<s>[INST] 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 $1+x^2+...+x^{2n}$.
 [/INST] For every $x\in(-1,1), \frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$. So we replace $x$ with $x^2$, and since $x\in(-1,1)\rightarrow x^2\in(-1,1)$ we get $\frac{1}{1-x^2}=\sum_{i=0}^{\infty}x^{2i}$.
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