Answer
$R=1$ ; interval of convergence is $[1,3]$
Work Step by Step
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 $[1,3]$
