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Math Help - Substitution into an infinite sum (Laurent series) question

  1. #1
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    Substitution into an infinite sum (Laurent series) question

    By substituting \frac{1}{1-z} for z in the expansion

    \frac{1}{(1-z)^2}=\sum_{n=0}^\infty(n+1)z^n

    derive the Laurent Series representation

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

    I've tried substituting and rearranging using algebra, but I can't get anything even close to that answer.. I just end up with really complicated polynomials. Is there a "trick" to this that I'm missing??
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  2. #2
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    May be this will be useful
    \frac{1}{z^2}=\frac{1}{({z+1-1})^2}=\frac{1}{(1-z)^2(1-\frac{1}{1-z})^2}
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