Answer
$\dfrac{d}{9}$
Work Step by Step
We have:
$0.dddd= \dfrac{d}{10}+\dfrac{d}{(10)^2}+\dfrac{d}{(10)^3}+....=\dfrac{d}{10}(1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..)$
The sum of a geometric series $1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..$ can be found as:
$S=\dfrac{a}{1-r}$
Here, $a=1, r=\dfrac{1}{10}=0.01$;
Thus,
$S=\dfrac{d}{10}(\dfrac{1}{1-0.01})=\dfrac{d}{9}$