How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$?

Question

I'm given a generating function $f(x)$: $$ f(x)=\frac{1}{1-x^2} $$ Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.

So far I figured out what follows: $$ f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac{1}{2}}{1-x} + \frac{\frac{1}{2}}{1+x}=\frac{1}{2}\cdot\left( \sum_{n=0}^\infty x^n+\frac{1}{1-(-x)} \right) $$$$ \frac{1}{1-(-x)}=1+(-x)+x^2+(-x^3)+x^4+\cdots $$$$ \frac{1}{1-(-x)}=(1+x^2+x^4+\cdots)-(x+x^3+x^5+\cdots)=\sum_{n=0}^\infty x^{2n}-x^{2n+1} $$ Therefore: $$ f(x)=\frac{1}{2} \left( \sum_{n=0}^\infty x^n+x^{2n}-x^{2n+1} \right) $$ Now I am lost. Is my train of thought good? Am I anywhere near finding the series which is generated by function $f(x)$?

Answer 1

To repair the approach you were taking: $$f(x) = \frac{1}{1-x^2} = \frac{1/2}{1-x} + \frac{1/2}{1+x} = \frac{1}{2} \sum_{n=0}^\infty x^n + \frac{1}{2} \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{1+(-1)^n}{2} x^n, $$ so $$ a_n = \frac{1+(-1)^n}{2} = \begin{cases} 1 &\text{if $2\mid n$} \\0 &\text{otherwise}\\ \end{cases} $$

Answer 2

$\frac 1 {1-x^2}=1+x^2+x^4+x^6+\cdots$ for $|x| <1$ since $\frac 1 {1-r}=1+r+r^2+\cdots$ for $|r|<1$. So $a_n=1$ for $n$ even and $a_n=0$ for $n$ odd.