Answer
The series has a radius of convergence $\frac{1}{2}$ and interval $[\frac{5}{2},\frac{7}{2})$.
Work Step by Step
Root test:$R=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{\frac{2(x-3)^{n+1}}{\sqrt {n+4}}}{\frac{2^{n}(x-3)^{n}}{\sqrt {n+3}}}|$
$=|2(x-3)[\lim\limits_{n \to \infty}|\sqrt {\frac{(n+3)}{(n+4)}}]|$
$=|2(x-3)[|\sqrt {\frac{(1+0)}{(1+0)}}]|$
$=|2(x-3)|\lt 1$
$=|(x-3)|\lt \frac{1}{2}$
and
$-\frac{1}{2}\lt (x-3) \lt \frac{1}{2}$
$\frac{5}{2} \lt x \lt \frac{7}{2}$
The series diverges for 7/2, but converges for 5/2. Thus, the series has a radius of convergence $\frac{1}{2}$ and interval $[\frac{5}{2},\frac{7}{2})$.
