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)).
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.