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.

$$ \begin{eqnarray} \coth(x) &=& \frac{\cosh(x)}{\sinh(x)} = \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{x + \frac{x^3}{6} + \frac{x^5}{120} + \mathcal{o}\left(x^5\right)} = \frac{1}{x} \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{1 + \frac{x^2}{6} + \frac{x^4}{120} + \mathcal{o}\left(x^4\right)} \\ &=& \frac{1}{x} \left( 1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right) \right) \left( 1 - \frac{x^2}{6} + \frac{7 x^4}{360} + \mathcal{o}\left(x^4\right) \right) \\ &=& \frac{1}{x} \left( 1 + \frac{x^2}{3} - \frac{x^4}{45} + \mathcal{o}\left(x^4\right) \right) = \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \mathcal{o}\left(x^3\right) \end{eqnarray} $$ where the reciprocation and multiplication of series used: $$ \frac{1}{1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right)} = 1 -a x^2 + \left( a^2-b \right) x^4 + \mathcal{o}\left(x^4\right) $$ $$ \left( 1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right) \right) \left( 1 + c x^2 + d x^4 \mathcal{o}\left(x^4\right) \right) = 1 + \left(a+c\right) x^2 + \left(b + d + a c\right) x^4 + \mathcal{o}\left(x^4\right) $$

Long division.

The problem:
Now $z$ into $1$ is $z^{-1}$
Multiply
Subtract
$z$ into $(1/3)z^2$ is $(1/3) z$
Multiply
Subtract
$z$ into $(-1/45)z^4$
If we want more terms in the quotent, we will have to fill in more terms in all of them where the dots are now.