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Chapter 11: Problem 23

Determine the radius and interval of convergence of the following power series. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}(x-1)^{k}}{k}$$

Short Answer

Expert verified
Question: Determine the radius and interval of convergence for the power series $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}(x-1)^{k}}{k}$. Answer: The radius of convergence is 1, and the interval of convergence is (0, 2).

Step by step solution

01

Identify the general term of the series

The general term of the series is given by: $$a_k = \frac{(-1)^{k+1}(x-1)^{k}}{k}$$
02

Apply the Ratio Test

The Ratio Test requires us to find the limit as k approaches infinity: $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \frac{|a_{k+1}|}{|a_k|}$$ Substitute the general term formula and simplify: $$\lim_{k\to\infty} \frac{\left|\frac{(-1)^{k+2}(x-1)^{k+1}}{k+1}\right|}{\left|\frac{(-1)^{k+1}(x-1)^{k}}{k}\right|}$$
03

Simplify the expression

We can further simplify the expression by noting that the (-1) terms in the numerator and denominator will cancel. This leaves us with: $$\lim_{k\to\infty} \frac{(x-1)^{k+1}}{(k+1)} \cdot \frac{k}{(x-1)^{k}}$$ Which simplifies to: $$\lim_{k\to\infty} \frac{k}{k+1} \cdot (x-1)$$
04

Evaluate the limit and check convergence

Evaluating the limit, we get: $$\lim_{k\to\infty} \frac{k}{k+1} \cdot (x-1) = (x - 1)$$ The Ratio Test states that the series converges if this limit is less than 1, diverges if the limit is greater than 1, and is inconclusive if the limit is equal to 1. So, we have the inequality: $$|(x - 1)| < 1$$
05

Solve the inequality and find the radius of convergence

Solve the inequality: $$-1<(x - 1)<1$$ Adding 1 to all sides, we get: $$0<x<2$$ This is the interval of convergence. The radius of convergence (R) is half of the width of the interval, so we have: $$R = \frac{2-0}{2} = 1$$ So, the radius of convergence is 1, and the interval of convergence is (0, 2).

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