Problem 22 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The radius of convergence is 1, and the interval of convergence is (0, 2).

Step 1: Use the Ratio Test for Convergence

To find the radius of convergence, we'll apply the Ratio Test to the power series. The Ratio Test states that for a series $$\sum a_k$$, the series converges if: $$\lim_{k\to\infty}\lvert\frac{a_{k+1}}{a_k}\rvert = L < 1$$ Our given power series is $$\sum k(x-1)^{k}$$, so let's set up the ratio test: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)^{(k+1)}}{k(x-1)^k}\rvert$$

Step 2: Simplify the Limit Expression

Now, simplify the expression within the absolute value: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)}{k}\rvert$$

Step 3: Determine the Limit

Calculate the limit as k goes to infinity: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)}{k}\rvert = |x-1|\lim_{k\to\infty}\lvert\frac{k+1}{k}\rvert = |x-1|$$

Step 4: Apply the Ratio Test

For the series to converge using the Ratio Test, |x-1| < 1. Therefore, we get -1 < x-1 < 1, which means that the interval of convergence is: $$0 < x < 2$$ The radius of convergence, R, can be determined by half the length of the interval. As we have an interval between 0 and 2, the radius of convergence R is: $$R = \frac{2 - 0}{2} = 1$$

Step 5: Test the Endpoints

Now, test the endpoints of the interval of convergence, x = 0 and x = 2: For x = 0, the series is $$\sum k(-1)^k$$, which is an alternating series and does not converge. For x = 2, the series is $$\sum k(1)^k$$ or $$\sum k$$, which is the harmonic series and also does not converge.

Step 6: Determine the Interval of Convergence

Since neither endpoint converges, the interval of convergence remains unchanged: $$0 < x < 2$$ In summary, the radius of convergence is 1, and the interval of convergence is (0, 2).