Answer
The radius of convergence is $R=\dfrac{1}{2}$ and the interval of convergence is $[\dfrac{5}{2}, \dfrac{7}{2})$ .
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\dfrac{2^k (x-3)^k}{k}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{\dfrac{2^{k+1} (x-3)^{k+1}}{k+1}}{\dfrac{2^k (x-3)^k}{k}}|=2 |x-3| \lim\limits_{k \to \infty} \dfrac{k}{k+1}=2|x-3|$
This means that the series converges for $r=2|x-3| \lt 1$ . Then $-1 \lt 2(x-3) \lt 1 \implies -1/2 \lt x-3 \lt 1/2$ or, $ \dfrac{5}{2} \lt x \lt \dfrac{7}{2}$
So, the radius of convergence for the power series is $R=\dfrac{1}{2}$
Next, we will compute the interval of convergence for the two points. Those points are $0$ and $2$.
Case 1: For $x=\dfrac{5}{2}$
$\sum_{k=1}^{\infty} \dfrac{2^k (\dfrac{5}{2}-3)^k}{k}=\sum_{k=1}^{\infty} \dfrac{(-1)^k}{k}$; this implies that the series converges.
Case 2: For $x=\dfrac{7}{2}$
$\sum_{k=1}^{\infty} \dfrac{2^k (\dfrac{7}{2}-3)^k}{k}=\sum_{k=1}^{\infty} \dfrac{1}{k}$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=\dfrac{1}{2}$ and the interval of convergence is $[\dfrac{5}{2}, \dfrac{7}{2})$ .
