the radius of convergence for the power series ∑(n = 1 to ∞) ((x - 3)^(2n))/n is equal to 1. what is the…

the radius of convergence for the power series ∑(n = 1 to ∞) ((x - 3)^(2n))/n is equal to 1. what is the interval of convergence? (a) - 4 ≤ x < - 2 (b) - 1 < x < 1 (c) - 1 ≤ x < 1 (d) 2 < x < 4 (e) 2 ≤ x < 4

the radius of convergence for the power series ∑(n = 1 to ∞) ((x - 3)^(2n))/n is equal to 1. what is the interval of convergence? (a) - 4 ≤ x < - 2 (b) - 1 < x < 1 (c) - 1 ≤ x < 1 (d) 2 < x < 4 (e) 2 ≤ x < 4

Answer

Explanation:

Step1: Recall power - series form

A power series is of the form $\sum_{n = 1}^{\infty}a_n(x - c)^n$. Here, $c = 3$ and the radius of convergence $R=1$. The general form of the interval of convergence for a power series centered at $c$ with radius of convergence $R$ is $(c - R,c + R)$ for the open - interval part.

Step2: Find the open - interval

Substitute $c = 3$ and $R = 1$ into the open - interval formula $(c - R,c + R)$. We get $(3-1,3 + 1)=(2,4)$.

Step3: Check the endpoints

Let $x = 2$. Then the series becomes $\sum_{n=1}^{\infty}\frac{(2 - 3)^{2n}}{n}=\sum_{n = 1}^{\infty}\frac{(-1)^{2n}}{n}=\sum_{n=1}^{\infty}\frac{1}{n}$, which is the harmonic series and diverges. Let $x = 4$. Then the series becomes $\sum_{n=1}^{\infty}\frac{(4 - 3)^{2n}}{n}=\sum_{n=1}^{\infty}\frac{1}{n}$, which is the harmonic series and diverges.

Answer:

D. $2<x<4$