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Show convergence using comparison test on sin(1/n)

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data

    a) Test the following series for convergence using the comparison test

    Explain your conclusion.

    2. Relevant equations

    3. The attempt at a solution
    i must show f(x)<g(x) in order for it to converge other wise divergence.

    g(x) = 1/n

    sin(1/n) > 1/n always therefore the series is divergent.
    is this correct?
  2. jcsd
  3. Jan 9, 2012 #2
    How did you obtain this inequality?? It isn't correct.
  4. Jan 9, 2012 #3
    it was an attempt. i was looking back on my notes and thought that might be right :/ do you have any idea how to complete this?
  5. Jan 9, 2012 #4
    The inequality sin(1/n) > 1/n might not hold, but it can be modified in something that does hold.
  6. Jan 9, 2012 #5
    does this prove that it is divergent? or how do i go about proving if its convergent/divergent using the comparison test?
  7. Jan 9, 2012 #6
    For large n, sin(1/n) ≈ 1/n. So use that to find your inequality.
  8. Jan 9, 2012 #7
    does that give sin(1/n) > 1/2n?
  9. Jan 9, 2012 #8
    It does for me.
  10. Jan 9, 2012 #9
    so using that its divergent? thanks :)
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