Answer
$\dfrac{2}{9}$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
Thus, we have $\sum_{n =1}^{ \infty} \dfrac{2}{10^n}$
We have a convergent geometric series with first term, $a=\dfrac{2}{10}$ and common ratio $r =\dfrac{1}{10}$
Thus, $S=\dfrac{\dfrac{2}{10}}{1-\dfrac{1}{10}}=\dfrac{2}{9}$
