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: 
$$\coth x=\frac{e^x+e^{-x}}{e^x-e^{-x}}\\\sim\frac{1+x^2/2!+x^4/4!+...}{x+x^3/3!+x^5/5!+...}\\\sim\frac1x(1+x^2/2!+x^4/4!+...)(1-(x^2/3!+x^4/5!+..))\\=\frac1x+\frac x3+...$$
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.