Problem 17 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The radius of convergence is 3, and the interval of convergence is (-3, 3).

Step 1: Apply Ratio Test

The Ratio Test is a technique that calculates the limit of the absolute value of the ratio of consecutive terms of a series when k tends towards infinity. The Ratio Test states that a series converges if the limit L is less than 1, diverges if the limit L is greater than 1, and is inconclusive if the limit is equal to 1. For the given power series, we will determine the ratio first: $$\frac{a_{k+1}}{a_k} = \frac{\left(\frac{x}{3}\right)^{k+1}}{\left(\frac{x}{3}\right)^k} = \frac{x}{3}$$ Then, we will find the limit of this ratio as k tends to infinity: $$L = \lim_{k\to\infty} \left|\frac{x}{3}\right| = \frac{|x|}{3}$$

Step 2: Determine Radius of Convergence

According to the Ratio Test, the series converges if L < 1: $$\frac{|x|}{3} < 1$$ Now, we will solve this inequality for x to find the radius of convergence: $$|x| < 3$$ The inequality states that the series converges when the absolute value of x is less than 3, which corresponds to a radius of convergence of 3.

Step 3: Test the Endpoints

The inequality gives us the interval of convergence without including the endpoints: $$-3 < x < 3$$ Now, we need to test the endpoints to determine if they belong to the interval of convergence. We will analyze the given power series when x = -3 and x = 3: 1. x = -3: $$\sum\left(\frac{-3}{3}\right)^{k} = \sum\left(-1\right)^{k}$$ This is an alternating series, and since its terms do not approach 0 when k tends to infinity, it does not converge. 2. x = 3: $$\sum\left(\frac{3}{3}\right)^{k} = \sum 1^{k} = \sum 1$$ The power series elements do not approach 0, and the series is merely a sum of 1's, which diverges.

Step 4: Determine the Interval of Convergence

As the given power series does not converge at the endpoints, the interval of convergence remains the same: $$-3 < x < 3$$ Therefore, the radius of convergence is 3, and the interval of convergence is (-3, 3).