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Math Help - A Problem about series

  1. #1
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    Cool A Problem about series

    Let's consider a series defined by x_{n+1} = sin(x_n). x_0\in(0,pi)
    then construct a series : \sum{x_n}^p, n from 0 to \infty
    Now discuss the convergence of the above series.
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    Thus, |x_n|\leq x_1 < 1 but since x_1^{1/n} \to 0 it means |x_n|^{1/n} \to 0 by the root test it means the series converges.
    Umm, unfortunately x_1^{1/n} \to 1, not 0, as n\to\infty.

    I haven't been able to think of any way of attacking this problem. It's clear that x_n\to 0 as n\to\infty, but it approaches 0 so desperately slowly that I very much doubt whether \sum x_n converges. Maybe the series of p'th powers will converge if p is large enough, but I don't see why even that should hold.

    It's a very interesting problem! I hope someone can find a way to tackle it.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Umm, unfortunately x_1^{1/n} \to 1, not 0, as n\to\infty.
    Sorry for the mistake.
    (My approach would be to ignore the p and try to do the simplified problem first.)
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  4. #4
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    I can see now how to do this for p=1.

    Since x_0 > 0 there is an integer k such that x_0 > 1/k. The strategy is to prove by induction that x_n >1/(k+n). It will then follow (by comparison with the harmonic series) that \sum x_n diverges.

    The proof relies on the fact that \textstyle \sin x > x-\frac16 x^3 = x\left(1-\frac16x^2\right) (for 0<x<1). This is because the power series for sin(x) is an alternating series with decreasing terms, so the partial sums are alternately greater than, and less than, the whole sum.

    So suppose that x_n>1/(k+n). Then x_{n+1} = \sin(x_n) > \sin(1/(k+n)) > \frac1{k+n}\left(1-\frac1{6(k+n)^2}\right). To complete the inductive step, we want to show that this is greater than \frac1{k+n+1} = \frac1{k+n}\left(1-\frac1{k+n+1}\right). But that is equivalent to 6(k+n)^2 > k+n+1, which is certainly true.

    A slightly more elaborate form of this same argument shows that the series \sum x_n^p diverges whenever p<2. I still have no idea what happens when p \geqslant2.
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