Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 798: 33

$arctan(0.2)=\Sigma_{n=0}^{2}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}\approx 0.19740$

$arctanx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}$ $arctan(0.2)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}$ So, $s_{2}=a_{0}+a_{1}+a_{2} \approx 0.197.40$ Thus, $arctan(0.2)=\Sigma_{n=0}^{2}(-1)^{n}\frac{(0.2)^{2n+1}}{2n+1}\approx 0.19740$