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Math Help - Prove that the sequence is monotone and bounded above

  1. #1
    MHF Contributor alexmahone's Avatar
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    Prove that the sequence is monotone and bounded above

    Using calculus, prove (for n\ge 2) that A_n=2^{n-1}\sin \frac{\pi}{2^{n-1}} is monotone (increasing) and bounded above. Find its limit.
    Last edited by alexmahone; January 15th 2012 at 03:33 PM.
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    Re: Prove that the sequence is monotone and bounded above

    Expand sin(x) as x + O(x); then the limit of A_n becomes obvious. To prove monotonicity and boundedness, expand A_n as a_0-a_1(n)+a_2(n)-\dots. This series satisfies the conditions to the alternating series test. From the info on that page, A_n\le a_0. Also, A_n\in[a_0-a_1(n),a_0-a_1(n)+a_2(n)]. You can show that a_0-a_1(n+1)>a_0-a_1(n)+a_2(n), so A_{n+1}>A_n.

    Another way to show monotonicity is to consider f(x)=x\sin(1/x) and show that its derivative is eventually positive (also using series).
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    MHF Contributor chisigma's Avatar
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    Re: Prove that the sequence is monotone and bounded above

    Setting x=2^{1-n} You have...

    \lim_{n \rightarrow \infty} 2^{n-1}\ \sin \frac{\pi}{2^{n-1}} = \lim_{x \rightarrow 0} \frac{\sin \pi x}{x} = \pi

    Kind regards

    \chi \sigma
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