Answer
$\dfrac{2}{9}$
Work Step by Step
A geometric series $\Sigma_{k=0}^{\infty} ar^k$ is said to be converges when $|r| \lt 1 $ and diverges when $|r| \geq 1$
and its sum is given by: $\Sigma_{k=0}^{\infty} ar^k=\dfrac{a}{1-r}$
We find that the given series is a convergent geometric series with $r=\dfrac{2}{3} \lt 1$.
So, its sum can be computed as: $\dfrac{1}{3^2} \Sigma_{k=1}^{\infty} (\dfrac{2}{3})^k=\dfrac{1}{9} \times\dfrac{2/3}{1-\dfrac{2}{3}}\\=\dfrac{2}{9}$
