Hi all,
I've been trying to show the convergence or divergence of
∞∑n=1sinn1n=sin11+sinsin12+sinsinsin13+ ⋯
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,
sinn1→0 as n→∞ (which I eventually proved by assuming a positive limit and having sinn1 fall below it, after getting its existence) helps the series to converge,
but at the same time
sinn+11=sinsinn1≈sinn1 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 sinn1=O(n−ϵ) and use the p-series. I'm not sure that's even true.
- Computer tests and looking at partial sums. Unfortunately, ∑1/n diverges very slowly, which is hard to distinguish from convergence.
- Somehow work in the related series
∞∑n=1cosn1n=cos11+coscos12+coscoscos13+ ⋯
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,
sinn(1)=√3√n(1+O(log(n)n)),
which implies that your series converges.
Edit: Aryabhata has pointed out in a comment that the problem of showing that √nsinn(1) converges to √3 already appeared in the question Convergence of √nxn where xn+1=sin(xn) (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.
No comments:
Post a Comment