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Limits of sums

  1. Feb 5, 2010 #1
    I'm stuck on how to approach the following problem.

    lim_{x \rightarrow \infty} \sum_{j=0} ^x e^{-j/x}

    Does anyone have any ideas?
  2. jcsd
  3. Feb 5, 2010 #2


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    Science Advisor

    It looks to me that the jth term -> 1 for any j, which makes the sum divergent to ∞.
  4. Feb 5, 2010 #3


    Staff: Mentor

    Looks like infinity to me. Expanding your sum gives
    [tex]e^0 + e^{-1/N} +e^{-2/N} + ... + e^{-N/N} [/tex]
    In this expression there are N + 1 terms, the smallest of which is 1/e.

    [tex]\sum_{i = 0}^N e^{-i/N} \geq (N + 1)(1/e)[/tex]

    As N gets larger, so does (N + 1)/e.
  5. Feb 5, 2010 #4
    Thanks you both for your replies. You are right, the expression does converge, as the integral test seems to varify.

    Mark, I have one quick question. Could you explain why
    \sum_{i = 0}^N e^{-i/N} \geq (N + 1)(1/e)

    Is this a general statement or does it only apply in this case.

    Thank you again.
  6. Feb 5, 2010 #5


    Staff: Mentor

    No, what mathman and I are saying is that the expression diverges.
    This statement applies to your sum. For each value of N, there are N + 1 terms being added. The smallest of these terms is 1/e, so we know that the sum has to be at least (N + 1) times 1/e, which is (N + 1)/e.
  7. Feb 6, 2010 #6
    Yes, I meant to say diverges.
    Thanks again for the thorough explanation.
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