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How would you develop the function $$f(z) = \frac{1}{z(z-1)(z-2)}$$ in the annulus $A = \{z: 1<|z|<2\}$?

I have split the function into partial fractions to obtain the result $$ -\sum_0^\infty (1/2)^{k+1}z^{k-1} - \sum_0^\infty z^{-k-2}$$

but is it alright that this series has two $\sum$'s in the equation? Is this still a valid Laurent series expansion?

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There's nothing wrong with using two summation symbols in any equation, but you want to make a few things clear:

First, keep your index of summation consistent: re-use it if it is a single series, don't if it is not.

Second, use parentheses if there is any doubt. Do you mean $\sum_k a_k + \sum_j b_j$, or $\sum_k \left( a_k + \sum_j b_j\right)$? Keeping your index of summation consistent helps make this clear.

Third, Laurent series are usually represented as a sum of non-negatively indexed terms that converges, along with a sum of negatively indexed terms which also converges. It's more conventional to write $\sum_{k=-\infty}^{-1} a_k z^k + \sum_{k=0}^\infty a_k z^k = \sum_{-k=\infty}^\infty a_kz^k$.

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  • $\begingroup$ Note: I haven't checked to see if your result is correct. $\endgroup$ – Emily May 6 '13 at 20:21
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It is absolutely ok, recall that a Laurent series is a double series (negative AND positive powers), if you want you can rearrange the terms of two series in order to make it more visible, as an example you can write

$$-\displaystyle\sum_{-\infty}^{-2}z^k-\displaystyle\sum_{-1}^{+\infty}\frac{z^k}{2^{k+2}}$$

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