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Permuting the Terms

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the series [tex]\displaystyle\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n}[/tex] is not absolutely convergent. Do so by permuting the terms of the series one can obtain different limits.

    2. Relevant equations

    3. The attempt at a solution
    I don't have a total solution; because I am not familiar with the terminology of "permuting".
    I assume (for example): a permute [tex]\pi_1 = \{1,3,5,7...\}[/tex] all of the odd values. And another permute [tex]\pi_2 = \{2,4,6,8...\}[/tex] the even values. You could show that...

    [tex]\displaystyle\Sigma_{\pi_1(n)}^\infty \frac{(-1)^n}{n} \to x[/tex]
    Where as
    [tex]\displaystyle\Sigma_{\pi_2(n)}^\infty \frac{(-1)^n}{n} \to y \neq x[/tex]

    Is that on the right track?
    Last edited: Sep 1, 2010
  2. jcsd
  3. Sep 1, 2010 #2


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    A permutation is just re-arranging the sum. Just taking all the odd terms is not a permutation of the sum because you left out the even terms. An example of a permutation would be

    [tex]1+\frac{-1}{3}+\frac{1}{2}+\frac{-1}{5}+\frac{1}{4}...[/tex] where just every other term is swapped
  4. Sep 1, 2010 #3
    So since...

    [tex]1+\frac{-1}{3} + \frac{-1}{2} + \frac{-1}{5} + \frac{1}{4}[/tex] is not equal to the original series it's not absolutely convergent. That makes sense. I am familiar with

    Is there an appropriate nomenclature for describing this?

    Could you point me to a more formalized proof?
    I realize you could construct a permutation such that the first element [tex]\alef_0[/tex] of [tex]\pi_n[/tex] equals the greatest upper bound of [tex]n[/tex] But since there is no greatest upper bound, the [tex]\lim sup (n) = \infty[/tex] That is...if we assume [tex] n \in \mathbb{N}[/tex]. Therefore there will always exist a value of [tex]n[/tex] for [tex]\alef_0[/tex] of [tex]\pi_n[/tex] such that the sequence could have a limit of infinity.

    This is not necessarily an attempt at a form proof, but a series of ideas.
  5. Sep 1, 2010 #4


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    Ahh.... the series I gave is equal to the original series (you can see that every other partial sum is the same as the partial sums from the alternating harmonic series in the original post).

    A series that is not absolutely convergent is called conditionally convergent. Conditionally convergent series can be re-arranged in some ways without changing the sum.

    The idea behind how the value changes in general is by noticing that if you just take the positive terms, and add them together, you will get infinity. If you just take the negative terms and add them together, you get negative infinity. So if you want to re-arrange your series to give a value of 7 say, just take positive terms until you have added enough to get above 7, then add negative terms until you're below 7, then add positive terms until you're above 7, then add negative terms until you're below 7, etc.

    Of course to do that, you have to prove that the positive and negative terms each give non-converging series, which for this series is the same thing as just showing it's not absolutely convergent directly... notice the even terms [tex]\sum \frac{-1}{2n}=\frac{-1}{2}\sum \frac{1}{n}[/tex] which is the sum you're proving indirectly doesn't converge.

    All of this doesn't really help answer the original question though. As a hint, there's a way to relatively simple re-arrange your series so that you get half of the original value, by taking elements of the series with odd denominators and adding them with certain elements that have even denominators so that you end up with only even denominators left
  6. Sep 1, 2010 #5
    Thank you for your help. After I realized that it was the negative of the Harmonic series, I realized I could use the same form of adding values consecutively to be a negative with an even denominator. So that I could simply divide by 2 to get the original sequence rendering

    (...permuted modified sequence....) = (1/2)(....original sequence...)
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