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Math Help - sum of the complex series

  1. #1
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    sum of the complex series

    Hello everybody, I am wondering how can I find the sum of the series \sum_{n=1}^{\infty}\frac{ni^n2^n}{(z+i)^{n+1}} ? I found out that the series converges for |z+i|>2, but I am not able to find the sum of the series , thank you very much
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    Re: sum of the complex series

    If it weren't for that "n" at the beginning, that would be a geometric series. But that n makes me think of a derivative. If we set f(z)= -(2i)^n (z+ i)^{-n} then f'(z)= n(2i)^n (z+ i)^{-n- 1}= \frac{n 2^n i^n}{(z+ i)^{n+1}}. Does that give you any ideas?
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    Re: sum of the complex series

    Thanks for the reply ..
    So you're saying that I could at first find the sum
      f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n and I get the function f(z)
    and just count the derivative of f'(z) ?
    this series looks not as difficult as the first one, but I still can't find the sum ..
    would you please give me another hint?
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    Re: sum of the complex series

    Quote Originally Posted by alteraus View Post
    Thanks for the reply ..
    So you're saying that I could at first find the sum
      f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n and I get the function f(z)
    and just count the derivative of f'(z) ?
    this series looks not as difficult as the first one, but I still can't find the sum ..
    would you please give me another hint?
    It's a geometric series...
    Thanks from topsquark
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    Re: sum of the complex series

    okey,
    can you please explain how does the geometric series work for complex numbers?
    I dealt only with real geometric series and the sum is then easy
    is there any sum formula for complex terms?
    in my series number z \in \mathbb{C}

    thank you
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    Re: sum of the complex series

    no more suggestions?
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    Re: sum of the complex series

    It works exactly the same way as a real geometric series.
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    Re: sum of the complex series

    thank you very much, I found the result which looks to be right
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    Re: sum of the complex series

    Hello, alteraus!

    I think I found the sum.
    But check my work . . . please!



    \sum_{n=1}^{\infty}\frac{n(2i)^n}{(z+i)^{n+1}}

    I found out that the series converges for |z+i|>2,
    but I am not able to find the sum of the series.

    \begin{array}{cccccc} \text{We are given:} & S &=& \dfrac{1\cdot(2i)}{(z+i)^2} + \dfrac{2\cdot(2i)^2}{(z+1)^3} + \dfrac{3\cdot (2i)^3}{(z+i)^4} + \cdots \\ \text{Multiply by }\frac{2i}{z+i}\!: & \dfrac{2i}{z+i}S &=& \qquad\qquad\;\; \dfrac{1\cdot (2i)^2}{(z+i)^3} + \dfrac{2\cdot (2i)^3}{(z+i)^4} + \cdots \end{array}

    \text{Subtract: }\:\left(1 - \frac{2i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2} + \frac{(2i)^2}{(z+i)^3} + \frac{(2i)^3}{(z+i)^4} + \cdots

    . . . . . . . . . . \left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\underbrace{\left[1 + \frac{2i}{z+i} + \frac{(2i)^2}{(z+i)^2} + \cdots \right]}_{\text{geometric series}}

    The geometric series has first term a = 1 and common ratio r = \frac{2i}{z+i}
    . . Its sum is: . \frac{1}{1-\frac{2i}{z+i}} \:=\:\frac{z+i}{z-i}

    We have: . \left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}

    . . . . . . . . . . . . . . S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}\cdot\frac{z+i}{z-i}

    . . . . . . . . . . . . . . S \;=\;\frac{2i}{(z-i)^2}
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    Re: sum of the complex series

    Thank you very much Soroban,
    nice way how to avoid the term-by-term differentiation,
    by which I got the same sum of the series, so it looks to be right
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