Problem 15 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The radius of convergence is 1, and the interval of convergence is \((-1, 1)\).

Step 1: Applying the Ratio Test

Let's apply the Ratio Test to the given power series: $$\lim_{k\to\infty} \left| \frac{a_{k+1}}{a_k} \right|$$ where \(a_k = \sin^{k}\left(\frac{1}{k}\right) x^{k}\).

Step 2: Calculate the Limit

Now, we calculate the limit: $$\lim_{k\to\infty} \left| \frac{\sin^{k+1}\left(\frac{1}{k+1}\right) x^{k+1}}{\sin^{k}\left(\frac{1}{k}\right) x^{k}} \right| = \lim_{k\to\infty} \left| \frac{\sin\left(\frac{1}{k+1}\right)}{\sin\left(\frac{1}{k}\right)}\right|^k \cdot |x| $$ By applying the Squeeze theorem, we know that \(\lim_{k\to\infty} \frac{\sin\left(\frac{1}{k+1}\right)}{\sin\left(\frac{1}{k}\right)} = 1\). So, we have: $$\lim_{k\to\infty} \left| \frac{\sin^{k+1}\left(\frac{1}{k+1}\right) x^{k+1}}{\sin^{k}\left(\frac{1}{k}\right) x^{k}} \right| = |x|$$

Step 3: Compare to 1

For convergence using the Ratio Test, the limit of the ratio should be less than 1. In our case, we get: $$|x| < 1$$ Thus, the radius of convergence is 1.

Step 4: Check Endpoints

We will now check the endpoints of the interval of convergence, which are -1 and 1, to see if they are included in the final interval. Case 1: When \(x = -1\) $$\sum \sin^{k}\left(\frac{1}{k}\right) (-1)^{k}$$ This series does not converge, as it does not meet the conditions for the Alternating Series Test. Case 2: When \(x = 1\) $$\sum \sin^{k}\left(\frac{1}{k}\right)$$ This series also does not converge, as it does not meet the conditions for convergence based on the Limit Comparison Test.

Step 5: Determine the Interval of Convergence

Based on our analysis of the endpoints, the interval of convergence is \((-1, 1)\).