## Calculus 8th Edition

$R=1$ ; interval of convergence is $[-1,1)$
Let $a_{n}=\frac {x^{n}}{2n-1}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac {x^{n+1}}{2(n+1)-1}}{\frac {x^{n}}{2n-1}}|$ $=|x|\lt 1$ At $x=-1$ , alternating series, thus converges and $x=1$ divergent. Hence, $R=1$ ; interval of convergence is $[-1,1)$