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Question about limit of series

  1. Oct 21, 2008 #1
    This isn't a homework question, it's a question I encountered during self-study.

    It said prove that the limit as n approaches infinity of 1/n^2 + 1/(n+1)^2 +...+ 1/(2n)^2 = 0

    Is it sufficient to say that each term converges to 0 as n approaches infinity, and so the entire sum must also approach infinity?
  2. jcsd
  3. Oct 21, 2008 #2

    Ben Niehoff

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    No, because infinity*0 is an indeterminate form. You'll have to sum the series and take the limit properly (or perhaps find something more creative).
  4. Oct 21, 2008 #3
    I think you can compare all terms in the series to a P series 1/n^2, which diverges.

    Each successive term in the series, an < an+1 < an+2, etc... is declining and all are <= the first term.

    So that argument should be OK, without knowing anything else about the series.

  5. Oct 21, 2008 #4
    I understand that infinity*0 is indeterminate, but if each term converges to 0, wouldn't the sum look like 0 + 0 + ... + 0? Making the entire thing 0?
  6. Oct 22, 2008 #5
    While that appears to be true at first, it is not always true. Consider the function:

    f(n) = 1/n + 1/(n+1) + 1/(n+2) + ... + 1/(2n)

    The smallest term in this sum is 1/(2n). There are a total of n+1 terms. This means that the sum f(n) > (n+1)/(2n) = 1/2 + 1/(2n) > 1/2 for n >= 1. Thus, it doesn't matter what value of n we have, f(n) is always greater than 1/2.

    Yet, in this example, each of the terms tends to 0 as n tends to infinity. What matters is how fast the terms go to 0 compared to how many terms there are.
  7. Oct 22, 2008 #6


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    You meant "converges" here, didn't you?

  8. Oct 22, 2008 #7

    Gib Z

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    Ben's already implied this, but I though i'd weigh in on this: Try Squeezing it.
  9. Oct 22, 2008 #8

    Could you please expand on this?
  10. Oct 22, 2008 #9
    Alright. How about this.

    0 < 1/n^2 + 1/(n+1)^2 + ... + 1/(2n)^2 < 1/n^2 + ... + 1/n^2 [There are n+1 terms]

    1/n^2 + ... + 1/n^2 = (n+1)/n^2 = 1/n + 1/n + 1/n^2, which converges to 0. Thus the original sequence must also converge to 0.
  11. Oct 23, 2008 #10

    Gib Z

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    Yes, you nailed it.
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