Answer
$\frac{1}{2}$
Work Step by Step
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
$e^{-ln2}=(e^{ln2})^{-1}=2^{-1}=\frac{1}{2}$
