Answer
$$\sum_{n=0}^{\infty}(-1) ^nx^ {2n}$$
$ |x|\lt1$
Work Step by Step
Given $$ f(x)=\frac{1}{1+ x^2}$$
Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$
By using (1), we get
\begin{align*} \frac{1}{1+x^2}&=\frac{1}{1-(- x^2)}\\
&= \sum_{n=0}^{\infty}(- x^2)^{n}\\
&= \sum_{n=0}^{\infty}(-1) ^nx^ {2n}
\end{align*} Such that $ |x|\lt1$
