$\displaystyle f(x)=\frac{4}{4+x^2}$, c=0

$\displaystyle f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}$

$\displaystyle \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n$

If I did this right, where do I go from here?? Thank you.