Answer
$R=1$ ; interval of convergence is $[-1,1)$
Work Step by Step
Let $a_{n}=\frac {x^{n}}{2n-1}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac {x^{n+1}}{2(n+1)-1}}{\frac {x^{n}}{2n-1}}|$
$=|x|\lt 1$
When $x=-1$, the series converges by the alternating series test. It diverges at $x=1$.
Hence, $R=1$ ; interval of convergence is $[-1,1)$
