Answer
$\dfrac{157}{111}$
Work Step by Step
We have $1.414414414=1+\dfrac{414}{1000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$
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,
$S=1+\dfrac{414}{1000}(\dfrac{1}{1-0.001})=\dfrac{157}{111}$
