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How to write the sum of series as Laurent series ?

$-\frac12\sum\limits_{n=0}^{\infty}(\frac z2)^n+-\frac4z\sum\limits_{n=0}^{\infty}(-z^{-2})^n$

I have somehow a blackout, how can I combine $2$ pieces together, I can change the term in the second one as $(-z^2)^n$ and $n$ from $-\infty$ to $0$ but I can't still write it as only one sum

the original exercise is from here

Laurent Series Expansion of $\frac{-3z^2+8z+1}{(z-2)(z^2+1)}$

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  • $\begingroup$ You can write it in the form "$\sum_{n=-\infty}^{+\infty} a_n z^n$, where [description of the coefficients]". But typically, $$\sum_{n=0}^\infty -\frac{1}{2^{n+1}}z^n + \sum_{n=0}^\infty (-1)^{n+1}\cdot 4 z^{-(2n+1)}$$ or the form you have is considered "Laurent enough". $\endgroup$ – Daniel Fischer May 20 '15 at 10:32
  • $\begingroup$ @Daniel Fischer Thanks I almost freaked out $\endgroup$ – derivative May 20 '15 at 10:35
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    $\begingroup$ Just sayin': For the series here, if you really want, you can write the coefficients as a closed formula. But such a closed formula would be much less readable. $\endgroup$ – Daniel Fischer May 20 '15 at 10:39

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