## Calculus: Early Transcendentals 8th Edition

$\Sigma_{n=0}^{\infty}(-2)(x)^{2n+1}$ The interval of convergence is: $(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ The radius of convergence is: $\frac{1}{\sqrt 2}$
The sum of a geometric series with initial term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(-2)^n(x)^{2n+1}$ This is the power series representation of $f(x)$. We know that the power series converges when $r=|-2x^{2}|\lt 1$ $x^{2}\lt \frac{1}{2}$ The interval of convergence is: $(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ The radius of convergence is: $\frac{1}{\sqrt 2}$