$$ thus, one knows that $b_n\to+\infty$, and the same expansion of sine yields $$b_{n+1}^{-1}=\sin(b_n^{-1/2})^2=\left(b_n^{-1/2}-\frac16b_n^{-3/2}+o(b_n^{-3/2})\right)^2=b_n^{-1}-\frac13b_n^{-2}+o(b_n^{-2}),$$ which implies $$b_{n+1}=b_n+\frac13+o(1),$$ hence we are done.

  • $\begingroup$ Thanks for this answer. I'm gonna have to work through that thouroughly if I'm going to have a chance to understand it, I'm afraid. How about my idea about using Leibniz convergence test? $\endgroup$ – Jarvi79 Sep 23 '14 at 20:52
  • $\begingroup$ Explain it in detail and we will see if it works... $\endgroup$ – Did Sep 23 '14 at 20:53
  • $\begingroup$ Well, I need to show that the terms alternate in sign. This is easy to see since sin(-v) = -sin v so for example $-\sin(-\sin(m))=sin(sin(m))$. This way you can see that the signs alternate. I also need to show that $a_k\rightarrow0$ as $k\rightarrow\infty$. This is more difficult. I tried to convince myself using a geometric approach where I drew the unit circle and explored the angles. Do you have a better idea? I also need to show that $|a_{k+1}|<|a_k|$. Maybe this follows from $|\sin{x}|<|x|$? My reasoning is that that implies that $|\sin{a_k}|<|a_k|$ which shows that $|a_{k+1}<|a_k|$ $\endgroup$ – Jarvi79 Sep 23 '14 at 21:09
  • $\begingroup$ @Did: Nice! I wonder if you can get an asymptotic expansion for $a_n$. $\endgroup$ – Orest Bucicovschi Sep 23 '14 at 21:34
  • $\begingroup$ "Do you have a better idea?" To show convergence to zero? Well... this is kind of... like... in my answer, no? $\endgroup$ – Did Sep 23 '14 at 22:16

We will prove that $$\lim n \cdot a_n^2 \to 3$$ where $a_{n+1} = \sin a_n$ and $a_0 = a \ne k \pi$.

It is easy to see that $a_n \ne 0$ for all $n$. ( the only root of $\sin x$ in the interval $\sin (\mathbb{R}) = [-1,1]$ is $x=0$) Moreover, since $|\sin x| \le |x|$ for all $x \ne 0$ we conclude that $(a_n)_{n \ge 1}$ is strictly decreasing and $>0$ if $a_1 \in (0,1]$ and strictly increasing and $<0$ if $a_1 \in [-1,0)$. Therefore the sequence $(a_n)$ is convergent. Let $l$ be the limit. Since $x\mapsto \sin x$ is a continuous function from $a_{n+1} = \sin a_n$ we conclude $\sin l = l$ and therefore $l = 0$.

We will show that $n\cdot a_n^2 \to 3$, or, equivalently $$\lim \frac{1/a_n^2}{n} =1/3$$ We'll prove a stronger(use Stolz–Cesàro) $$\lim_{n\to \infty} (1/a^2_{n+1} - 1/a_n^2) = 1/3$$ Since $a_n\to 0$ we'll prove instead $$\lim_{x->0} (1/\sin^2 x- 1/x^2) = 1/3$$

Indeed we have the expansion $$ 1/\sin ^2 x =1/x^2 + 1/3 + x^2/15 + O(x^4)$$

and the statement is proved.

It is apparent that the speed of the convergence of a sequence of iterates $f^n(a)$ to the (locally) unique fixed point $0$ of $f(x)$ has to do with the expansion of $f(x)$ at $0$. Write: $$f(x) =x( 1 + c_s x^s + c_{s'}x^{s'}+ \ldots )$$ Then we have $$ \lim_{x\to 0} \frac{1}{f(x)^s} - \frac{1}{x^s} = -s \cdot c_s$$ Conclusion: If $0$ is an isolated fixed point and $a_n \ne 0$ is a sequence of iterates converging to $0$ then $$n\cdot a_n^s\to - s\cdot c_s$$

Observation: If the expansion at $0$ is more general $c_0 x + \ldots $ then for $|c_0|<1$ any sequence of iterates has convergence at least exponential while for $|c_0|>1$ it seems that $0$ will not be an attractive fixed point.


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