Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 497: 24

$\dfrac{157}{111}$

Work Step by Step

We have $1.414414414=1+\dfrac{414}{1000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ The sum of the geometric series $1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..$ can be found as: $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{1000}=0.001$; Thus, $S=1+\dfrac{414}{1000}(\dfrac{1}{1-0.001})=\dfrac{157}{111}$

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