Chapter 11 - Section 11.8 - Power Series - 11.8 Exercises - Page 751: 15

$R=1$ ; interval of convergence is $[1,3]$

Let $a_{n}=\frac{(x-2)^{n}}{n^{2}+1}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(x-2)^{n+1}}{(n+1)^{2}+1}}{\frac{(x-2)^{n}}{n^{2}+1}}|$ $=|x-2|$ Hence, $R=1$ ; interval of convergence is $[1,3]$