Answer
Interval of Convergence: $-\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$
and $\Sigma_{n=1}^{\infty} (3x)^n=\dfrac{1}{1-3x}$
Work Step by Step
We have $\Sigma_{n=1}^{\infty} 3^nx^n= \Sigma_{n=0}^{\infty} (f(x))^n$
The series will converge when $|f(x)|=|3x| \lt 1$
So, $|x| \lt \dfrac{1}{3} \implies -\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$
Now, the sum of a given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-3x}$
Thus, the Interval of Convergence is: $-\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$
and $\Sigma_{n=1}^{\infty} (3x)^n=\dfrac{1}{1-3x}$
