Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 21

$R=b$ ; interval of convergence is $(a-b,a+b)$

Let $a_{n}=\frac{n}{b^{n}}(x-a)^{n}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{n+1}{b^{n+1}}(x-a)^{n+1}}{\frac{n}{b^{n}}(x-a)^{n}}|$ $=\frac{|x-a|}{b}$ $=\frac{|x-a|}{b}\lt 1$ $a-b\lt x\lt a+b$ Hence, $R=b$ ; interval of convergence is $(a-b,a+b)$