#### Answer

$\dfrac{1}{11}$

#### Work Step by Step

Consider the series, $ \dfrac{9}{100}(1+\dfrac{1}{100}+\dfrac{1}{100^2}+...)$
This is a geometric series.
Here, common ratio, $r=\dfrac{1}{100}$
Formula to find Nth partial sum of a geometric series is
$s_n=\dfrac{a(1-r^n)}{1-r}$
and $s_n=\dfrac{\dfrac{9}{100}(1-(\dfrac{1}{100})^n)}{1-\dfrac{1}{100}}$
Formula to find Sum of a geometric series is
$S=\dfrac{a}{1-r}=\dfrac{\dfrac{9}{100}}{1-(\dfrac{1}{100})}=\dfrac{1}{11}$