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Math Help - series

  1. #1
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    series

    determine the convergence:

     <br />
\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )<br /> <br />

    thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by karimath View Post
    determine the convergence:

     <br />
\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )<br /> <br />

    thanks
    This does not converhe for any x \in (0,\pi) . The proof hinges on showing that \sin(nx) \not\to 0 as n \to \infty . Then the limit of the terms does not go to zero, and so the series cannot converge.

    A proof that \sin(nx) \not\to 0 as n \to \infty can be constructed by supposing otherwise, that is we suppose (for fixed x ) that:

    \sin(nx) \to 0 as n \to \infty .

    This means that for all \varepsilon>0 there exists an N_{\varepsilon} such that for all n>N_{\varepsilon} :

    |\sin(nx)|<\varepsilon

    Now:

    |\sin((n+1)x)|=|\sin(nx)\cos(x)+\cos(nx)\sin(x)|

    but by making \varepsilon sufficiently small we can make the right hand side of the last equation as close to |\sin(x)| as we like for n>N_{\varepsilon}, but this contradicts the consequence of our assumption that |\sin((n+1)x)|<\varepsilon for sufficently small \varepsilon


    UNDER CONSTRUCTION
    Last edited by CaptainBlack; August 16th 2008 at 10:52 AM.
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