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

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
$$
a_n = \frac{1+(-1)^n}{2} = \begin{cases} 1 &\text{if $2\mid n$} \\0 &\text{otherwise}\\ \end{cases}
$$
