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: 
Here's a somewhat different approach, which can be extended by using the power series of $\cosh$ and $\sinh$ as others have.

$$\coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{1}{x} \frac{\cosh(x)}{\frac{\sinh(x)}{x}} \\
= \frac{1}{x} \frac{\cosh(x)}{1+\left ( \frac{\sinh(x)}{x} - 1 \right )} \\
= \frac{1}{x} \cosh(x) \sum\_{n=0}^\infty \left ( 1-\frac{\sinh(x)}{x} \right )^n$$

provided $\left | 1 - \frac{\sinh(x)}{x} \right | < 1$ (and of course provided $x \neq 0$, otherwise neither side is defined). The last line applies the geometric series. You can turn this into a power series by substituting the power series of $\cosh$ and $\sinh$ and then pulling some tricks (a Cauchy convolution product, among others).

A similar approach to this one would be to write

$$\coth(x) = \frac{1}{x} \left ( x \coth(x) \right )$$

and then work out the Maclaurin series of $x \coth(x)$ by directly calculating derivatives. Of course this, too, quickly becomes a complicated mess as the number of terms grows.
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.