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

$\dfrac{17}{6}$
We have two geometric series: $(1+1)+(\dfrac{1}{2} -\dfrac{1}{5})+(\dfrac{1}{4} +\dfrac{1}{25})+(\dfrac{1}{8} -\dfrac{1}{125})+...$ Here, $a=5, r=\dfrac{1}{2}$ and $a=1, r=\dfrac{-1}{5}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ or, $=s_1 +s_2=\dfrac{1}{1-\dfrac{1}{2}}+\dfrac{1}{1+\dfrac{1}{5}}=2+\dfrac{5}{6}=\dfrac{17}{6}$