Answer
$\dfrac{1}{11}$
Work Step by Step
Consider the series, $ \dfrac{9}{100}(1+\dfrac{1}{100}+\dfrac{1}{100^2}+...)$
The given sequence shows a geometric series with the common ratio, $r=\dfrac{1}{100}$
The Nth partial sum of a geometric series can be found as:
$s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{\dfrac{9}{100}(1-(\dfrac{1}{100})^n)}{1-\dfrac{1}{100}}$
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}=\dfrac{\dfrac{9}{100}}{1-\dfrac{1}{100}}=\dfrac{1}{11}$