Question: I'm trying to approximate $\coth(x)$ around $x = 0$, up to say, third order in $x$. Now obviously a simple taylor expansion doesn't work, as it diverges around $x = 0$. I'm not quite sure how to proceed from here. I had a look at 
[Series expansion of $\coth x$ using the Fourier transform](https://math.stackexchange.com/questions/845506/series-expansion-of-coth-x-using-the-fourier-transform)

but this is a bit technical for my purposes, and I also don't need the entire series. Is there an easy way of getting some terms, or do I really have to solve for the entire series? And is there a simpler way to do so than with the Fourier transform?

Answer: 
There is a simple way of approximating $\coth$ by noticing that it is a logarithmic derivative. Since:
$$ \frac{\sinh z}{z}=\prod\_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1} $$
by the Weierstrass product for the (hyperbolic) sine function, we have:
$$ \log\sinh z-\log z = \sum\_{n=1}^{+\infty}\log\left(1+\frac{z^2}{\pi^2 n^2}\right),\tag{2}$$
so, by differentiating both sides:
$$ \coth z-\frac{1}{z} = \sum\_{n=1}^{+\infty}\frac{2z}{\pi^2 n^2+z^2}.\tag{3} $$
By expanding the terms in the RHS of $(3)$ as geometric series we can also derive the full Laurent expansion of $\coth$ in a puntured neighbourhood of the origin, in terms of $\zeta(2m)$. This is also the standard technique for showing that $\zeta(2m)$ just depends on the Bernoulli number $B\_{2m}$ (formula $(41)$ [here](http://mathworld.wolfram.com/BernoulliNumber.html)).
My copy of the classic "Handbook of Mathematical Functions" by Abramowitz and Stegun says

$$ \mathrm{coth}(z) = \frac{1}{z} + \frac{z}{3} - \frac{z^3}{45} + \frac{2}{945} z^5 - \ldots + \frac{2^{2n}B\_{2n}}{(2n)!} z^{2n-1} $$

where $B\_n$ is the $n^{th}$ Bernoulli number.

Note in particular the difference in the $z^3$ term compared with the answer above.