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Homework Help: Tricky limit problem

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem is as follows:


    2. Relevant equations

    3. The attempt at a solution
    I feel like this should converge since sin oscillates between -1 and 1 and 2k keeps getting larger and larger.

    Do I have to use power series somehow?

    Any ideas, anyone?
  2. jcsd
  3. Oct 19, 2011 #2


    Staff: Mentor

    Are you trying to determine whether this series converges?
    [tex]\sum_{k = 1}^\infty \frac{sin(k\theta)}{2^k}[/tex]

    What tests do you know?
  4. Oct 19, 2011 #3
    Thanks for your response.

    The problem can be rewritten as sum (ak*(1/2)^k) where ak=sin(kθ)
    Then I can use limsup ( n√abs(ak) = 1
    Therefore my radius of convergence is 1. But now how do I use that to find the limit?
  5. Oct 19, 2011 #4


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    Your series is related to the geometric series a^k where a=e^(i*theta)/2. How is it related?
  6. Oct 19, 2011 #5

    I see what you mean. If I do ((e^i*theta)/2)^k I would get

    I just need to get the numerator equal to sin(kθ)
    How do I do that?? I need to get rid of the cos (kθ) and the i in front of sin

    Could I do(e^i*theta - e^-i*theta / 2i)^k = sin(k*theta) ?
  7. Oct 19, 2011 #6


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    No, you don't want to do that. Just take the imaginary part of the complex series.
  8. Oct 19, 2011 #7
    Aha. And then I have

    lim n →∞ of Im ( sum (e^iθ/2)^k , k = 1 to n)
    So this is a geometric series and would be equal to
    lim n →∞ of Im ( e^iθ/2 * (1-((e^iθ)/2)^n)/(1-(e^iθ)/2)
  9. Oct 19, 2011 #8
    I got sin(theta)/(2-sin(theta)) as my answer. Sound right?
  10. Oct 19, 2011 #9


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    Nope. I don't think you were careful enough separating real and imaginary. Try and think of a way to check it. Suppose you put theta=pi/2 in your series. Can you sum it? Then put pi/2 in your formula.
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