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Chapter 10: Problem 16

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{2^{k}(x-3)^{k}}{k}$$

Short Answer

Expert verified
The radius of convergence for the power series \(\sum_k \frac{2^{k}(x-3)^{k}}{k}\) is \(\frac{1}{2}\), and the interval of convergence is \([\frac{5}{2}, \frac{7}{2})\).

Step by step solution

01

Apply the Ratio Test

To apply the Ratio Test, find the limit as \(k\to\infty\) of \(\left|\frac{a_{k+1}}{a_{k}}\right|\), where \(a_k = \frac{2^{k}(x-3)^{k}}{k}\). Calculate \(a_{k+1}\): \(a_{k+1} = \frac{2^{k+1}(x-3)^{k+1}}{k+1} \) Now, compute \(\left|\frac{a_{k+1}}{a_{k}}\right|\): \(\left|\frac{a_{k+1}}{a_{k}}\right| = \left|\frac{\frac{2^{k+1}(x-3)^{k+1}}{k+1}}{\frac{2^{k}(x-3)^{k }}{k}}\right|\)
02

Simplify the Limit Expression

Simplify the limit expression by canceling terms and applying limit rules: \(\left|\frac{a_{k+1}}{a_{k}}\right| = \left|\frac{2}{\frac{k+1}{k}} \cdot \frac{(x-3)}{1}\right|\) As \(k\to\infty\), \(\frac{k+1}{k} \to 1\). Thus, the limit is: \(\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_{k}}\right| = |2(x-3)|\)
03

Determine the Radius of Convergence

For the power series to converge, the limit must be less than 1: \(|2(x-3)| < 1\) Divide both sides by 2: \(|x-3| < \frac{1}{2}\) Hence, the radius of convergence (\(R\)) is \(\frac{1}{2}\).
04

Test the Endpoints

To determine the interval of convergence, test the endpoints of the interval \((3-\frac{1}{2}, 3+\frac{1}{2})\), or \((\frac{5}{2}, \frac{7}{2})\): 1. For \(x = \frac{5}{2}\): The series becomes $$\sum_k \frac{2^{k}(x-3)^{k}}{k} = \sum_k \frac{2^{k}(-\frac{1}{2})^k}{k}$$ This is an alternating series. Apply the Alternating Series Test; since it satisfies the decreasing and limit conditions, the series converges at this endpoint. 2. For \(x = \frac{7}{2}\): The series becomes $$\sum_k \frac{2^{k}(x-3)^{k}}{k} = \sum_k \frac{2^{k}(\frac{1}{2})^k}{k}$$ This is a harmonic series times another convergent series. Since the harmonic series \(\sum_k \frac{1}{k}\) diverges, the series diverges at this endpoint.
05

Determine the Interval of Convergence

The interval of convergence is all values of \(x\) for which the series converges, including the convergent endpoint. Therefore, the interval of convergence is \([\frac{5}{2}, \frac{7}{2})\).

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