Answer
The given series converges only at $x=2$, and the radius of convergence is $R=0$.
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{(2k+3)!}{(k+1)^3}(x-2)^{k+1} \times \dfrac{k^3}{(2k+1)!(x-2)^k}| \\=\lim\limits_{k \to \infty}|x-2| \dfrac{(2k+3)(2k+2)k^3}{(k+1)^3}\\=+ \infty$
So, we can conclude that the given series converges only at $x=2$, and the radius of convergence is $R=0$.
