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Series convergence by Parts

  1. Jun 10, 2010 #1
    Series convergence "by Parts

    Supose:

    [tex] \sum c_n = \sum (a_n+b_n) [/tex] (*1)

    [tex] \sum a_n [/tex] is conditionaly convergent (*2)

    [tex] \sum b_n [/tex] is absolutly convergent (*3)

    And I have seen this proof: [Proving [tex] \sum c_n[/tex] is conditionally convergent]

    From (*1) and (*2) [tex]\Rightarrow[/tex] [tex] \sum c_n[/tex] its convergent [this one I understand, basic properties of series]

    But now they do something like this: [proof by contradiction]

    Supose [tex] \sum |c_n| [/tex] is convergent

    so [tex] |a_n|\leq|c_n-b_n|\leq|c_n|+|b_n| [/tex] (How they "jump" to this conclusion?!:confused:)

    and now the use comparison test to show that [tex]\sum c_n[/tex] is conditionally convergent. [No problems from here]
     
    Last edited: Jun 10, 2010
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  3. Jun 10, 2010 #2

    LCKurtz

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    Re: Series convergence "by Parts

    It's easy:

    cn = an + bn
    an = cn - bn
    |an| = |cn - bn| ≤ |cn| + |bn|

    This would imply the an series converges absolutely, which is a contradiction.
     
  4. Jun 10, 2010 #3
    Re: Series convergence "by Parts

    I don't understand this:

    [tex]If\ \sum c_n = \sum (a_n+b_n)\ convergent,[/tex]

    [tex]why\ c_n=a_n+b_n[/tex].

    Don't understand the theory.
     
  5. Jun 10, 2010 #4

    LCKurtz

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    Re: Series convergence "by Parts

    You are given two series, ∑an and ∑bn and examining their sum

    ∑an + ∑bn = ∑(an+bn)

    to see if it is conditionally convergent. It is just a convenience to call the term

    an+bn on the right side cn. There is nothing to prove about that.
     
  6. Jun 10, 2010 #5
    Re: Series convergence "by Parts

    Oh, bad idea to do such things in a book. (confusing)

    Thanks Again!
     
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