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 } 2 ^ { n } x ^ { n }
the interval of convergence is
(simplify your answer. type your answer in interval notation.)

Explanation:

Step1: Recognize the series as a geometric series

The given series \(\sum_{n = 0}^{\infty}2^{n}x^{n}=\sum_{n=0}^{\infty}(2x)^{n}\). A geometric series has the form \(\sum_{n = 0}^{\infty}r^{n}\), where \(r = 2x\) in this case.

Step2: Apply the ratio test (or geometric series convergence condition)

For a geometric series \(\sum_{n=0}^{\infty}r^{n}\), the series converges when \(|r|\lt1\). Substituting \(r = 2x\), we get \(|2x|\lt1\).
Solve the inequality \(|2x|\lt1\):

• First, rewrite it as \(- 1\lt2x\lt1\).
• Divide each part of the inequality by \(2\): \(-\frac{1}{2}\lt x\lt\frac{1}{2}\).

When \(x =-\frac{1}{2}\), the series becomes \(\sum_{n = 0}^{\infty}(2\times(-\frac{1}{2}))^{n}=\sum_{n=0}^{\infty}(- 1)^{n}\), which diverges (by the divergence test, \(\lim_{n
ightarrow\infty}(-1)^{n}\) does not exist).
When \(x=\frac{1}{2}\), the series becomes \(\sum_{n = 0}^{\infty}(2\times\frac{1}{2})^{n}=\sum_{n=0}^{\infty}1^{n}\), which diverges (by the divergence test, \(\lim_{n
ightarrow\infty}1^{n}=1
eq0\)).

Step3: Find the sum of the series within the interval of convergence

For a geometric series \(\sum_{n = 0}^{\infty}r^{n}\) with \(|r|\lt1\), the sum \(S=\frac{1}{1 - r}\). Here \(r = 2x\), so the sum \(S(x)=\frac{1}{1-2x}\) for \(x\in(-\frac{1}{2},\frac{1}{2})\).

Answer:

The interval of convergence is \((-\frac{1}{2},\frac{1}{2})\) and the sum of the series as a function of \(x\) is \(S(x)=\frac{1}{1 - 2x}\) for \(x\in(-\frac{1}{2},\frac{1}{2})\).