Answer
The interval of convergence for the series is: $(1,5)$ and the radius of convergence is $R=2$.
Work Step by Step
Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$
1. For $l \lt 1$, the series is absolutely convergent.
2. For $l \gt 1$, the series is divergent.
3. For $l = 1$, the series is inconclusive.
Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|\\=\lim\limits_{k \to \infty} \dfrac{(x-3)^{k+1}}{2^{k+1}} \times \dfrac{2^k}{(x-3)^k} \\=\lim\limits_{k \to \infty} \dfrac{|x-3|}{2}\\=\dfrac{|x-3|}{2}$
So, we can conclude that the given series converges absolutely if $l=\dfrac{|x-3|}{2} \lt 1$ or $-1 \lt \dfrac{(x-3)}{2} \lt 1\implies 1 \lt x\lt 5 $. The test is inconclusive if $|x-3|/2 =1$ for $x=1$ or $x=5$. It diverges if $l=\dfrac{|x-3|}{2} \gt 1$.
Therefore, the interval of convergence for the series is: $(1,5)$ and the radius of convergence is $R=2$.
