Correct Answer - Option 3 :
\(-\dfrac{1}{2} < x < \dfrac{1}{2}\)
Given series, \(\mathop \sum \limits_{n = 0}^\infty {\left( {2x} \right)^n}\)
Let, \({a_n} = \mathop \sum \limits_{n = 0}^\infty {\left( {2x} \right)^n}\)
\({a_{n + 1}} = \mathop \sum \limits_{n = 0}^\infty {\left( {2x} \right)^{n + 1}}\)
Using ratio test for convergence;
Given series converges if;
\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{\left( {n + 1} \right)}}}}{{{a_n}}}} \right| < 1\)
= \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {2x} \right)}^{n + 1}}}}{{{{\left( {2x} \right)}^n}}}} \right|\;\)
= \(\mathop {\lim }\limits_{n \to \infty } \left| {\left( {2x} \right)} \right| < 1\)
|2x| < 1
\(\left| x \right| < \frac{1}{2}\)
∴ \( - \frac{1}{2} < x < \frac{1}{2}\)
