Answer
$S_{∞}=\frac{4}{5}$
Work Step by Step
Given geometric sequence
$\frac{3}{5},\frac{3}{20},\frac{3}{80},...$
$a_{1}= \frac{3}{5}$
Common ratio $r = \frac{a_{n}}{a_{n-1}}$
$r= \frac{a_{2}}{a_{1}} =\frac{\frac{3}{20}}{\frac{3}{5}} = \frac{3}{20} \times \frac{5}{3} = \frac{1}{4}$
$|r| \lt 1$, So $S_{∞}$ exists.
Sum of the terms of an infinite geometric sequence is
$S_{∞}=\frac{a_{1}}{1-r}$
Substituting $a_{1}$ and $r$
$S_{∞}=\frac{\frac{3}{5}}{1-\frac{1}{4}}$
$S_{∞}=\frac{\frac{3}{5}}{\frac{3}{4}}$
$S_{∞}=\frac{3}{5} \times \frac{4}{3}$
$S_{∞}= \frac{4}{5}$
