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Homework Help: Coverging or diverging series

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data
    the sum of (n sin n)/(n^3+1)
    n from 1 to infinity

    2. Relevant equations



    3. The attempt at a solution
    what I have done so far is using the comparison method by comparing this with n sin n/ n^3, but I don't know what's the next step and even if my comparison is correct or not
     
  2. jcsd
  3. Mar 5, 2007 #2

    JasonRox

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    Have you heard of L'Hopital's Rule?

    The comparing method your using is also another way.

    So, what happens to [n*sin(n)]/n^3 when n goes to infinity?
     
  4. Mar 5, 2007 #3
    that's where I was stuck. I think it doesn't go anywhere therefore it doesn't exist. Am I right??
     
  5. Mar 5, 2007 #4

    JasonRox

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    What's your argument for it not going anywhere?
     
  6. Mar 5, 2007 #5
    well because sin n varies between -1 and 1. This what makes me think that the limit does not converges to some value
     
  7. Mar 5, 2007 #6

    JasonRox

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    Yes, but where does the denominator go?
     
  8. Mar 5, 2007 #7
    it goes to infinity, so therefore it goes to 0?? Am I right?
     
  9. Mar 5, 2007 #8

    JasonRox

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    Not so quick.

    So, the denominator goes to infinity, and the numerator bounces around -1 and 1.

    Can you replace sin(n) by something else? Like say, -1 or 1?
     
  10. Mar 5, 2007 #9
    yeah If I replace it with either -1 or 1 both will most likely to reach 0 from both sides, from the negative sides and the positive sides.
     
  11. Mar 5, 2007 #10

    JasonRox

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    So, why not just pick one?

    So, we have 1/n^2, and [n*sin(n)]/n^3, and the original one.

    Write down the comparisons, and you should be good to go.
     
  12. Mar 5, 2007 #11
    well we can simplify [n * sin n]/n^3 into sin n/n^2 right?? therefore the answer is 0?
     
  13. Mar 5, 2007 #12

    JasonRox

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    Yeah, the answer is 0, but you have to deduce it now. Or I think you do. Sometimes they just ask for the solution.
     
  14. Mar 5, 2007 #13
    what do you mean by deduce?
     
  15. Mar 5, 2007 #14

    JasonRox

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    Like, actually show the limit is 0.
     
  16. Mar 6, 2007 #15

    HallsofIvy

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    Excuse me, but the original question asked for the SUM of [itex]\frac{nsin(n)}{n^3+1}[/itex], not the limit of the sequence.
     
  17. Mar 6, 2007 #16

    JasonRox

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    Whoa! My mistake.
     
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