I'm working on a problem that asks me to determine the convergence center, radius, and interval of the following power series:

$$\sum^{\infty }\_{k=2} \left( k+3\right)^{2} \left( 2x-3\right)^{k}$$

Here's what I've attempted so far:

To find the convergence center, I set $$(2x-3)^k = 0$$ and solved for x. This gives me x = 3/2, which is the convergence center.

To find the convergence radius, I used the formula $$R = \lim\_{k\to\infty}|\frac{a\_{k}}{a\_{k+1}}|$$, where $$a\_k$$ is the coefficient of the $k$th term. So I calculated:

\begin{align\*}
\lim\_{k\to\infty}|\frac{(k+3)^2}{(k+4)^2}| &= \lim\_{k\to\infty}|\frac{k^2+6k+9}{k^2+8k+16}|\\
&= \lim\_{k\to\infty}|\frac{1+6/k+9/k^2}{1+8/k+16/k^2}|\\
&= \frac{1}{1}\\
&= 1
\end{align\*}

Therefore, the convergence radius is 1.

Now, to find the convergence interval, I know that the series converges for $$|x - 3/2| < R$$, and diverges for $$|x - 3/2| > R$$. So, plugging in the values I found, I get that the series converges for $$|x - 3/2| < 1$$

However, I'm not quite sure if my approach is correct, and I'd appreciate any feedback or guidance on how to correctly find the convergence center, radius, and interval.