Answer
$R=2$ ; the interval of convergence is $[-2,2)$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{n+1}{n}\frac{x^{n+1}}{x^n}\frac{2^n}{2^{n+1}}\frac{n^2+1}{n^2+2n+2}|$
$=\frac{|x|}{2}$
Hence, $R=2$ ; the interval of convergence is $[-2,2)$
