Thomas' Calculus 13th Edition

$\dfrac{1}{11}$
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 $S=\dfrac{a}{1-r}=\dfrac{\dfrac{9}{100}}{1-(\dfrac{1}{100})}=\dfrac{1}{11}$