## Calculus 8th Edition

$\frac{416909}{99900}$
Convert repeating part $326$ into a series. $4.17326326326 . . .=(417+ \Sigma _{n=1}^{\infty}\frac{326}{(1000)^{n}})10^{-2}$ $=(417+ \Sigma _{n=1}^{\infty}\frac{326}{1000}\frac{1}{(1000)^{n-1}})10^{-2}$ The sum of a geometric series $\Sigma _{n=1}^{\infty}ar^{n-1}$ equals $a/1-r$ Thus, $=(417+\frac{\frac{326}{1000}}{1-\frac{1}{1000}})10^{-2}$ $=(417+\frac{326}{999})10^{-2}$ $=(\frac{416583+326}{999})10^{-2}$ $=\frac{416909}{99900}$