Find the interval of convergence for the given power series. \( \sum_{n=1}^{\infty} \frac{(x-7)^{n}}{n(-4)^{n}} \) The series is convergent on the interval:

To find the interval of convergence for the power series \[ \sum_{n=1}^{\infty} \frac{(x-7)^{n}}{n(-4)^{n}}, \] we can use the ratio test. The general term of the series is \[ a_n = \frac{(x-7)^{n}}{n(-4)^{n}}. \] We will apply the ratio test, which involves finding the limit of the absolute value of the ratio of consecutive terms: \[ \left| \frac{a_{n+1}}{a_n} \right|. \] Calculating \(a_{n+1}\): \[ a_{n+1} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n+1}} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n} \cdot (-4)}. \] Now, we compute the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n+1}} \cdot \frac{n(-4)^{n}}{(x-7)^{n}} = \frac{(x-7) \cdot n}{(n+1)(-4)}. \] Taking the absolute value: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-7) \cdot n}{(n+1)(-4)} \right| = \frac{|x-7| \cdot n}{(n+1) \cdot 4}. \] Now, we take the limit as \(n\) approaches infinity: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{|x-7| \cdot n}{(n+1) \cdot 4} = \lim_{n \to \infty} \frac{|x-7| \cdot n}{4n + 4} = \lim_{n \to \infty} \frac{|x-7|}{4} \cdot \frac{n}{n + 1} = \frac{|x-7|}{4}. \] For the series to converge, we require: \[ \frac{|x-7|}{4} < 1. \] Multiplying both sides by 4 gives: \[ |x-7| < 4. \] This inequality can be rewritten as: \[ -4 < x - 7 < 4. \] Adding 7 to all parts of the inequality results in: \[ 3 < x < 11. \] Thus, the interval of convergence is \( (3, 11) \). Next, we need to check the endpoints \(x = 3\) and \(x = 11\) to determine if they are included in the interval of convergence. 1. **At \(x = 3\)**: \[ \sum_{n=1}^{\infty} \frac{(3-7)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{1}{n}. \] This series diverges (harmonic series). 2. **At \(x = 11\)**: \[ \sum_{n=1}^{\infty} \frac{(11-7)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{4^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{n=1}^{\infty} \frac{1}{n}. \] This series also diverges. Since both endpoints do not converge, the interval of convergence is: \[ \boxed{(3, 11)}. \]