## University Calculus: Early Transcendentals (3rd Edition)

converges to $\dfrac{1}{1+x^2}$
The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series $\sum_{n=0}^\infty (-1)^n x^{2n}=\sum_{n=0}^\infty (-x^2)^n$ shows a convergent geometric series with first term, $a=1$ and common ratio $r =-x^2$ $S=\dfrac{1}{1-(-x^2)}=\dfrac{1}{1+x^2}$ Hence, the given series converges to $\dfrac{1}{1+x^2}$ for $|x| \lt 1$