Using the geometric series 11−u=∑n=0∞un\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n1−u1=∑n=0∞un, find the power series for 11+x2\frac{1}{1+x^2}1+x21.
∑n=0∞(−1)nx2n\displaystyle\sum_{n=0}^{\infty}(-1)^n x^{2n}n=0∑∞(−1)nx2n, for ∣x∣<1|x|<1∣x∣<1
∑n=0∞x2n\displaystyle\sum_{n=0}^{\infty} x^{2n}n=0∑∞x2n, for ∣x∣<1|x|<1∣x∣<1
∑n=0∞(−1)nxn\displaystyle\sum_{n=0}^{\infty}(-1)^n x^{n}n=0∑∞(−1)nxn, for ∣x∣<1|x|<1∣x∣<1
∑n=1∞(−1)nx2n\displaystyle\sum_{n=1}^{\infty}(-1)^n x^{2n}n=1∑∞(−1)nx2n, for ∣x∣<1|x|<1∣x∣<1