consider the series $\\sum_{n = 0}^{\\infty}\\frac{(x - 8)^{n}}{9^{n}}$\n(a) find the series radius and…
consider the series $\\sum_{n = 0}^{\\infty}\\frac{(x - 8)^{n}}{9^{n}}$\n(a) find the series radius and interval of convergence\n(b) for what values of x does the series converge absolutely?\n(c) for what values of x does the series converge conditionally?\n(a) the interval of convergence is $(-1,17)$ (simplify your answer type your answer in interval notation)\nthe radius of convergence is 9\n(b) the series converges absolutely on the interval
Answer
Explanation:
Step1: Recall the concept of absolute convergence
A power series (\sum_{n = 0}^{\infty}a_{n}(x - c)^{n}) converges absolutely if (\sum_{n=0}^{\infty}\vert a_{n}(x - c)^{n}\vert) converges. For the series (\sum_{n = 0}^{\infty}\frac{(x - 8)^{n}}{9^{n}}), we can rewrite it as (\sum_{n=0}^{\infty}(\frac{x - 8}{9})^{n}), which is a geometric series of the form (\sum_{n = 0}^{\infty}r^{n}) with (r=\frac{x - 8}{9}).
Step2: Use the convergence condition for geometric series
A geometric series (\sum_{n=0}^{\infty}r^{n}) converges if (\vert r\vert<1). For our series, we have (\vert\frac{x - 8}{9}\vert<1).
Step3: Solve the inequality
[ \begin{align*} \vert\frac{x - 8}{9}\vert&<1\
- 1&<\frac{x - 8}{9}<1\ -9&<x - 8<9\ -9+8&<x<9 + 8\ -1&<x<17 \end{align*} ]
Answer:
The series converges absolutely on the interval ((-1,17))