Answer
$\left( { - 4,4} \right)$
Work Step by Step
$$\eqalign{
& \sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{4}} \right)}^n}} \cr
& {\text{Rewrite}} \cr
& \sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{4}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\underbrace {\left( 1 \right)}_a\underbrace {{{\left( {\frac{x}{4}} \right)}^n}}_{{r^n}}} \cr
& {\text{This is a geometric series with }}a = 1{\text{ and }}r = \frac{x}{4} \cr
& {\text{The geometric series converges when }}\left| r \right| < 1,{\text{ then}} \cr
& \left| {\frac{x}{4}} \right| < 1 \cr
& {\text{Solving the inequality}} \cr
& - 1 < \frac{x}{4} < 1 \cr
& - 4 < x < 4 \cr
& \cr
& {\text{At the endpoints}}{\text{, we have}} \cr
& x = - 4 \to \sum\limits_{n = 0}^\infty {{{\left( {\frac{{ - 4}}{4}} \right)}^n} = } \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n},{\text{Alternating series }}\left( {{\text{Diverges}}} \right)} \cr
& x = 4 \to \sum\limits_{n = 0}^\infty {{{\left( {\frac{4}{4}} \right)}^n} = } \sum\limits_{n = 0}^\infty {1,{\text{ }}\left( {{\text{Diverges}}} \right)} \cr
& {\text{Diverges on both endpoints}}{\text{, so the interval of convergence is}} \cr
& \left( { - 4,4} \right) \cr} $$
