## Calculus 8th Edition

series has radius of convergence is $\frac{1}{2}$ and interval $\infty$ or $[\frac{5}{2},\frac{7}{2})$
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 $-\frac{1}{2}\lt (x-3) \lt \frac{1}{2}$ $\frac{5}{2} \lt x \lt \frac{7}{2}$ Thus, the series has radius of convergence is $\frac{1}{2}$ and interval $\infty$ or $[\frac{5}{2},\frac{7}{2})$