consider the series \n sum _ { n = 0 } ^ { infty } \frac { ( x - 8 ) ^ { n } } { 9 ^ { n } } \n(a) find the…

Explanation:

Step1: Apply the ratio test

Let (a_{n}=\frac{(x - 8)^{n}}{9^{n}}). Then (a_{n+1}=\frac{(x - 8)^{n+1}}{9^{n+1}}). The ratio is (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{(x - 8)^{n+1}}{9^{n+1}}}{\frac{(x - 8)^{n}}{9^{n}}}\right|). Simplify the ratio: (\lim_{n\rightarrow\infty}\left|\frac{(x - 8)^{n+1}}{9^{n+1}}\cdot\frac{9^{n}}{(x - 8)^{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{x - 8}{9}\right|=\left|\frac{x - 8}{9}\right|). For convergence, (\left|\frac{x - 8}{9}\right|<1).

Step2: Solve the inequality for (x)

(\left|\frac{x - 8}{9}\right|<1) implies (- 1<\frac{x - 8}{9}<1). Multiply each part by (9): (-9<x - 8<9). Add (8) to each part: (-9 + 8<x<9+8), so (-1<x<17).

Step3: Check the endpoints

When (x=-1): The series becomes (\sum_{n = 0}^{\infty}\frac{(-9)^{n}}{9^{n}}=\sum_{n=0}^{\infty}(-1)^{n}), which diverges (by the divergence test, (\lim_{n\rightarrow\infty}(-1)^{n}) does not exist). When (x = 17): The series becomes (\sum_{n=0}^{\infty}\frac{9^{n}}{9^{n}}=\sum_{n = 0}^{\infty}1), which diverges (by the divergence test, (\lim_{n\rightarrow\infty}1\neq0)).

Answer:

The interval of convergence is ((-1,17))