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Infinite sum question

  1. Jan 1, 2013 #1
    Happy new year. All the best.

    I have one question. Is it true?
    I saw in one book relation
    Can you give me some explanation for this step?
  2. jcsd
  3. Jan 1, 2013 #2


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    hi matematikuvol! :smile:

    the RHS is a function of n (ie, it's different for different n), but the LHS isn't …

    so that equation can't possibly be correct :redface:

    perhaps they mean
    (which obviously is true)
  4. Jan 1, 2013 #3
    Hi matematikuvol,
    Hi tiny-tim
    I think what is missing in your equations is the term limit.
    That is, the infinite sum would be (might be) the limit of the other sum when n goes to infinity.
    But in this case, beware that although what tiny-tim says about it being obviously true (you just reverse the summation order) this is not necessarily true anymore when you take the limit.
    For it to hold even when n goes to infinity your sum must be absolutely convergent, otherwise changing terms order in interesting ways could lead you to find the limit to be just about anything (0, ∏, whatever)
  5. Jan 1, 2013 #4

    This particular derangement doesn't affect the sum when passing to the limit because [itex] \forall n \in \mathbb{N} , \sum_{k=0}^n a_k x^k = \sum_{k=0}^n a_{n-k} x^{n-k} [/itex]. Thus [itex] \sum_{k=0}^{\infty} a_k x^k = \sum_{k=0}^{\infty} a_{n-k} x^{n-k} [/itex]
  6. Jan 1, 2013 #5


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    Your last remark is incorrect, since you are still looking at an n in each term of the infinite sum.
    You would need ∞-k, which is meaningless in this context.
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