Answer
series has radius of convergence is $\infty$ and interval $\infty$ or $(-\infty, \infty)$
Work Step by Step
Root test:$R=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{\frac{2^{n+1}(x-2)^{n+1}}{(n+3)!}}{\frac{2^{n}(x-2)^{n}}{(n+2)!}}|$
$=\lim\limits_{n \to \infty}|\frac{2(x-2)}{n+3)}|$
$=|\frac{2(x-2)}{\infty}|=0\lt 1$
Thus, the series has radius of convergence is $\infty$ and interval $\infty$ or $(-\infty, \infty)$
