Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 17

$R=2$ ; interval of convergence is $[-4,0)$

Let $a_{n}=\frac{(x+2)^{n}}{2^{n}lnn}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(x+2)^{n+1}}{2^{n+1}ln(n+1)}}{\frac{(x+2)^{n}}{2^{n}lnn}}|$ $=\frac{|x+2|}{2}$ Hence, $R=2$ ; interval of convergence is $[-4,0)$