# Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 497: 7

$\dfrac{4}{5}$

#### Work Step by Step

Here, $r=\dfrac{-1}{4}, a=1$ The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{1(1-(\dfrac{-1}{4})^n)}{1-(\dfrac{-1}{4}})$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}$

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