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Rearangement of series

  1. Jun 8, 2007 #1

    quasar987

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    My book proves that if a series is absolutely convergent, then every rearangement is absolutely convergent also.

    They then argue that the thm does not hold for conditionally convergeant series and give as a counter exemple the following thing: Let [itex]\sum a_n[/itex] be a conditionally convergent series and define the positive part of {a_n} by p_n = a_n if a_n > 0 and =0 otherwise and the negative part of {a_n} by q_n = a_n if a_n < 0 and =0 otherwise. Then observe that

    [tex]p_n=\frac{a_n+|a_n|}{2}[/tex]

    and

    [tex]q_n=\frac{a_n-|a_n|}{2}[/tex]

    or that inversely,

    [tex]|a_n|=p_n-q_n[/tex] and [tex]a_n=p_n+q_n[/tex]

    But what does this prove? Where is the rearangement?
     
  2. jcsd
  3. Jun 8, 2007 #2
    all i can imagine is that neither of those new smaller series is convergent

    edit

    i think only [tex] p_n [/tex] is divergent so i don't know what that does

    mathworld says

    If [tex]\Sigma u_k[/tex] and [tex]\Sigma v_k[/tex] are convergent series, then [tex]\Sigma (u_k+v_k)[/tex] and [tex]\Sigma (u_k-v_k)[/tex] are convergent.

    but i don't know if the reverse ( or w/e the proper word is ) has to be true

    where if [tex]\Sigma u_k[/tex] is divergent then [tex]\Sigma (u_k+v_k)[/tex] is also divergent
     
    Last edited: Jun 8, 2007
  4. Jun 9, 2007 #3

    quasar987

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    Ugh. Sorry ice109 if you spent time thinking about this. I just realized that the definitions and observations made in my OP are just a kind of lemma to Riemann's theorem that every conditionally convergent series can be rearranged to converge to any real number or to diverge.
     
  5. Jun 9, 2007 #4
    which theorem is that?
     
  6. Jun 9, 2007 #5

    quasar987

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    well I just stated it:

    "every conditionally convergent series can be rearranged to converge to any real number or to diverge."
     
  7. Jun 9, 2007 #6
    i meant which theorem so i coud look up a proof
     
  8. Jun 9, 2007 #7

    quasar987

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    What more can I say other than that it's a thm of Riemann and give you the exact statement? :confused:
     
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