Answer
$\dfrac{1}{63}$
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 have: $\Sigma_{k=1}^{\infty} 4^{-3k}=\Sigma_{k=1}^{\infty} \dfrac{1}{(4^{3})^k}\\=\Sigma_{k=1}^{\infty} \dfrac{1}{64^k}$
We find that $r=\dfrac{1}{64} \lt 1$.
So, its sum can be computed as:
$\Sigma_{k=1}^{\infty} 4^{-3k}=\dfrac{a}{1-r}\\=\dfrac{1/64}{1-\dfrac{1}{64}}\\=\dfrac{1}{63}$
