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Convergence of complex series

  1. Jul 8, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum_{n=0}^\infty} (n+1)z^n[/tex]

    We have to find for which values of z is the series converging and also, we should find the sum of such a complex series.

    2. Relevant equations

    [tex]R = \frac{1}{\lim a_n^(1/n)}[/tex]

    [tex] r<\rho<R [/tex]

    [tex] q = \frac{r}{\rho} [/tex]

    3. The attempt at a solution

    First of all, I searched for a series expansion similar to our sum (I first thought it would be log (n+1) ) but it wasn't the case.

    Then, I looked up for my lecture notes on uniform convergence and continuity, and I found these equations stated above. I think this should definitely help since it can also be applied to complex series.

    The problem, though, is that I don't really know how to use such radii of convergence. I also wrote down the factor q, since it rather seems that to solve this sum I will most likely need the geometric series (sum q^n = 1/(1-q) ). So my problem really is how to deal with the (n+1) factor, otherwise, if it was just z^n it would just converge to 1/(1-z) for all |z| < 1 = r.

    I would appreciate any help.
     
    Last edited: Jul 8, 2010
  2. jcsd
  3. Jul 8, 2010 #2

    LCKurtz

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    Hint: Does that look like the derivative of anything?
     
  4. Jul 9, 2010 #3
    Yeah, I guess that just looks as if we took the derivative of z^(n+1), but I don't see how is this going to help us anyway *shrug*.

    Maybe I'm not counting some fundamental definition of integrals, series and its relations?
     
  5. Jul 9, 2010 #4
    Oh got it!
    You were right about looking at the derivative. It was just that the sum of z^n (from 0 to inf) equals to a geometric series, then, if we take the derivative of this geometric series, we get exactly our sum. I.e. taking the derivative of the geometric series sum (1/(1-z)) will be our result, which ended up to be 1/(z-1)^2

    Thanks!
     
  6. Jul 9, 2010 #5

    LCKurtz

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    Be careful about the first term and your sign when you differentiate.
     
  7. Jul 11, 2010 #6
    Oh, sure. In that case though, (z-1)^2 = (1-z)^2. But for odd powers we should put a minus sign. So actually, if you think of it, we can come up with a nice formula, such as:

    [tex]\sum_{n=0}^\infty} (n+1)(n+2)...(n+\alpha)z^n = \frac {\alpha !}{(1-z)^{\alpha+1}}[/tex]
     
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