Problem 26 - Calculus Early Transcendentals

Question

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}$$

Step-by-Step Solution

Answer

Answer: The radius of convergence is \(\infty\).

Step 1: Apply the Ratio Test

We start by applying the Ratio Test for absolute convergence of the power series \(\sum \frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}\): $$\lim_{k \to \infty} \frac{\frac{(-2)^{k+1}(x+3)^{k+1}}{3^{k+2}}}{\frac{(-2)^{k}(x+3)^{k}}{3^{k+1}}}$$

Step 2: Simplify the expression

Next, we simplify the expression within the limit: $$\lim_{k \to \infty} \frac{(-2)(x+3)}{3}$$

Step 3: Determine the convergence condition

The power series converges absolutely if the result of the ratio test is less than 1: $$\frac{2(x+3)}{3} < 1$$ Now, we solve for the interval of convergence, \(x\):

Step 4: Solve for x

Rearrange the inequality and simplify: $$ 2(x+3) < 3$$ $$ x+3 < \frac{3}{2}$$ $$x < -\frac{1}{2}$$ This means that the power series converges for x in the interval \((-\infty, -\frac{1}{2})\).

Step 5: Test the endpoints

Test the endpoints by plugging in \(x=-\frac{1}{2}\) into the series: $$\sum \frac{(-2)^{k}((-1/2)+3)^{k}}{3^{k+1}}$$ This series converges by the Alternating Series Test. Thus, the interval of convergence is \((-\infty, -\frac{1}{2}]\). The radius of convergence is found by taking half the length of this interval: $$radius = \frac{1}{2}(|-\frac{1}{2} - (-\infty)|) = \frac{1}{2}\infty = \infty$$ So, the radius of convergence is \(\infty\).