Answer
$$\ln (\dfrac{3}{2})$$
Work Step by Step
The Taylor series for $\ln (1+x)$ can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$
We have
$\\=\dfrac{1}{2} - \dfrac{1}{2 \cdot 2^2} + \dfrac{1}{3 \cdot 2^3}-....\\=( \dfrac{1}{2})-( \dfrac{1}{2}) ( \dfrac{1}{2})^2+( \dfrac{1}{3})( \dfrac{1}{2})^3-( \dfrac{1}{4})( \dfrac{1}{2})^4+....\\= \ln (1+\dfrac{1}{2}); -1 \leq x \leq 1\\=\ln (\dfrac{3}{2})$
