Question

find the values of x for which the geometric series converges. also, find the sum of the series (as a function of x) for those values of x.
\\( \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } ( x - 2 ) ^ { n } \\)
the series converges for \\( \square \\).
(type an inequality or a compund inequality.)

Explanation:

Step1: Identify the common ratio

For the geometric series \(\sum_{n = 0}^{\infty}(-1)^{n}(x - 2)^{n}=\sum_{n=0}^{\infty}[-(x - 2)]^{n}\), the common ratio \(r=-(x - 2)\).

Step2: Apply the convergence condition

A geometric series \(\sum_{n = 0}^{\infty}r^{n}\) converges when \(|r|\lt1\). So, \(|-(x - 2)|\lt1\).
Since \(|-(x - 2)|=|x - 2|\), we have \(|x - 2|\lt1\).

Step3: Solve the inequality

\(-1\lt x - 2\lt1\).
Adding \(2\) to all parts of the compound - inequality: \(-1+2\lt x-2 + 2\lt1 + 2\).
So, \(1\lt x\lt3\).

Step4: Find the sum of the geometric series

The sum of an infinite geometric series \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), when \(|r|\lt1\). Here \(r=-(x - 2)\), so the sum \(S=\frac{1}{1-(-(x - 2))}=\frac{1}{1+(x - 2)}=\frac{1}{x - 1}\).

Answer:

The series converges for \(1\lt x\lt3\). The sum of the series is \(S(x)=\frac{1}{x - 1}\), for \(x\in(1,3)\).