## University Calculus: Early Transcendentals (3rd Edition)

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