Answer
$S_{\infty}=\dfrac{-64}{3}$
Work Step by Step
Using $S_{\infty}=\dfrac{a_1}{1-r}$, the sum of the infinite geometric sequence $
-16,-4,-1,...
$ is
\begin{array}{l}
S_{\infty}=\dfrac{-16}{1-\dfrac{1}{4}}\\\\
S_{\infty}=\dfrac{-16}{\dfrac{3}{4}}\\\\
S_{\infty}=\dfrac{-64}{3}
.\end{array}