Taylor expansion around infinity of a fraction

I want to Taylor expand the following function:

$f(x)=\frac{1}{a-x^2}$ when $x \rightarrow \infty$.

I know the result (from Wolfram Alpha) to be $-\frac{1}{x^2} - \frac{a}{x^4} + O(\frac{1}{x^6})$

but I have no idea how to calculate it. Any help is greatly appreciated!

-
Use the binomial theorem: en.wikipedia.org/wiki/Binomial_series with $\alpha=-1$ after collecting $-x^{-2}$, it should work. – Giuseppe Negro Jun 1 '12 at 18:37
The first term should read $\dfrac{-1}{x^2}$. – user17762 Jun 1 '12 at 18:37
@Marvis, I've fixed the question. Thanks for the great answer! – Ben Ruijl Jun 1 '12 at 20:24

One way is as follows. Set $x= \dfrac1y$ and expand it around $y=0$. $$g(y) = \dfrac{1}{a - \dfrac1{y^2}} = \dfrac{y^2}{ay^2-1} = - y^2 \dfrac1{1-ay^2} = -y^2 \left( 1 + ay^2 +a^2y^4 + a^3 y^6 + \cdots\right)$$ Now replace $y$ by $1/x$ to get that $$f(x) = -\dfrac1{x^2} \left( 1 + \dfrac{a}{x^2} + \dfrac{a^2}{x^4} + \dfrac{a^3}{x^6} + \cdots\right)$$

-