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Homework Help: Sum question.

  1. Feb 27, 2006 #1

    MathematicalPhysicist

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    sum question.........

    i have b_k=a_(n_(k)+1)+...+a_(n_(k+1))

    c_k=b_1+...+b_k
    let us suppose that for every term in b_k has the same sign and that the sum b_k (from n=1 to infinity) converges.
    S_n=a_1+...+a_n
    and n belongs to {n_k+1,...,n_(k+1)}

    then i need to show that
    |S_n-c_n|=|a_(n_(k)+1)+...+a_n|

    i tried just opened it, and it looked disastrous ( if that's even a word in the anglo-american lexicon).
    anyway, your help is appreciated.
     
  2. jcsd
  3. Feb 27, 2006 #2

    benorin

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    Do you mean [tex]b_k=a_{n_{k}+1}+\cdots +a_{n_{k+1}}[/tex] ?

    And, if so, I don't "get it".
     
  4. Feb 27, 2006 #3

    VietDao29

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    It's like your 700-th post here, and I guess it does not hurt much if you try to learn how to LaTeX properly, right?
    I don't really get what you mean, either... :frown:
     
  5. Feb 28, 2006 #4

    MathematicalPhysicist

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    yes, your latex is correct.
    and to the other user, now it's 701, but who counts it anyway? (-:
     
  6. Feb 28, 2006 #5

    qtp

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    funny thing is you are already using latex lol just put the [ tex] [ /tex] on either end and change ( ) for { } and it's pretty much the same thing
     
  7. Mar 1, 2006 #6

    MathematicalPhysicist

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    now after we cleared it off, can somone please help me on this?
     
  8. Mar 2, 2006 #7

    VietDao29

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    Well, I don't think I understand the problem fully, but you can try to open everything:
    [tex]b_k = a_{n_k + 1} + ... + a_{n_{k + 1}}[/tex]
    So:
    [tex]c_k = \sum_{i = 1} ^ k b_i = a_{n_1 + 1} + a_{n_1 + 2} + ... + a_{n_2} + a_{n_2 + 1} + ... + a_{n_3} + ... + a_{n_{k + 1}}[/tex]
    [tex]S_n = \sum_{i = 1} ^ n a_i[/tex]
    So that means:
    [tex]S_n - c_n = ...[/tex]
    EDIT: But are you sure you posted this problem correctly? I don't think it's correct... :frown:
     
    Last edited: Mar 2, 2006
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