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: 
The series representation **must** reflect that divergence at $x=0$. Thus,

$$\begin{align}\coth{x} &= \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)} \\ &= \frac1{x} \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{1+\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6)}\\&= \frac1{x} \left (1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \right ) \left [1 - \left (\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6) \right )+\left (\frac{x^2}{3!}+O(x^4) \right )^2+O(x^6) \right ] \\ &= \frac1{x} \left [1+\left (\frac1{2!}-\frac1{3!} \right ) x^2 +\left (\frac1{(3!)^2} + \frac1{4!}-\frac1{5!} - \frac1{2! 3!}\right ) x^4+O(x^6) \right ]\\ &= \frac1{x}+\frac{x}{3} - \frac{x^3}{45}+O(x^5) \end{align}$$

Note that the divergence at $x=0$ is reflected in the $1/x$ term.
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)).