## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{41,333}{33,300}$
We have: $1.24123123123=\dfrac{124}{100}+\dfrac{123}{100000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ The sum of the geometric series $1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..$ can be found as: $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{1000}=0.001$; Thus, $S=\dfrac{124}{100}+\dfrac{123}{100000}(\dfrac{1}{1-0.001})=\dfrac{41,333}{33,300}$