Given an alternating series, $ \sum (-1)^n a_n$, where $a_n$ is a real valued sequence, we know that the sum converges if $\underset{n\to\infty}{\lim} a_n = 0$ and $ \left| a_n \right| \geq \left| a_{n+1} \right| $ via the alternating series test.
_Example:_
$$ f(x) = \sum \frac{(-1)^n}{x^n} $$ so $a_n = \frac{1}{x^n}$, which means that $f(x)$ converges if $\underset{n\to\infty}{\lim} \frac{1}{x^n} = 0$. This is true for $|x| > 1$. The next test requires that $\left|\frac{1}{x^{n+1}} \right| \leq \left| \frac{1}{x^n} \right|$, which holds for $ |x| \geq 1$.
So the interval of convergence is $ x \in (-\infty, -1)\cup (1, \infty) $