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Tricky Complex Series

  1. Feb 14, 2012 #1

    I am trying to determine the convergence/divergence of

    Ʃn=1 in/n.

    I have tried all the tests I could think of (Comparison, Ratio, Root, nth term) and cannot determine it's convergence.

    If there was a formula, for say, the Mth partial sum SM then, if the limit as M → ∞ of SM is L, we have convergence to L but I can't seem to arrange for the thing to add up the first M terms.

    Clearly, I think something can be done with the fact that in = {i, -1, -i, 1} repeatably with periodicity 4. I'm not sure how this can exactly be of help though!

    Any help appreciated, thanks.
  2. jcsd
  3. Feb 14, 2012 #2
    Do you know the power series for [itex]\log(1-z)[/itex]? It has a raidus of convergence equal to one. However convergence on the boundary [itex]|z|=1[/itex] is well, what? Take a look at Wikipedia for "Radius of Convergence".
  4. Feb 14, 2012 #3
    Ah! I see, and we have |i| = 1 but i ≠ 1 so there we go!

    Thanks a bunch. Factoring out the -1 seems to make it all go quite nicely.
  5. Feb 14, 2012 #4
    Ʃn=1 in/n = -log(1-i), woo hoo!
  6. Feb 14, 2012 #5
    You know, that's really not good enough. Why does it converge on the boundary except for z=1? Or rather prove that it does. If you wish anyway.
  7. Feb 14, 2012 #6
    Ah, Thanks for the heads up. I suppose I sort of took that for granted. Would http://en.wikipedia.org/wiki/Abel's_test#Abel.27s_test_in_complex_analysis works to show convergence for all z ≠ 1 on the boundary (in particular, at our point i ≠ 1), as the an's are monotonically decreasing? Some the an, however, are negative...

    Trying to take some zo ≠ 1 with |zo| = 1 and using some other test (root, comparison, etc.) doesn't seem to work either, though...
  8. Feb 15, 2012 #7
    Yeah, I don't know off-hand how to show that. There are tests to check for convergence on the boundary. Would need to look into thoses. Don't thinik Abel test would apply though.
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