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Math Help - Is properly divergent series still properly divergent under rearrangement?

  1. #1
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    Is properly divergent series still properly divergent under rearrangement?

    For series \sum\limits_{n = 1}^\infty a_n, it is said to be properly divergent if its partial sum \sum\limits_{i = 1}^n a_i approaches +\infty (or -\infty) as n\to \infty. Then, is it true that any properly divergent series is still properly divergent under arbitrary rearrangement? Could you please given a proof if it is true or a counterexample if it does not always hold? Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by zzzhhh View Post
    For series \sum\limits_{n = 1}^\infty a_n, it is said to be properly divergent if its partial sum \sum\limits_{i = 1}^n a_i approaches +\infty (or -\infty) as n\to \infty. Then, is it true that any properly divergent series is still properly divergent under arbitrary rearrangement? Could you please given a proof if it is true or a counterexample if it does not always hold? Thanks!
    Let's consider the two series...

    \displaystyle 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}} + ... (1)

    \displaystyle 1 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{7}} - \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{4n -3}} + \frac{1}{\sqrt{4n-1}} - \frac{1}{\sqrt{2n}} + ... (2)

    ... each of them can be derived from the other rearranging the terms. The series (1) is convergent according to the Leibnitz's rule. Now we indicate with \sigma_{2n} the sum of the first 2n terms of (1) and with \sigma_{3n} the sum of the first 3n of (2), so that is...

    \displaystyle \sigma_{3n} = \sigma_{2n} + \frac{1}{\sqrt{2n+1}}+ \frac{1}{\sqrt{2n+3}} +...+ \frac{1}{\sqrt{4n-1}} (3)

    ... and from (3) we derive that is...

    \displaystyle \sigma_{3n} > \sigma_{2n} + \frac{n}{\sqrt{4n-1}} (4)

    But is...

    \displaystyle \lim_{n \rightarrow \infty} \frac{n}{\sqrt{4n-1}} = \infty (5)

    ... so that the series (2) diverges...

    Kind regards

    \chi \sigma
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    An ingenious answer! thank you very much, chisigma! :-)
    This also shows that the countable additivity condition of signed measure is really a stringent requirement.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Although the question has been solved, I'd like to comment that for any S\in [-\infty,+\infty] and for any conditionally conergent series there exists an arrangement whose sum is S (Riemann's arrangement theorem) .


    Fernando Revilla
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