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$$\sum _{ n=1 }^{ \infty }{ { a }_{ n } } $$ where $${ a }_{ n }=sin({ a }_{ n-1 })$$ and $${ a }_{ 0 }=1$$ I know the limit of an is zero but how do I know if the series converges or diverges? Can I show it using one of the tests?

John
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  • I posted a similar question on MathOverflow a while back. The answer is that it diverges. You can see it here for a very intricate proof, which is non obvious.

    https://mathoverflow.net/questions/245425/summability-of-iterates-of-analytic-function

    Note that I asked a more general question, that your question falls into.

     –  Apr 04 '18 at 03:54
  • @james.nixon I can't really understand what the other post is saying. Can you please simplify it for this example? – John Apr 04 '18 at 04:09
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    Possible duplicate of https://math.stackexchange.com/questions/2721047/how-to-calculate-limit-for-raabes-test-of-sum-n-1-infty-a-n-where-a-1-1 – saulspatz Apr 04 '18 at 04:31
  • Perhaps I'm missing something about the complexity in the other answers. Can't you just inductively prove a lower bound to show that it diverges? For instance, $a_{n} \ge 1/(n+1)$? – mjqxxxx Apr 04 '18 at 06:24
  • @mjqxxxx My answer wasn't really an answer, it was just a comment with a reference that happened to resolve the problem. There is probably an easier way, my comment was using an atom bomb to kill a fly. –  Apr 04 '18 at 17:57
  • @John I would explain it in simpler terms, but that's rather difficult in this case... Sadly I don't have a good mind for simplifying the complex. The related questions prove it rather wonderfully though. There's a reason I posted my case as a comment and not an answer... –  Apr 04 '18 at 21:22

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Let's first prove that $a_n \ge 1/(n+1)$ by induction. It's clearly true for $n=0$. Now assume it's true for some $k>0$, i.e., $1 \ge a_k \ge 1/(k+1)$. Since $\sin x$ is increasing on $[0,1]$, we have $a_{k+1}=\sin a_k \ge \sin \frac{1}{k+1}.$ And we know that $\sin x \ge x - \frac{1}{3!}x^3$ for positive $x$; so $$\sin\frac{1}{k+1}\ge \frac{1}{k+1}-\frac{1}{6(k+1)^3} > \frac{1}{k+1} - \frac{1}{(k+1)(k+2)}=\frac{1}{k+2}.$$ Therefore, $a_k \ge 1/(k+1) \implies a_{k+1} \ge 1/(k+2)$.

We conclude that $a_n \ge 1/(n+1)$ for all $n\ge 0$; and hence the series of iterated sines diverges by comparison with the harmonic series.

mjqxxxx
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