Answer
$arctan(0.2)=\Sigma_{n=0}^{2}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}\approx 0.19740$
Work Step by Step
$arctanx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}$
$arctan(0.2)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}$
So, $s_{2}=a_{0}+a_{1}+a_{2} \approx 0.19740$
Thus, $arctan(0.2)=\Sigma_{n=0}^{2}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}\approx 0.19740$