Answer
$ a.\quad R=\displaystyle \frac{1}{3}.\ \quad $Interval of convergence:$\quad \displaystyle \frac{1}{3} \leq x \lt 1$
$b. \quad$ Interval of absolute convergence:$\displaystyle \quad \frac{1}{3} \lt x \lt 1$
$ c.\quad$ At $x=\displaystyle \frac{1}{3},\quad $ the series converges conditionally
Work Step by Step
(See text: "How to Test a Power Series for Convergence".)
$\text{Step 1.} $
Use the Ratio Test to find the interval where the series converges absolutely.
Ordinarily, $|x-a|\lt R\quad $ or $\quad a-R\lt x\lt a+R.$
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(3x-2)^{n+1}}{n+1}\div \frac{(3x-2)^{n}}{n}\\
&= \displaystyle \frac{n(3x-2)^{n+1}}{(n+1)(3x-2)^{n}}\\
&= \displaystyle \frac{n(3x-2)}{(n+1)}\\
\displaystyle \lim_{n\rightarrow\infty}|\frac{n(3x-2)}{(n+1)}|&\lt 1 \\
& \text{ (...the parentheses are constant in terms of n )}\\
|3x-2|\displaystyle \lim_{n\rightarrow\infty}(\frac{n}{n+1})|&\lt 1 \\
& \text{ (...the limit equals 1)}\\
|3x-2|&\lt 1 \\
-1& \lt 3x-2 \lt 1 \\
1& \lt 3x \lt 3 \\
\displaystyle \frac{1}{3}& \lt x \lt 1 \end{align*}.$
Determine the center and radius:
$a -R\lt x\lt a+R$
$\displaystyle \frac{2}{3} -\frac{1}{3}\lt x\lt \frac{2}{3} +\frac{1}{3}$
$ a=\displaystyle \frac{2}{3},\quad$
the radius is $R=\displaystyle \frac{1}{3},$
the interval of absolute convergence is $\displaystyle \frac{1}{3} \lt x \lt 1$
$\text{Step 2.} $
If the interval of absolute convergence is finite,
test for convergence or divergence at each endpoint.
$ x=\displaystyle \frac{1}{3}\quad \Rightarrow$
$ \displaystyle \sum_{n=1}^{\infty}\frac{(1-2)^{n}}{n}=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} \qquad$ ... alternating harmonic series (convergent).
$x=$1$\quad \Rightarrow$
$ \displaystyle \sum_{n=1}^{\infty}\frac{(3-2)^{n}}{n}=\sum_{n=1}^{\infty}\frac{1}{n} \qquad$ ... harmonic series (divergent).
... At $x=\displaystyle \frac{1}{3}$, the series is conditionally convergent
the interval of convergence is $\displaystyle \frac{1}{3} \leq x \lt 1$
$\text{Step 3.} $
If the interval of absolute convergence is $a -R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$ a.\quad R=\displaystyle \frac{1}{3}.\ \quad $Interval of convergence:$\quad \displaystyle \frac{1}{3} \leq x \lt 1$
$b. \quad$ Interval of absolute convergence:$\displaystyle \quad \frac{1}{3} \lt x \lt 1$
$ c.\quad$ At $x=\displaystyle \frac{1}{3},\quad $ the series converges conditionally
