Problem 25 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

** The interval of convergence for the given power series is (-1, 2].

Step 1: Apply the Ratio Test

Apply the Ratio Test to the series by dividing the (k+1)-th term by the k-th term and taking the limit as k approaches infinity: $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_{k}}\right| = \lim_{k \to \infty} \left|\frac{\frac{(x-1)^{k+1}(k+1)^{k+1}}{((k+1)+1)^{k+1}}}{\frac{(x-1)^{k}k^{k}}{(k+1)^{k}}}\right| $$

Step 2: Simplify the Ratio

Simplify the terms inside the absolute value: $$\lim_{k \to \infty} \left|\frac{(x-1)(k+1)}{k+2}\right|$$

Step 3: Determine the Radius of Convergence

For the series to converge, the limit must be less than 1: $$\lim_{k \to \infty} \left|\frac{(x-1)(k+1)}{k+2}\right| < 1$$ The limit does not depend on k, so we have: $$|(x-1)| < 1$$ To find the radius of convergence (R), we set: $$R = |x-1|$$ So, the radius of convergence, R, is 1.

Step 4: Test the Endpoints

Test the endpoints of the interval (-R, R) by plugging them into the original series and checking for convergence or divergence: For the left endpoint, x = 0: $$\sum \frac{(-1)^{k} k^{k}}{(k+1)^{k}}$$ This series does not converge, because the terms do not approach 0, and the series is not an alternating series. For the right endpoint, x = 2: $$\sum \frac{k^{k}}{(k+1)^{k}}$$ We can rewrite this series as: $$\sum \left(\frac{k}{k+1}\right)^{k}$$ This series converges by the Ratio Test.

Step 5: Determine the Interval of Convergence

Since the radius of convergence is 1, and the series converges for the right endpoint x=2 but not for the left endpoint x=0, the interval of convergence is: $$(-1, 2]$$