7. try again get help practice similar attempt 1: one or more answers is incorrect. you will be told if you…

Explanation:

Step1: Apply ratio - test

Let $a_k=\frac{(x - 4)^k}{k(-4)^k}$. Then $a_{k + 1}=\frac{(x - 4)^{k+1}}{(k + 1)(-4)^{k+1}}$. Calculate the ratio $\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{\frac{(x - 4)^{k+1}}{(k + 1)(-4)^{k+1}}}{\frac{(x - 4)^k}{k(-4)^k}}\right|=\left|\frac{(x - 4)^{k+1}}{(k + 1)(-4)^{k+1}}\times\frac{k(-4)^k}{(x - 4)^k}\right|=\left|\frac{x - 4}{-4}\right|\times\frac{k}{k + 1}$.

Step2: Find the limit

$\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\rightarrow\infty}\left|\frac{x - 4}{-4}\right|\times\frac{k}{k + 1}=\left|\frac{x - 4}{4}\right|\lim_{k\rightarrow\infty}\frac{k}{k + 1}$. Since $\lim_{k\rightarrow\infty}\frac{k}{k + 1}=1$, the limit is $\left|\frac{x - 4}{4}\right|$.

Step3: Determine convergence condition

For convergence, $\left|\frac{x - 4}{4}\right|\lt1$. Solve the inequality: $- 1\lt\frac{x - 4}{4}\lt1$. Multiply each part by 4: $-4\lt x - 4\lt4$. Add 4 to each part: $0\lt x\lt8$.

Step4: Check endpoints

When $x = 0$: The series becomes $\sum_{k = 1}^{\infty}\frac{(0 - 4)^k}{k(-4)^k}=\sum_{k=1}^{\infty}\frac{(-4)^k}{k(-4)^k}=\sum_{k = 1}^{\infty}\frac{1}{k}$, which is the harmonic series and diverges. When $x = 8$: The series becomes $\sum_{k = 1}^{\infty}\frac{(8 - 4)^k}{k(-4)^k}=\sum_{k=1}^{\infty}\frac{4^k}{k(-4)^k}=\sum_{k = 1}^{\infty}\frac{(-1)^k}{k}$, which is the alternating - harmonic series and converges.

Answer:

$(0,8]$