Answer
$-1 \lt x \lt 1$
$f(x)=\displaystyle \frac{1}{1+x^{2}}, \quad -1 \lt x \lt 1$
Work Step by Step
We aim for terms of Th.9.6 (Convergence of a Geometric Series).
A geometric series with ratio $r$ diverges when $|r| \geq 1$.
If $0 \lt |r| \lt 1$, then the series converges to the sum
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
------------
$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}x^{2n}=\sum_{n=0}^{\infty}(-x^{2})^{n}$
Geometric series, $a=1, r=-x^{2}$, converges for
$|-x^{2}| \lt 1$
$x^{2} \lt 1$
$-1 \lt x \lt 1$
$f(x)=\displaystyle \sum_{n=0}^{\infty}(-x^{2})^{n}=\frac{1}{1-(-x^{2})}$
$=\displaystyle \frac{1}{1+x^{2}}, -1 \lt x \lt 1$