Answer
$-1 \lt x \lt 1$
$f(x)=\displaystyle \frac{1}{1+x},\ \quad -1 \lt x \lt 1$
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
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
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$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}x^{n}=\sum_{n=0}^{\infty}(-x)^{n}$
Geometric series, $a=1, r=-x$, converges for
$|-x| \lt 1$
$|x| \lt 1$
$-1 \lt x \lt 1$
$f(x)=\displaystyle \sum_{n=0}^{\infty}(-x)^{n}=\frac{1}{1+x},\ -1 \lt x \lt 1$
