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Following interesting trigonometric series came into my mind when I solve another series problem.

$\sum\limits_{n=1}^{\infty}\underbrace{\sin \left ( \sin \left ( \sin \left ( \dots \sin \left ( \dfrac {\pi}{2} \right ) \dots \right ) \right ) \right )}_{n \text { number of } \sin \text { terms}}\\=\sin \left ( \dfrac {\pi}{2} \right )+\sin \left ( \sin \left ( \dfrac {\pi}{2} \right ) \right )+\sin \left ( \sin \left ( \sin \left ( \dfrac {\pi}{2} \right ) \right ) \right )+\dots\\=1+\sin \left ( \sin \left ( \dfrac {\pi}{2} \right ) \right )+\sin \left ( \sin \left ( \sin \left ( \dfrac {\pi}{2} \right ) \right ) \right )+\dots $

Since for each $x\in \mathbb {R}$, $-1\le \sin x \le 1$, $n$th term of the series is defined and lies between minus one and plus one.

Do you have any idea about convergence of the series?

I believe that many of you have nice ideas and arguments. So please share with us. Thank you.

ASB
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    It diverges, the $n^{\text{th}}$ term is $\sim \sqrt{\frac{3}{n}}$. Let me search a bit. – Daniel Fischer Jun 23 '17 at 12:52
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    Here – Daniel Fischer Jun 23 '17 at 12:54
  • Denote by $u$ the sequence defined by $u_0 =1$ and $u_{n+1} = \operatorname{sin} u_n$. Clearly your question amounts to the following : does $\displaystyle\sum u_n$ converge ? To answer this, you can try 1. Proving that $u_n\to 0$ 2. Finding an equivalent of $u_n$ (for 2., try finding $\alpha$ such that $\frac{1}{u_{n+1}^\alpha} - \frac{1}{u_n^\alpha}$ behaves nicely, then you can apply Cesaro's theorem) – Maxime Ramzi Jun 23 '17 at 12:55
  • Thank you, Daniel Fischer. – ASB Jun 23 '17 at 13:26
  • @DanielFischer : If you formalized your comments as an answer, we could vote it up, and so clear this question from the list of “unanswered” questions. – John Bentin Apr 30 '20 at 17:23
  • @JohnBentin Done. It may be faster if you convert comments to an answer yourself, however. – Daniel Fischer Apr 30 '20 at 20:01

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As shown here, for every $x_0$ such that $\sin x_0 > 0$ the sequence defined by the recursion $$x_{n+1} = \sin x_n$$ satisfies the asymptotic relation $$x_n \sim \sqrt{\frac{3}{n}}\,,$$ in other words $\sqrt{n}\cdot x_n \to \sqrt{3}$. When $\sin x_0 < 0$ we have the similar $\sqrt{n}\cdot x_n \to -\sqrt{3}$ of course, since the sine is odd. And if $\sin x_0 = 0$, then $x_n = 0$ for all $n \geqslant 1$.

It follows that $$\sum_{n = 0}^{\infty} x_n$$ converges if and only if $x_0 \in \pi\mathbb{Z}$.

Daniel Fischer
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