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Find the Laurent expansion of the following around $z_0 = 0$

\begin{align*} f(z) &= \frac{1}{z(e^z - 1)} \\ \end{align*}

The answer is

\begin{align*} f(z) &= \frac{1}{z^2} - \frac{1}{2z} + \frac{1}{12} - \frac{z^2}{720} + \frac{z^4}{30240} + \cdots \end{align*}

But I don't see how to find that by hand. I tried:

\begin{align*} f(z) &= -\frac{1}{z} \cdot \frac{1}{(1 - e^z)} = -\frac{1}{z} \cdot \sum\limits_{n=0}^\infty e^{zn}\\ \end{align*}

or

\begin{align*} f(z) &= \frac{1}{z} \cdot \frac{1}{e^z - 1} = \frac{1}{z} \cdot \frac{1}{\left( \sum\limits_{n=0}^\infty \frac{z^n}{n!} \right) - 1} \\ &= \frac{1}{\sum\limits_{n=1}^\infty \frac{z^{n+1}}{n!}} \\ \end{align*}

Neither of the two approaches I tried seem to get me to the solution. Any ideas?

clay
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    Getting the whole Laurent expansion might be involved but for the first terms, use indeed the identity $$z^2f(z)=\left(\sum_{n=0}^\infty\frac{z^n}{(n+1)!}\right)^{-1}$$ and expand $$\left(1+\frac12z+\frac16z^2+\frac1{24}z^3+\frac1{120}z^4\right)^{-1}$$ neglecting every term $O(z^5)$. After you are done with the computations, you might wish to look for Bell numbers on WP and for duplicates of your question on the site. – Did Apr 26 '17 at 07:07
  • *Correction: one needs the expansion $1+\frac12z+\ldots$ up to order $z^6$ included, not only to order $z^4$. The rest of my comment stands. – Did Apr 26 '17 at 07:29
  • @Did I think you mean Bernoulli numbers: http://129.81.170.14/~vhm/bernoulli.pdf – Olivier Oloa Apr 26 '17 at 07:39
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    @OlivierOloa Definitely Bernoulli numbers, yes, thanks for catching the typo. – Did Apr 26 '17 at 08:03

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