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 notice that $\Sigma_{n=1}^{\infty} 3^nx^n= \Sigma_{n=0}^{\infty} (f(x))^n$
Remember that the series will converge when $|f(x)|=|3x| \lt 1$
This implies that
$|x| \lt \dfrac{1}{3}$ and $ -\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$
Next, we will find the sum of a given geometric series.
$S=\dfrac{a}{1-r}=\dfrac{1}{1-3x}$
This shows that the interval of convergence is: $-\dfrac{1}{3} \lt x \lt \dfrac{1}{3}$ and $\Sigma_{n=1}^{\infty} (3x)^n=\dfrac{1}{1-3x}$
