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Math Help - Comparision test for convergence

  1. #1
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    Comparision test for convergence

    I was wondering whether one showing series is greater than another is sufficent to show if the former converges, so does the latter. Thats what I took the proof in the link to imply. Cauchy condensation test - Wikipedia, the free encyclopedia. But I've figured that must be wrong. So how does this proof justify convergence simply by noting one series is bigger than the other?

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    Re: Comparision test for convergence

    What do you mean exactly by "a series is greater than another". It works if the a_n are nonnegative numbers.
    Last edited by girdav; August 24th 2011 at 07:12 AM.
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    Re: Comparision test for convergence

    If for all n, a_n\le b_n ("for all but a finite n" is sufficient) and a_n\ge 0, then if \sum b_n converges, so does \sum a_n.
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    Re: Comparision test for convergence

    I mean, as it shows on the link, the infinite sum. The difficulty I have with the proof is that it doesn't compare a_n to b_n, it just refers to the size of the series.
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    Re: Comparision test for convergence

    So if you show that the \sum b_n is greater than or equal to \sum a_n , does this imply b_n\ge a_n for n\ge N?
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    Re: Comparision test for convergence

    Quote Originally Posted by Duke View Post
    So if you show that the \sum b_n is greater than or equal to \sum a_n , does this imply b_n\ge a_n for n\ge N?
    No it does not. That is backwards.
    0 \leqslant a_n  \leqslant b_n \, \Rightarrow \,\sum\limits_n {a_n }  \leqslant \sum\limits_n {b_n } . Not the other way round.
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    Re: Comparision test for convergence

    ok thanks for that. I am still confused about 2 things. One is that \sum a_n does not always exist so how does the implication work then. Second, the proof of the cauchy condensation theorem (on wikipedia) does not compare each term in a_n with the corresponding term in b_n so how does the proof work?
    Last edited by Duke; August 26th 2011 at 03:55 AM.
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