Answer
The interval of convergence for the series is: $(\dfrac{-13}{2}, \dfrac{19}{2})$, and the radius of convergence is $R=8$.
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{(2x-3^{k+1}}{4^{2k+2}} \times \dfrac{4^{2k}}{(2x-3)^k}|\\=\dfrac{|2x-3|}{16}$
So, we can conclude that the given series converges absolutely if $\dfrac{|2x-3|}{16} \lt 1 \implies \dfrac{-13}{2} \lt x \lt \dfrac{19}{2}$ and diverges if $\dfrac{|2x-3|}{16} \gt 1$
Therefore, the interval of convergence for the series is: $(\dfrac{-13}{2}, \dfrac{19}{2})$, and the radius of convergence is $R=8$.
