Problem 28 - Calculus Early Transcendentals

Question

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum(-1)^{k} \frac{x^{3 k}}{27^{k}}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the general term of the power series

First, identify the general term of the power series (a_k): $$a_k=(-1)^k \frac{x^{3k}}{27^k}$$

Step 2: Apply the Ratio Test

Apply the Ratio Test: Calculate the limit as k approaches infinity of the absolute value of the ratio of consecutive terms, i.e.: $$\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k}\right|$$ Substitute the expression for a_k and simplify: $$\lim_{k \to \infty} \left| \frac{(-1)^{k+1} \frac{x^{3(k+1)}}{27^{k+1}}}{(-1)^k \frac{x^{3k}}{27^k}}\right|$$ Simplify the expression by cancelling out some terms: $$\lim_{k \to \infty} \left| \frac{x^3}{27} \right|$$ Observe that there is no k term left in the expression, so we can conclude: $$\left| \frac{x^3}{27} \right| < 1$$ This expression is valid for the convergence of the power series.

Step 3: Determine the radius of convergence

We can deduce the radius of convergence (R) by solving the inequality: $$\left| \frac{x^3}{27} \right| < 1$$ Rearrange the expression: $$\left| x^3 \right| < 27$$ Take the cube root on both sides: $$\left| x \right| < 3$$ So the radius of convergence, R, is 3.

Step 4: Test the endpoints of the interval

To determine the interval of convergence, we need to test the endpoints of the interval (-3, 3). For x = -3: $$\sum (-1)^k \frac{(-3)^{3k}}{27^k} = \sum (-1)^k 3^{3k-3k} = \sum (-1)^k$$ This series diverges due to the Alternating Series Test. For x = 3: $$\sum (-1)^k \frac{(3)^{3k}}{27^k} = \sum (-1)^k 3^{3k-3k} = \sum (-1)^k$$ This series also diverges due to the Alternating Series Test.

Step 5: Find the interval of convergence

Since the series converges for |x| < 3 but not at the endpoints, the interval of convergence is: $$-3 < x < 3$$