Answer
$\Sigma_{n=0}^{\infty}(\frac{2}{3^{n+1}}){x}^{n}$
The interval of convergence is: $(-3,3)$
The radius of convergence is $3$
Work Step by Step
The sum of a geometric series with initial term $a$ and common ratio $r$ is
$S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
Therefore,
$f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(\frac{2}{3})(\frac{x}{3})^{n}=\Sigma_{n=0}^{\infty}(\frac{2}{3^{n+1}}){x}^{n}$
This is the power series representation of $f(x)$.
We know that the power series converges when $r=|\frac{x}{3}|\lt 1$
$|x|\lt 3$
Hence, the interval of convergence is: $(-3,3)$
The radius of convergence is: $3$
