Answer
$\dfrac{23}{99}$
Work Step by Step
We have:
$0.232323= \dfrac{23}{100}+\dfrac{23}{(100)^2}+\dfrac{23}{(100)^3}+....=\dfrac{23}{100}(1+ \dfrac{1}{100}+ \dfrac{1}{(100)^2}+..)$
The sum of a geometric series $1+ \dfrac{1}{100}+ \dfrac{1}{(100)^2}+..$ can be found as:
$S=\dfrac{a}{1-r}$
We have: $a=1, r=\dfrac{1}{100}=0.01$;
Thus,
$S=\dfrac{23}{100}(\dfrac{1}{1-0.01})=\dfrac{23}{99}$
