<s>[INST] Power (Laurent) Series of $\coth(x)$ I need some help to prove that the power series of $\coth x$ is:
$$\frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + O(x^5) \ \ \ \ \ $$
I don't know how to derive this, should I divide the expansion of $\cosh(x)$ by the expansion of $\sinh(x)$? (I've tried but without good results :( )
Or I have to use residue calculus? 
Anyone can suggest me a link where I can find a detailed explanation of this expansion?
Thanks.
 [/INST] $$ \begin{eqnarray}
   \coth(x) &=& \frac{\cosh(x)}{\sinh(x)} = \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{x + \frac{x^3}{6} + \frac{x^5}{120} + \mathcal{o}\left(x^5\right)} = \frac{1}{x} \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{1 + \frac{x^2}{6} + \frac{x^4}{120} + \mathcal{o}\left(x^4\right)} \\ &=&  
  \frac{1}{x}  \left( 1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right) \right) \left( 1 - \frac{x^2}{6} + \frac{7 x^4}{360} + \mathcal{o}\left(x^4\right) \right) \\ &=& 
  \frac{1}{x} \left( 1 + \frac{x^2}{3} - \frac{x^4}{45} + \mathcal{o}\left(x^4\right) \right) = \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \mathcal{o}\left(x^3\right)
\end{eqnarray}
$$
where the reciprocation and multiplication of series used:
$$
 \frac{1}{1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right)} = 1  -a x^2 + \left( a^2-b \right) x^4 + \mathcal{o}\left(x^4\right)
$$
$$
  \left( 1 + a x^2 + b x^4 +  \mathcal{o}\left(x^4\right) \right) \left( 1 + c x^2 + d x^4 \mathcal{o}\left(x^4\right) \right) = 1 + \left(a+c\right) x^2 + \left(b + d + a c\right) x^4 + \mathcal{o}\left(x^4\right)
$$

The result for the reciprocation is obtained using the geometric series:
$$
    \frac{1}{1-w} = 1 + w + w^2 + \mathcal{o}(w^2)
$$
Now substitute in the above $w = a x^2 + b x^4 + \mathcal{o}(x^4)$, and use $$w^2 = \left( a x^2 + b x^4 + \mathcal{o}(x^4) \right)^2 = a^2 x^4 + \mathcal{o}(x^4)$$
</s>