Answer
$\tan^{-1} (\dfrac{1}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^3}{3}+\dfrac{(\dfrac{1}{2})^5}{5}-\dfrac{(\dfrac{1}{2})^7}{7}+....$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as: $\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+......$
Plug $x=\dfrac{1}{2}$ in the above equation to obtain:
$\tan^{-1} (\dfrac{1}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^3}{3}+\dfrac{(\dfrac{1}{2})^5}{5}-\dfrac{(\dfrac{1}{2})^7}{7}+....$
Therefore, the first $4$ non-zero terms of an infinite series is equal to:
$\tan^{-1} (\dfrac{1}{2})=\dfrac{1}{2}-\dfrac{(\dfrac{1}{2})^3}{3}+\dfrac{(\dfrac{1}{2})^5}{5}-\dfrac{(\dfrac{1}{2})^7}{7}+....$
