Question

find the series interval of convergence and, within this interval, the sum of the series as a function of x

sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n }

the interval of convergence is ( left( - \frac { 1 } { 10 } \frac { 1 } { 10 }
ight) )
(simplify your answer type your answer in interval notation )

within the interval ( sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n } = )

Explanation:

Step1: Recognize the geometric series form

The series \(\sum_{n = 0}^{\infty}16^{n}x^{n}=\sum_{n = 0}^{\infty}(16x)^{n}\). A geometric series has the form \(\sum_{n=0}^{\infty}r^{n}\), where \(r = 16x\) in this case.

Step2: Use the formula for the sum of a geometric series

The sum of a geometric series \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), when \(|r|\lt1\). Substituting \(r = 16x\) into the formula, we get \(\frac{1}{1-16x}\), since for convergence \(|16x|\lt1\) (which gives the interval \(-\frac{1}{10}\lt x\lt\frac{1}{10}\) as already found for convergence).

Answer:

\(\frac{1}{1 - 16x}\)