Answer
$\frac{1}{11}$
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{2^{3n}}=\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{8^{n}}$
$=\Sigma_{n=1}^{\infty}\frac{1}{8}(\frac{-3}{8})^{n-1}$
It appears to be geometric series $\Sigma_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}$
where $a$ is common ratio.
Here, $a=\frac{1}{8}$ and $r=-\frac{3}{8}$
Thus,
$\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{2^{3n}}=\frac{a}{1-r}$
$=\frac{1/8}{1-(-3/8)}$
$=\frac{1}{11}$
