I thought about sequences, and I had an idea.

Let's define a sequence $(a_i)$ such that $$a_0 = c \qquad\text{and}\qquad a_{n+1}=\sin a_n$$ where $c$ is arbitrary constant. Let's also define $$S_N := \sum_{i=0}^Na_i=c\;+\;\sin c\;+\;\sin\sin c\;+\;\cdots\;+\;\underbrace{\sin\sin\cdots\sin}_{N\text{ times}} c$$

Is $\lim_{N\to\infty} S_N$ convergent or divergent?

The problem is that I have no idea how I can, or if it is even possible to, determine $a_n$. Or if it is possible to show whether the sum converges.

  • $\begingroup$ I think the limit $\lim_{n \to \infty} a_n$ exists and is in general not equal to 0 ($\implies$ the sum does not converge for arbitrary $c$ it trivially does converge for $c=\pi k$, with $k \in \mathbb{Z}$), I think you can show this by fixed-point iteration. $\endgroup$
    – MrYouMath
    Oct 21, 2017 at 10:59
  • 1
    $\begingroup$ Compare math.stackexchange.com/questions/276135/… for the asymptotics of the iterated sine. $\endgroup$
    – Martin R
    Oct 21, 2017 at 11:09
  • $\begingroup$ Hence it's convergent if and only if $\sin c = 0$. $\endgroup$ Oct 21, 2017 at 11:23
  • $\begingroup$ If you examine the answer in the link Martin R provided, you will see that it shows $\lim a_n = 0$ always (the answer only proves it for $c > 0$, but since sine is odd, it holds for all starting $c$), so MrYouMath is incorrect. However, as $a_n$ gets small, $\sin$ becomes nearly linear, which means the convergence of $a_n$ becomes very slow. That likely means that $S_N$ still does not converge, but it will have to be more carefully examined than it has been so far to prove it. $\endgroup$ Oct 21, 2017 at 17:31
  • $\begingroup$ The linked analysis shows that $a_n$ approaches $c\sqrt{1/n}$ which converges to $0$ slower than the harmonic series so the sum diverges. $\endgroup$ Oct 21, 2017 at 20:27

1 Answer 1


Let denote $\sin^{[n]}c$ to be the $n$-times iteration of $\sin$. It is quite clear we may well assume $c \in (0,1]$, since the first iteration $\sin c$ is in $[-1,1]$. If $\sin c = 0$ it is trivial, if $\sin c = [-1,0)$, just flip the sign in the subsequent arguments.

We first note that by concavity of $\sin$, $\sin x \geq x\sin 1$ for $x \in (0,1]$.

Now given any $\epsilon \in (0,\sin c)$, since $a_n$ is strictly decreasing, the set $\{n:a_n \geq \epsilon\}$ (non-empty) has a largest element $N$, in particular it holds that we may always find $N$ such that $\epsilon \sin 1 \leq \sin^{[N+1]} c < \epsilon$.

By concavity of $\sin$ again $\sin^{[N+1+k]}(c) \geq \epsilon^{-k} (\sin \epsilon)^k \sin^{[N+1]} c \geq \epsilon^{-k} (\sin \epsilon)^k \epsilon \sin 1$ holds for all $k \in \mathbb{N}$, so that in particular $\sum_{k=0}^\infty a_{N+1+k} \geq \sin (1) \epsilon (1 - \epsilon^{-1} \sin \epsilon)^{-1}$.

Since $\epsilon (1 - \epsilon^{-1} \sin \epsilon)^{-1} \to \infty$ as $\epsilon \to 0$, we can conclude that for any $K > 0$, we may find $N$ such that $\sum_{n=N+1}^\infty a_n > K$, hence the sum diverges.

  • $\begingroup$ I don't understand how you got to $\sin^{[N+1+k]}(c)\ge \epsilon^{-k}...$ and how did you got to the last inequality of $\sum\limits_{k=0}^{\infty}...\ge\sin(1)\epsilon....$. Can you explain those parts? $\endgroup$
    – ℋolo
    Oct 22, 2017 at 4:07
  • $\begingroup$ @Holo For $0 < x < y < \pi/2$, $\sin x > x y^{-1} \sin y$. One can easily see this by consider the graph of $\sin x$ on the domain and draw a straight line between $(0,0)$ and $(y, \sin y)$ (it's below the graph) or prove directly from the definition of concavity. Note also that $\sin x < x < y$, so we can repeat $\sin (\sin x) > \sin (x) y^{-1} \sin y > x y^{-2} (\sin y)^2$. By induction, we have $\sin^{[N]} x > x y^{-N} (\sin y)^N$. Define $\sin^{[0]} x = x$, then we obtain $\sum_{n=0}^\infty \sin^{[n]} x \geq \sum_{n=0}^\infty x y^{-n} (\sin y)^n$, which is an infinite geometric sum. $\endgroup$ Oct 22, 2017 at 7:01
  • $\begingroup$ @Holo $\sum_{n=0}^\infty r^{n} = \frac{1}{1 - r}$ for $r \in (0,1)$. $\endgroup$ Oct 22, 2017 at 7:03

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