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Let $a_{n+1} = \sin{a_n}$ and $a_0 = 1$. Does the series $\displaystyle \sum_{i=0}^\infty a_i$ converge?

Here is my solution.

Let's prove by induction, that $a_n > \frac{1}{n}$.

We see that $\sin{(1)} > \frac{1}{2}$.

Suppose, that $a_n > \frac{1}{n}$. Then:

$$\underbrace{\sin{\sin{(...\sin{(1)}...)}} )}_\text{$n+1$ sines}\ ?\ \frac{1}{n+1}$$

$$\underbrace{\sin{\sin{(...\sin{(1)}...)}} )}_\text{$n$ sines}\ ?\ \mbox{arcsin}(\frac{1}{n+1})$$

If $\mbox{arcsin}(\frac{1}{n+1}) < \frac{1}{n}$ (becuase left part is bigger then it by induction hypothesis) then we are done. And it really is, because $\sin{\frac{1}{n}} > \frac{1}{n+1} \Leftrightarrow \frac{1}{n} - \frac{1}{n^33!} > \frac{1}{n+1}$ (here I used Teylor expansion for sine).

As $a_n > \frac{1}{n}$ our series is divergent by comparison test.

Is my reasoning correct?

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marked as duplicate by J. M. is a poor mathematician, Martin R, Willie Wong, Lord Shark the Unknown, Henrik May 26 '17 at 21:26

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    $\begingroup$ Yes. See here math.stackexchange.com/questions/2293125/…. $\endgroup$ – eyeballfrog May 26 '17 at 18:42
  • $\begingroup$ You seem to show $a_n>\frac1{n+1}$ for $n\ge 1$. $\endgroup$ – Hagen von Eitzen May 26 '17 at 18:46
  • $\begingroup$ @eyeballfrog The question was: Does the series converge? $\endgroup$ – zhw. May 26 '17 at 19:35
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    $\begingroup$ @zhw. "Is my reasoning correct?" seemed to be the more relevant question. $\endgroup$ – eyeballfrog May 26 '17 at 19:41
  • $\begingroup$ @eyeballfrog Ah you're right, I didn't read that far. $\endgroup$ – zhw. May 26 '17 at 19:45
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Given $a_0=1$ and $a_{n+1}=\sin(a_n)$, it is pretty clear that $a_n\to 0$ as $n\to +\infty$.
On the other hand we may show that $a_n$ behaves like $\frac{C}{\sqrt{n}}$ for large values of $n$, hence the series $\sum_{n\geq 1}a_n$ is divergent by asymptotic comparison and the $p$-test. In a neighbourhood of the origin we have $\sin(x)=x-\frac{x^3}{6}+o(x^4)$, hence $a_{n+1}-a_n\sim -\frac{a_n^3}{6}$. The positive solutions of the differential equation $f'(x)=-\frac{f(x)^3}{6}$ have the form $f(x)=\frac{\sqrt{3}}{\sqrt{x+K}}$, hence $$ \lim_{n\to +\infty} \sqrt{n} a_n = \sqrt{3} \tag{1}$$ and $\sum_{n=1}^{N}a_n$ diverges like $\sum_{n=1}^{N}\frac{\sqrt{3}}{\sqrt{n}} \approx 2\sqrt{3N}$. The previous limit is indeed pretty well-known and it can be proved through the Stolz-Cesàro theorem too.

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  • $\begingroup$ @FelixMarin: you are correct (as usual), now fixed. $\endgroup$ – Jack D'Aurizio May 28 '17 at 13:48

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