1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tricky limit problem

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

    The problem is as follows:

    lim(n→∞)(sum(sin(kθ)/2k,k=1,n)


    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

    Mark44

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    (cos(kθ)+isin(kθ))/2^k
    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tricky limit problem
  1. Tricky Limits (Replies: 11)

  2. A tricky limit (Replies: 2)

  3. Tricky limit (Replies: 10)

Loading...