Problem 19 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Solution: Using the Ratio Test, we determined that the radius of convergence, R, is e. By testing the endpoints, we found that the interval of convergence is $[-e, e)$.

Step 1: Apply the Ratio Test

First, we'll apply the Ratio Test to determine the radius of convergence. The Ratio Test involves evaluating the limit as k goes to infinity of the absolute value of the ratio of consecutive terms of the series: $$\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k}\right|$$ In this case, our series is given by: $$\sum \frac{x^{k}}{k^{k}}$$ So, the terms of our series are: $$a_k = \frac{x^{k}}{k^{k}}$$ Let's find the ratio of consecutive terms:

Step 2: Find the ratio of consecutive terms

Now, we will find the ratio of consecutive terms, which is: $$\frac{a_{k+1}}{a_k} = \frac{\frac{x^{k+1}}{(k+1)^{k+1}}}{\frac{x^{k}}{k^{k}}}$$ We can simplify this expression: $$\frac{a_{k+1}}{a_k} = \frac{x^{k+1}k^{k}}{x^{k}(k+1)^{k+1}} = \frac{xk^{k}}{(k+1)^{k}}$$ Now let's evaluate the limit as k goes to infinity:

Step 3: Evaluate the limit

Evaluate the limit as k goes to infinity: $$\lim_{k \to \infty} \left| \frac{xk^{k}}{(k+1)^{k}}\right|$$ To better evaluate this limit, we can rewrite the expression as: $$\lim_{k \to \infty} \left| x\frac{k^{k}}{(k+1)^{k}}\right| = \lim_{k \to \infty} \left| x\frac{1^k}{(1+\frac{1}{k})^k}\right|$$ As k goes to infinity, the limit inside the absolute value goes to: $$x\frac{1^{\infty}}{e} = xe^{-1}$$ So, we have: $$\lim_{k \to \infty} \left| \frac{xk^{k}}{(k+1)^{k}}\right| = |xe^{-1}|$$ The Ratio Test tells us that the series converges if this limit is less than 1: $$|xe^{-1}| < 1$$

Step 4: Find the radius of convergence

Solving the inequality, we find the radius of convergence: $$|x| < e$$ So the radius of convergence, R, is e.

Step 5: Test the endpoints

Finally, we must test the endpoints of the interval to determine the interval of convergence. We have two endpoints, x = -e and x = e: For x = -e: $$\sum_{k=1}^{\infty} \frac{(-e)^{k}}{k^{k}}$$ This series converges by the Alternating series test. For x = e: $$\sum_{k=1}^{\infty} \frac{e^{k}}{k^{k}}$$ This series diverges (e.g., by comparing it to the integral of the same function). Therefore, the interval of convergence is: $$[-e, e)$$