Answer
$${\text{The terms of the sequence will not have a sum}}$$
Work Step by Step
$$\eqalign{
& \sum\limits_{i = 1}^\infty { - 2{{\left( {\frac{6}{5}} \right)}^i}} \cr
& = \sum\limits_{i = 1}^\infty { - 2\left( {\frac{6}{5}} \right){{\left( {\frac{6}{5}} \right)}^{i - 1}}} \cr
& = \sum\limits_{i = 1}^\infty {\left( { - \frac{{12}}{5}} \right){{\left( {\frac{6}{5}} \right)}^{i - 1}}} \cr
& {\text{The }}n{\text{th term is }}{a_n} = {a_1}{r^{n - 1}},\, \cr
& {a_n} = \left( { - \frac{{12}}{5}} \right){\left( {\frac{6}{5}} \right)^{i - 1}}{\text{ then, }}{a_1} = - \frac{{12}}{5}{\text{ and }}r = \frac{6}{5} \cr
& {\text{The sum of the Terms of an Infinite Geometric Sequence is}} \cr
& {S_\infty } = \frac{{{a_1}}}{{1 - r}},{\text{ }}\left( {{\text{where }}\left| r \right| < 1} \right) \cr
& \left| r \right| = \left| {\frac{6}{5}} \right| > 1,{\text{then the terms of the sequence will not have a sum}}. \cr} $$
