Answer
$\dfrac{-2}{15}$
Work Step by Step
Let us consider a geometric series $\Sigma_{n=k}^{n=\infty} ar^n$ when $a \ne 0$. This series will diverge when $|r| \gt1$ and the sum of this series can be computed as: $S=\dfrac{ar^k}{1-r}$
We are given the series $\Sigma_{k=1}^{\infty} \dfrac{(-2)^k}{3^{k+1}}$
Here, we have $a=\dfrac{1}{3}$ and $r=\dfrac{-2}{3}$ and $k=1$
The sum of the given geometric series is:
$S=\dfrac{ar^k}{1-r}=\dfrac{\dfrac{1}{3} (\dfrac{-2}{3})^1}{1-(\dfrac{-2}{3})}$
or, $=\dfrac{(\dfrac{1}{3})(\dfrac{-2}{3})}{\dfrac{5}{3}}$
or, $=\dfrac{-2}{15}$
