## Calculus: Early Transcendentals 8th Edition

$e^{-ln2}=1/2$
Given: $1-ln2+\frac{(ln2)^{2}}{2!}-\frac{(ln2)^{3}}{3!}+....$ $=\Sigma_{n=1}^{\infty}\dfrac{(-ln2)^{n}}{n!}$ The above series resembles: $e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{(n!)}$ with $x=-ln2$ Therefore, this series is equal to $e^{-ln2}$. which can be simplified to $e^{-ln2}=(e^{ln2})^{-1}=2^{-1}=\frac{1}{2}$