Answer
The interval of convergence for the series is: $[3,5]$ and the radius of convergence is $R=1$.
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-4)^{k+1}}{(k+2)} \times \dfrac{(k+1)^2}{(x-4)^k} \\=\lim\limits_{k \to \infty} \dfrac{|x-4|(k+1)^2}{(k+2)^2}\\=|x-4|$
So, we can conclude that the given series converges absolutely if $l=|x-4| \lt 1$ or $-1 \lt (x-4) \lt 1\implies 3 \lt x\lt 5 $. The test is inconclusive if $|x-4 | = 1$ for $x=3$ or $x=5$. It diverges if $l=|x-4| \gt 1$.
Therefore, the interval of convergence for the series is: $[3,5]$ and the radius of convergence is $R=1$.
