Problem 21 - Calculus Early Transcendentals

Question

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{k^{2} x^{2 k}}{k !}$$

Step-by-Step Solution

Answer

Answer: The radius of convergence for the given power series is ∞, and the interval of convergence is (-∞, ∞).

Step 1: Apply the Ratio Test

Recall the Ratio Test: If the limit as k goes to infinity of the absolute value of the ratio of successive terms is less than 1, then the series converges. First, let's write down the general term of the given power series: $$a_k = \frac{k^2 x^{2k}}{k!}.$$ Next, we calculate the ratio of successive terms: $$\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)^2 x^{2(k+1)}}{(k+1)!}}{\frac{k^2 x^{2k}}{k!}}.$$

Step 2: Simplify the Ratio

Now, let's simplify the ratio of successive terms: $$\frac{a_{k+1}}{a_k} = \frac{(k+1)^2 x^{2k+2}}{(k+1)!} \cdot \frac{k!}{k^2 x^{2k}} = \frac{(k+1)^2}{(k+1)(k+1)} \cdot \frac{x^2}{k+1} = \frac{x^2}{k+1}.$$

Step 3: Calculate the Limit

Next, we need to calculate the limit as k goes to infinity of the absolute value of the ratio of successive terms: $$\lim_{k \to \infty} \left|\frac{x^2}{k+1}\right|.$$ This limit is equal to 0 for all x because the denominator goes to infinity.

Step 4: Determine the Radius of Convergence

Since the limit is less than 1 for all x, the power series converges for all x. This implies that the radius of convergence is infinity, i.e., R = ∞.

Step 5: Test the Endpoints

Since the radius of convergence is infinite, there are no endpoints to test. Thus, the interval of convergence is the whole real line, represented as (-∞, ∞). So, the radius of convergence for the given power series is ∞, and the interval of convergence is (-∞, ∞).