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Thread: Convergence or divergence

  1. #1
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    Convergence or divergence

    I am not sure how to work this problem out.

    Investigate the behavior (convergence or divergence ) of
    ∑an if
    an = 1/(1 + z^n)
    Any help will be appreciated
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  2. #2
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    Quote Originally Posted by poorna View Post
    I am not sure how to work this problem out.

    Investigate the behavior (convergence or divergence ) of
    ∑an if
    an = 1/(1 + z^n)
    Any help will be appreciated
    Is $\displaystyle z$ a complex number?

    If so, $\displaystyle z = r\cos{\theta} + i\,r\sin{\theta}$

    and $\displaystyle z^n = r^n\cos{n\theta} + i\,r^n\sin{n\theta}$.


    So $\displaystyle a_n = \frac{1}{1 + z^n}$

    $\displaystyle = \frac{1}{1 + r^n\cos{n\theta} + i\,r^n\sin{n\theta}}$

    $\displaystyle = \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{(1 + r^n\cos{n\theta})^2 + r^{2n}\sin^2{n\theta}}$

    $\displaystyle = \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}\cos^2{\theta} + r^{2n}\sin^2{n\theta}}$

    $\displaystyle = \frac{1 + r^n\cos{\theta} - i\,r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}$

    $\displaystyle = \frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}} + i \left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)$


    So $\displaystyle \sum_n{\left(\frac{1}{1 + z^n}\right)} = \sum_n{\left[ \frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}} + i \left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)\right]}$

    $\displaystyle = \sum_n{\left(\frac{1 + r^n\cos{\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)} + n\,i\sum_n{\left(-\frac{r^n\sin{n\theta}}{1 + 2r^n\cos{n\theta} + r^{2n}}\right)}$.


    Now investigate the behaviour of both of these sums.
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