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Math Help - z

  1. #1
    Super Member
    Joined
    Aug 2009
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    639

    z

    prove that the series summation from n=1 to infinity from (n+1) /(n+2) z^n converges if lzl<1 and diverges if lzl>1

    thanks
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  2. #2
    Junior Member
    Joined
    Apr 2010
    Posts
    61
    You can either use one of the many ways to find radius of convergence (e.g., root or ratio tests), or you might notice similarity to the geometric series.

    For example, \sum_{n=0}^\infty z^n is known to converge for \lvert z\rvert <1 and to diverge for \lvert z\rvert >1. You could say that since \frac{1}{2}\leq\frac{n+1}{n+2}\leq1, then \frac{1}{2}\sum_{n=0}^\infty z^n\leq\sum_{n=0}^\infty \frac{n+1}{n+2}z^n\leq\sum_{n=0}^\infty z^n. One of these inequalities shows convergence at \lvert z\rvert <1, and the other shows divergence at \lvert z\rvert >1 (itís not hard to figure out which is which).
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