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Thread: series

  1. #1
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    series

    determine the convergence:

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )

    $

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

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )

    $

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

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

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

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

    $\displaystyle |\sin(nx)|<\varepsilon$

    Now:

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

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


    UNDER CONSTRUCTION
    Last edited by CaptainBlack; Aug 16th 2008 at 09:52 AM.
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