Finding the Taylor series for $\frac{1}{1-x^2}$ Where I'm stuck
$f(x)=\frac{1}{1-x^2}=\frac{1}{1-x}\cdot\!\frac{1}{1+x}=(1-x)^{-1} \cdot(1+x)^{-1}$
$(1+x)^{-1}= 1-x+x^2-x^3+...+(-1)^{2n}x^{2n}+o(x^{2n})$
$(1-x)^{-1}= 1+x+x^2+x^3+...+x^{2n}+o(x^{2n})$
At this point I have no idea how to multiply all this together.Could you please help?
The answer is $1+x^2+...+x^{2n}$.

In the interest of proving $\sum_{n\ge0}x^{2n}$ is the Taylor series of $\frac{1}{1-x^2}$ expanded at $x=0$, which isn't quite the same point as the limit being correct, use $\frac{1}{1-x^2}=\frac12\left(\frac{1}{1-x}+\frac{1}{1+x}\right)$ to average their Taylor series, giving$$\sum_{n\ge0}\frac12(1+(-1)^n)x^n=\sum_{n\ge0}x^{2n}$$(by separately considering powers of each parity).
