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To what value do these series converge?

  1. Oct 16, 2009 #1
    Consider these series:
    sin(n)/n; and
    cos(n)/n.
    To what value do they converge? You can use Diriclet's test to show that they do converge, but to what? I think I should use a Fourier trigonometric series, but not certain how.
     
  2. jcsd
  3. Oct 16, 2009 #2
    Without ado, the answer must be "zero", because sin(n) and cos(n) can only have values between -1 to +1, while n tends to infinity.
     
  4. Oct 16, 2009 #3
    I meant series, not sequence. With summation n = 1 to infinity. Sorry for any confusion.
     
  5. Oct 16, 2009 #4
    Note: I'm not sure if this actually works, but it's an idea.

    _____________________________________________________

    [tex]\sum[/tex] cos kx = -.5 + (.5sin[(2n +1)x/2)] )/sin x/2

    So, if you integrate the sum, from 0 to 1, you should get

    [tex]\sum[/tex] (sin k)/k , which is the sum you want, so integrate the other side from 0 to 1.
     
    Last edited: Oct 16, 2009
  6. Oct 17, 2009 #5
    Thanks for the idea. Im having a hard time to find a way to integrate that expression. Meantime I was able to calculate the series sin(n)/n by differentiating the fouries cosine series cos(nx)/n^2 and substituting 1. It equates to pi/2-1
    /2. Im still stuck trying to evaluate cos(n)/n however.
     
  7. Oct 17, 2009 #6
    First evaluate [itex] \sum z^n/n[/itex] then substitute in that to evaluate [itex] \sum e^{int}/n[/itex], then take real and imaginary parts to get [itex] \sum \cos(nt)/n[/itex] and [itex] \sum \sin(nt)/n[/itex]
     
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