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