How would you calculate Laurent series of
$$f(z) = \frac{\cosh z - 1}{\sinh z - z} \ ?$$
I am struggling to find its residue actually and I believe there is a simple pole at $z = 0$ (as this is what I get by WolframAlpha), but I can't seem to figure out how to find the Laurent series. We could expand both the nominator and the denominator and that would give sth of the form
$$f(z) = \frac{\frac{z^2}{2} + O(z^2)}{\frac{z^3}{3!} + O(z^3)} = \frac{\frac{1}{2} + O(1)}{\frac{z}{6} + O(z)} = \frac{3}{z} + O(z)$$ which I presume would indicate the residue is 3, but is there a way to evaluate the whole Laurent series explicitly?
To answer the question you asked in your comments : So how does Wolfram calculates its first few terms?
I suppose that "they" wrote something like $$z (\cosh(z)-1)=(\sinh(z)-z) \sum_{m=0}^\infty a_m z^m$$ and, using Taylor expansions of $\cosh(z)$ and $\sinh(z)$, arrived to $$\sum_{n=1}^\infty \frac{z^{2n+1}}{(2n)!}=\sum_{n=1}^\infty \frac{z^{2n+1}}{(2n+1)!}\times\sum_{m=0}^\infty a_m z^m$$ and then identification of the $a_m$'s (which is not simple because of the product of the two summations in the rhs).
The beginning of the expansion is $$0=\left(\frac{1}{2}-\frac{a_0}{6}\right) x^3-\frac{1}{6} a_1 x^4+\frac{1}{120} (-a_0-20 a_2+5) x^5+\left(-\frac{a_1}{120}-\frac{a_3}{6}\right) x^6+\frac{(-a_0-42 a_2-840 a_4+7)}{5040} x^7+\frac{(-a_1-42 a_3-840 a_5) }{5040}x^8+\cdots$$
I must confess that I have not been able to generate the general form of the coefficients. But going it that way, I "quicky" arrived to $$z\,\frac{\cosh (z)-1}{\sinh (z)-z}=3+\frac{z^2}{10}-\frac{z^4}{4200}-\frac{z^6}{126000}+\frac{89 z^8}{388080000}-\frac{1579 z^{10}}{454053600000}+\frac{5069 z^{12}}{190702512000000}+\frac{1013 z^{14}}{4631346720000000}+O\left(z^{16}\right)$$