Answer
$$ - \frac{4}{5}$$
Work Step by Step
$$\eqalign{
& \sum\limits_{i = 1}^\infty {2{{\left( { - \frac{2}{3}} \right)}^i}} \cr
& = \sum\limits_{i = 1}^\infty {2\left( { - \frac{2}{3}} \right){{\left( { - \frac{2}{3}} \right)}^{i - 1}}} \cr
& = \sum\limits_{i = 1}^\infty {\left( { - \frac{4}{3}} \right){{\left( { - \frac{2}{3}} \right)}^{i - 1}}} \cr
& {\text{The }}n{\text{th term is }}{a_n} = {a_1}{r^{n - 1}},\, \cr
& {a_n} = \left( { - \frac{4}{3}} \right){\left( { - \frac{2}{3}} \right)^{i - 1}}{\text{ then, }}{a_1} = - \frac{4}{3}{\text{ and }}r = - \frac{2}{3} \cr
& {\text{The sum of the Terms of an Infinite Geometric Sequence is}} \cr
& {S_\infty } = \frac{{{a_1}}}{{1 - r}},{\text{ }}\left( {{\text{where }}\left| r \right| < 1} \right) \cr
& {\text{Then}} \cr
& {S_\infty } = \frac{{ - 4/3}}{{1 - \left( { - 2/3} \right)}} \cr
& {S_\infty } = - \frac{4}{5} \cr} $$
