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I'm trying to solve the following task

Sequence $\{a_n\}$ is given by the rule: $a_1 = 1,\: a_{n+1} = \sin (a_n)$. Does the series $\sum a_n$ converge?

Can you give me any hints how to solve it, cause i got totally stuck at the very beginning, please?

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  • $\begingroup$ @Misakov if series converges than $a_n \to 0$, but $a_n$can converge either monotonically either not, doesnt it? $\endgroup$
    – Alexander Markov
    May 23, 2017 at 9:09
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    $\begingroup$ It is indeed monotonically decreasing because of $\sin(x)<x$ (for $x>0$). But it is still not trivial to see whether it converges or not (imo). My first guess would be divergence. $\endgroup$
    – M. Winter
    May 23, 2017 at 9:11
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    $\begingroup$ See this. $\endgroup$
    – David Mitra
    May 23, 2017 at 9:23

3 Answers 3

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The series diverges. To see this, first note that $$ a_1 = 1\ge 1 $$ and that, if $a_n \ge 1/n$, then $$ a_{n+1} = \sin(a_n) \ge \sin(1/n) > 1/(n+1) $$ By induction, we have $a_n \ge 1/n$ for all $n$. Since $\sum\frac{1}{n}$ diverges, so does $\sum a_n$.

Note that $(n+1)\sin(1/n) > 1$ can be shown by Taylor expansion.

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Just an intuition:

Since $sin(x)$ almost equals $x$ for values of $x$ near $0$, so it's like adding the same term infinitely with hardly any change in it...so it would diverge for sure!

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If you look up "iterated sine" you will find that $\lim_{n \to \infty} \sqrt{n}\sin^{(n)}(x) =\sqrt{3} $ independent of $x$.

See, for example, http://jonathan.bergknoff.com/journal/iterated-sine

Therefore $\sin^{(n)}(x) \approx \sqrt{\dfrac{3}{n}} $. Since $\sum \dfrac1{\sqrt{n}}$ diverges, $\sum \sin^{(n)}(x) $ diverges.

Even more is true. Since $(\sin^{(n)}(x))^2 \approx \dfrac{3}{n} $, $\sum (\sin^{(n)}(x))^2 $ diverges.

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