Question

find the values of x for which the given geometric series converges. also, find the sum of the series (as a function of x) for those values of x.
sum _ { n = 0 } ^ { infty } 8 ^ { n } x ^ { n }
the series converges for x such that (square).
(type an inequality or a compound inequality.)

Explanation:

Step1: Identify the common ratio

For the geometric series \(\sum_{n = 0}^{\infty}8^{n}x^{n}=\sum_{n=0}^{\infty}(8x)^{n}\), the common ratio \(r = 8x\).

Step2: Apply the convergence condition

A geometric series \(\sum_{n = 0}^{\infty}r^{n}\) converges when \(|r|\lt1\). So, for \(r = 8x\), we have \(|8x|\lt1\).
Solve the inequality:

$$
LATEXBLOCK0
$$

Step3: Find the sum of the convergent geometric series

The sum of an infinite geometric series \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\) (when \(|r|\lt1\)). Here \(r = 8x\), so the sum \(S=\frac{1}{1 - 8x}\)

Answer:

The series converges for \(x\) such that \(-\frac{1}{8}\lt x\lt\frac{1}{8}\). The sum of the series is \(\frac{1}{1 - 8x}\) for \(x\in(-\frac{1}{8},\frac{1}{8})\)