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Convergence or not

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}[/tex]


    I dont have a solution, and wondered if the execution is correct.

    3. The attempt at a solution
    I thought that one can use comparison test where; [tex]\sum b_n[/tex]= [tex]\frac{1}{n^{1/2}}[/tex].
    Since p<1 ---> divergent. But many of the students says it converges. Some suggestions?

    We know that the series is alternating, and if I use its test I get that it converges. Reckon that perhaps the fault lies there?
     
  2. jcsd
  3. Dec 4, 2016 #2

    hilbert2

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    Let's define a function that's related to that series:

    ##f(x) = \frac{1}{2}\sum_{n=2}^{\infty}sin(\frac{n\pi x}{2})## .

    Now obviously the sum of the original series, if it exists, is ##f(1)##. Is the expression of ##f(x)## the Fourier series of some function that you know?

    EDIT1: This might help: http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/06/01/

    Is the Dirac delta an acceptable function in the sense of rigorous mathematics?

    EDIT2: Also, if a sum of terms ##a_k## is convergent, what can we tell about the limit of the sequence ##(a_k)## when ##k \rightarrow \infty## ?
     
    Last edited: Dec 4, 2016
  4. Dec 4, 2016 #3

    Ray Vickson

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    Well, the actual terms for ##n = 2,3,4, \ldots## are 0, -1/2, 0, 1/2, 0, -1/2, 0, 1/2, 0, ... . Do you think those terms give a convergent series?
     
  5. Dec 4, 2016 #4

    hilbert2

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  6. Dec 4, 2016 #5

    Ray Vickson

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    I would rather that the OP have a good grasp of ordinary convergence/divergence before exploring more arcane topics like cesaro or Abel or ... summability.
     
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