Answer
The radius of convergence is $R=2$ and the interval of convergence is $(2,6)$ .
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=(-1)^k \dfrac{k (x-4)^k}{2^k}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{(-1)^{k+1} \dfrac{k (x-4)^{k+1}}{2^{k+1}}}{(-1)^k \dfrac{k (x-4)^k}{2^k}}|= |\dfrac{x-4}{2}| \lim\limits_{k \to \infty} \dfrac{k+1}{k}=|\dfrac{x-4}{2}|$
This means that the series converges for $r=|\dfrac{x-4}{2}| \lt 1 \implies 2 \lt x \lt 6$. So, the power series is centered at $4$ and the radius of convergence for the power series is $R=2$.
Next, we will compute the interval of convergence for the two points. Those points are $2$ and $6$.
Case 1: For $x=2$
$\sum_{k=1}^{\infty} (-1)^k \dfrac{k (2-4)^k}{2^k}=\sum_{k=1}^{\infty} k$; this implies that the series diverges.
Case 2: For $x=6$
$\sum_{k=1}^{\infty} (-1)^k \dfrac{k (6-4)^k}{2^k}=\sum_{k=1}^{\infty} (-1)^k
k$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=2$ and the interval of convergence is $(2,6)$ .
