Answer
$-\dfrac{1}{171}$
Work Step by Step
We are given the geometric series:
$S=\sum_{k=1}^{\infty} 3\left(-\dfrac{1}{8}\right)^{3k}$
Rewrite the series:
$S=\sum_{k=1}^{\infty} 3\left(\left(-\dfrac{1}{8}\right)^3\right)^k=\sum_{k=1}^{\infty} 3\left(-\dfrac{1}{512}\right)^k=\sum_{k=0}^{\infty} \left(-\dfrac{3}{512}\right)\left(-\dfrac{1}{512}\right)^k$
We have:
$a=-\dfrac{3}{512}$
$r=-\dfrac{1}{512}$\\
Because $|r|<1$, the series converges and its sum is:
$\sum_{k=0}^{\infty} \left(-\dfrac{3}{512}\right)\left(-\dfrac{1}{512}\right)^k=\dfrac{a}{1-r}=\dfrac{-\dfrac{3}{512}}{1-\left(-\dfrac{1}{512}\right)}=\dfrac{-\dfrac{3}{512}}{\dfrac{513}{512}}=-\dfrac{3}{513}=-\dfrac{1}{171}$
