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Infinite series (i think it's riemann)

  1. Jul 4, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\sum_{1}^{inf}[/itex] k^2/(n^3+k^2)


    3. The attempt at a solution
    I think it's Riemann but i cannot find a suitable function to integrate.
     
  2. jcsd
  3. Jul 4, 2011 #2
    Isn't it suppose to be

    [tex]\sum_{n = 1}^{\infty} \frac{k^2}{n^3 + k^2}[/tex]
     
  4. Jul 4, 2011 #3

    Gib Z

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    Please take more care to expressing to others (and yourself) what it is you want to find. As written the sum doesn't make total sense. It could be what Susanne217 said above, or Riemann sum comment makes me think you could have also meant [itex]\displaystyle\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2}{n^3 + k^2} [/itex].
     
  5. Jul 4, 2011 #4
    I agree...

    I have been away for a while but I think it could be that the young man is suppose re-write the sum to an improper integral and thereby finding the sum of the series??
     
  6. Jul 4, 2011 #5

    Gib Z

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    In the case that it is what I thought, then it's not as simple as recognizing it as a pre-prepared Riemann sum. With some careful estimates to bound the sum, you should get the result to be [itex]1/3[/itex].
     
  7. Jul 4, 2011 #6
    Isn't it true that you need to rewrite this series into more easy to work with??
     
  8. Jul 4, 2011 #7
    First of all sorry for misspelling.

    I used the bounds k^2/(n^3+n^2) <= k^2/(n^3+k^2) <= k^2/(n^3+1) and i worked it out to 1/3.
     
  9. Jul 4, 2011 #8
    Then I have made a mistake cause it get to be 0.373551 and not a 1/3 :(
     
  10. Jul 4, 2011 #9
    I may have been wrong but wolframalpha says it's 1/3 so it must be correct :).
    Use those bounds and take the limits, you'll see they both converge to 1/3.
     
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