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How do I come to a series expansion of $\frac{1}{(e^z-1)^2}?$

$e^z-1$ can be expanded to: $$1 + z + \frac{1}{2} z^2 + \frac{1}{6} z^3 + \dots -1$$

so the series becomes: $$\frac{1}{(z^2 (1 + \frac{1}{2} z + \frac{1}{6} z^2 +\dots)^2}$$

I don't now how to make the step to: $$\frac{1}{z^2}(1 - z + \frac{5}{12} z^2 + \dots)$$

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  • $\begingroup$ I've got it. Use the binomial expansion: $(1+x)^n=1+nx+n(n-1)/2! x + ...$ With n=-2. Thanks for the effort tho! $\endgroup$ – Tijn Nov 4 '13 at 19:55
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Use long division or alternatively restart by considering instead $1/(1-z)^2=1+2z+3z^2+4z^3+\dots$ and substituting in from there.

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  • $\begingroup$ Okay, well now, how do I come from $1/(1-z)^2$ to $1+2z+3z^2+4z^3..$?... That would solve the whole problem for me. It is the method that is troubling me. $\endgroup$ – Tijn Nov 4 '13 at 19:05
  • $\begingroup$ Well surely you're familiar with $1/(1-z)=1+z+z^2+z^3+\dots$. Differentiate both sides, doing the series term by term. $\endgroup$ – oldrinb Nov 4 '13 at 22:19

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