Answer
$\left( { - \frac{1}{2},\frac{1}{2}} \right)$
Work Step by Step
$$\eqalign{
& \sum\limits_{n = 0}^\infty {{{\left( {2x} \right)}^n}} \cr
& {\text{Rewrite}} \cr
& \sum\limits_{n = 0}^\infty {{{\left( {2x} \right)}^n}} = \sum\limits_{n = 0}^\infty {\underbrace {\left( 1 \right)}_a\underbrace {{{\left( {2x} \right)}^n}}_{{r^n}}} \cr
& {\text{This is a geometric series with }}a = 1{\text{ and }}r = 2x \cr
& {\text{The geometric series converges when }}\left| r \right| < 1,{\text{ then}} \cr
& \left| {2x} \right| < 1 \cr
& {\text{Solving the inequality}} \cr
& - 1 < 2x < 1 \cr
& - \frac{1}{2} < x < \frac{1}{2} \cr
& \cr
& {\text{At the endpoints}}{\text{, we have}} \cr
& \sum\limits_{n = 0}^\infty {{{\left( {2\left( { - \frac{1}{2}} \right)} \right)}^n}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n},{\text{Alternating series }}\left( {{\text{Diverges}}} \right)} \cr
& \sum\limits_{n = 0}^\infty {{{\left( {2\left( {\frac{1}{2}} \right)} \right)}^n}} = \sum\limits_{n = 0}^\infty {1,{\text{ }}\left( {{\text{Diverges}}} \right)} \cr
& {\text{Diverges on both endpoints}}{\text{, so the interval of convergence is}} \cr
& \left( { - \frac{1}{2},\frac{1}{2}} \right) \cr} $$
