Using the geometric series 11−u=∑n=0∞un\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n1−u1=∑n=0∞un for ∣u∣<1|u| < 1∣u∣<1, what is the Maclaurin series for f(x)=11+x2f(x) = \frac{1}{1+x^2}f(x)=1+x21?
1−x2+x4−x6+x8−⋯1 - x^2 + x^4 - x^6 + x^8 - \cdots1−x2+x4−x6+x8−⋯
1+x2+x4+x6+x8+⋯1 + x^2 + x^4 + x^6 + x^8 + \cdots1+x2+x4+x6+x8+⋯
1−x+x2−x3+x4−⋯1 - x + x^2 - x^3 + x^4 - \cdots1−x+x2−x3+x4−⋯
1−x22+x44−x66+⋯1 - \frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{6} + \cdots1−2x2+4x4−6x6+⋯