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Suppose $a_1 = 1, a_{n+1} = \sin (a_n).$ Does the series $$\sum_{n=1}^\infty a_n$$ converge?

First of all I got that $\lim_{n \to \infty} a_n = 0$ it's easy. Then I tried using all tests I know (The Cauchy root test, The Ratio test, etc.) and I failed to prove either convergence or divergence. Some of my results: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$$ $$\lim_{n \to \infty} \sqrt[n]{a_n} = 1$$ $$a_{n+1} < a_n$$ So, I don't know what to do. Could you please give me any hints about this problem? Thanks in advance!

D F
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    This was posted several times on the site. Hint: Consider $$b_n=\frac1{a_n^2}$$ and, using the expansion $\sin x-x\sim-x^3/6$ when $x\to0$, show that $$b_{n+1}-b_n\to\frac13$$ hence $$a_n\sim\sqrt{\frac3n}$$ – Did Mar 07 '18 at 17:25
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    @Did Unfortunately, I haven't managed to find the same question on the site, but thank you a lot for your answer! – D F Mar 07 '18 at 17:31
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    https://math.stackexchange.com/q/3215/ – Did Mar 07 '18 at 17:42
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    https://math.stackexchange.com/q/276135/ – Did Mar 07 '18 at 17:42
  • @Did oh, why I didn't guess to look for $x_n$ instead of $a_n$... Thank you, once again! – D F Mar 07 '18 at 17:47

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