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Math Help - complex analysis series

  1. #1
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    complex analysis series

    Let a_n, b_n be nonzero complex numbers. Suppose lim_{n\to\infty}|\frac{a_n}{b_n}|=l exists, and l \not= 0, \infty.
    1. Show that if one of the series \sum_{n=0}^\infty a_n and \sum_{n=0}^\infty b_n converges absolutely, then so does the other.
    2. What if l=0 or l=\infty?

    For #1, I assume \sum_{n=0}^\infty a_n converges absolutely and trying to prove \sum_{n=0}^\infty b_n converges absolutely. That is, \forall \epsilon>0 there exists N \in \mathbb{N} such that |b_{n+1}|+|b_{n+2}|+...+|b_{n+p}|<\epsilon for p>0.
    I tried to use what I have to prove this, but not getting what I want. Can I get some help?

    I don't know #2...
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Let a_n, b_n be nonzero complex numbers. Suppose lim_{n\to\infty}|\frac{a_n}{b_n}|=l exists, and l \not= 0, \infty.
    1. Show that if one of the series \sum_{n=0}^\infty a_n and \sum_{n=0}^\infty b_n converges absolutely, then so does the other.
    2. What if l=0 or l=\infty?

    For #1, I assume \sum_{n=0}^\infty a_n converges absolutely and trying to prove \sum_{n=0}^\infty b_n converges absolutely. That is, \forall \epsilon>0 there exists N \in \mathbb{N} such that |b_{n+1}|+|b_{n+2}|+...+|b_{n+p}|<\epsilon for p>0.
    I tried to use what I have to prove this, but not getting what I want. Can I get some help?

    I don't know #2...

    As \lim_{n\to\infty}\frac{a_n}{b_n}=l , we have that |l|-\epsilon<\left|\frac{a_n}{b_n}\right|<|l|+\epsilon  \Longleftrightarrow (|l|-\epsilon)|b_n|<|a_n|<(|l|+\epsilon)|b_n| for all but a finite number of indexes n\in\mathbb{N} , so:

    1) if \sum\limits_{n=0}^\infty |a_n| converges then so converges \sum\limits_{n=0}^\infty (|l|-\epsilon)|b_n| by the comparison test and thus...

    2) if \sum\limits_{n=0}^\infty |b_n| converges then again by the comparison test converges the series ....

    End both arguments above, and note that we can choose \epsilon > 0 so as to get |l|-\epsilon > 0 (why is this important?)

    Tonio
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