Sunday, 17 November 2019

analysis - Convergence of series involving iterated sin









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,




  • sinn10 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=sinsinn1sinn1 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(nϵ) 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)=3n(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.


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