Hi all,
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.
Answer
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.
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