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I've been trying to show the convergence or divergence of

$$ \sum_{n=1}^\infty \frac{\sin^n 1}{n} = \frac{\sin 1}{1} + \frac{\sin \sin 1}{2} + \frac{\sin \sin \sin 1}{3} + \ \cdots $$

where the superscript means iteration (not multiplication, so it's not simply less than a geometric series -- I couldn't find the standard notation for this).

Problem is,

  • $ \sin^n 1 \to 0 $ as $ n \to \infty $ (which I eventually proved by assuming a positive limit and having $ \sin^n 1 $ fall below it, after getting its existence) helps the series to converge,

but at the same time

  • $ \sin^{n+1} 1 = \sin \sin^n 1 \approx \sin^n 1 $ for large $ n $ makes it resemble the divergent harmonic series.

I would appreciate it if someone knows a helpful convergence test or a proof (or any kind of advice, for that matter).

In case it's useful, here are some things I've tried:

  • Show $ \sin^n 1 = O(n^{-\epsilon}) $ and use the p-series. I'm not sure that's even true.
  • Computer tests and looking at partial sums. Unfortunately, $ \sum 1/n $ diverges very slowly, which is hard to distinguish from convergence.
  • Somehow work in the related series $$ \sum_{n=1}^\infty \frac{\cos^n 1}{n} = \frac{\cos 1}{1} + \frac{\cos \cos 1}{2} + \frac{\cos \cos \cos 1}{3} + \ \cdots $$ which I know diverges since the numerators approach a fixed point.
Vandermonde
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    @Qiaochu: this problem is not a duplicate, it is asking for something much weaker than the other problem, and there may be additional solutions that do not cite the other result. (Not that I know of any, but a suspicion that any solution of this problem has to repeat the other one is not enough to make the question a clone). – T.. Dec 17 '10 at 18:12
  • @T..: fair enough. – Qiaochu Yuan Dec 17 '10 at 18:55
  • Boy do I feel stupid for not noticing the other question. Sorry. – Vandermonde Dec 18 '10 at 02:42
  • You shouldn't feel bad about not seeing the other question. I don't see any obvious way for you to have searched for it. In any case, even if were exactly the same question, duplicates are not a big deal. Duplicates get closed because there is no reason to keep 2 threads open in such cases, not as punishment for offenders. – Jonas Meyer Dec 18 '10 at 03:18
  • See also here (num.U213) https://www.awesomemath.org/wp-content/uploads/reflections/2012_1/MR6solutions.pdf for a proof of the first two terms of the expansion – Unoqualunque Dec 14 '13 at 09:30
  • Iteration is typically marked as $\sin^{(n)}(1)$ adding it in the comment that it is iteration of the function as it can be confused with n-th derivative. Otherwise... it is very easy to make a confusion of other kind. –  Aug 03 '19 at 10:31

1 Answers1

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A Google search has turned up an analysis of the asymptotic behavior of the iterates of $\sin$ on page 157 of de Bruijn's Asymptotic methods in analysis. Namely,

$$\sin^n(1)=\frac{\sqrt{3}}{\sqrt{n}}\left(1+O\left(\frac{\log(n)}{n}\right)\right),$$

which implies that your series converges.

Edit: Aryabhata has pointed out in a comment that the problem of showing that $\sqrt{n}\sin^n(1)$ converges to $\sqrt{3}$ already appeared in the question Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ (asked by Aryabhata in August). I had missed or forgot about it. David Speyer gave a great self contained answer, and he also referenced de Bruijn's book. De Bruijn gives a reference to a 1945 work of Pólya and Szegő for this result.

Community
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Jonas Meyer
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    Wow, so apparently even $ \sum_{n=1}^\infty \frac{\sin^n 1}{n^{1/2 + \epsilon}} $ would converge. I've got to wonder: how do you search for math like that? – Vandermonde Dec 12 '10 at 04:56
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    The search was "iterations of sine asymptotic" without quotes, and the Google Books result for page 157 of de Bruijn's book was the first hit. – Jonas Meyer Dec 12 '10 at 05:12