Problem 16 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: 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 \(\left|\frac{a_{k+1}}{a_{k}}\right|\): \(\left|\frac{a_{k+1}}{a_{k}}\right| = \left|\frac{\frac{2^{k+1}(x-3)^{k+1}}{k+1}}{\frac{2^{k}(x-3)^{k }}{k}}\right|\)

Step 2: 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: \(\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_{k}}\right| = |2(x-3)|\)

Step 3: 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 \(\frac{1}{2}\).

Step 4: 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.

Step 5: 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 \([\frac{5}{2}, \frac{7}{2})\).