Answer
$\ln(\dfrac{3}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^2}{2}+\dfrac{(\dfrac{1}{2})^3}{3}-\dfrac{(\dfrac{1}{2})^4}{4}+....$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as:
$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+....$
Plug $x=\dfrac{1}{2}$ in the above equation to obtain:
$\ln(\dfrac{3}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^2}{2}+\dfrac{(\dfrac{1}{2})^3}{3}-\dfrac{(\dfrac{1}{2})^4}{4}+....$
Therefore, the first $4$ non-zero terms of an infinite series is equal to:
$\ln(\dfrac{3}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^2}{2}+\dfrac{(\dfrac{1}{2})^3}{3}-\dfrac{(\dfrac{1}{2})^4}{4}+....$
