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Math Help - Complex Series Convergence

  1. #1
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    Complex Series Convergence

    Let \sum_{n=1}^\infty z_n be a series of complex numbers.. Show that if \lim_{n \to \infty}|\frac{z_{n+1}}{z_n}|=L<1 then there exists N such that for all positive integers k: |z_{N+k}|\leq|z_N|(\frac{L+1}{2})^k.

    My first thoughts were to find a specific epsilon in terms of L and N to make the equation, but I couldn't find any. Could someone point me in the right direction?
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  2. #2
    Super Member girdav's Avatar
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    Put \varepsilon =\frac{1-L}2.
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  3. #3
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    So I then get \forall \epsilon >0 \exists N s.t. \forall n \geq N: |\frac{z_{n+1}}{z_n}-L|<\frac{1-L}{2}. Can I then say from this that:
    |\frac{z_{n+1}}{z_n}|<\frac{1+L}{2}?
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  4. #4
    Super Member girdav's Avatar
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    Since we have \frac{z_{n+1}}{z_n}-L<\frac{1-L}2, the answer is yes.
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