Answer
$R=\infty$ ; interval of convergence is $(-\infty,\infty)$
Work Step by Step
Let $a_{n}=\frac{x^{n}}{1.3.5.....(2n-1)}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n+1}}{1.3.5.....(2(n+1)-1)}}{\frac{x^{n}}{1.3.5.....(2n-1)}}|$
$=\lim\limits_{n \to \infty}\frac{|x|}{2n+1}\lt 1$
$=0\lt 1$
Hence, $R=\infty$ ; interval of convergence is $(-\infty,\infty)$
