Problem 12 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Apply the Ratio Test

First, apply the Ratio Test to the given power series by finding the limit as \(k\) goes to infinity of the absolute value of the ratio between the \((k+1)\)-th term and the \(k\)-th term: $$\lim_{k\to\infty} \left|\frac{\frac{(x-1)^{(k+1)}}{(k+1) !}}{\frac{(x-1)^{k}}{k !}}\right|$$

Step 2: Simplify the expression

Simplify the expression inside the limit: $$\lim_{k\to\infty} \frac{(x-1)^{(k+1)}k!}{(k+1)!(x-1)^k}$$ We can see that \((x-1)^k\) appears in both the numerator and denominator. Cancel out these common factors: $$\lim_{k\to\infty} \frac{(x-1)k!}{(k+1)!}$$

Step 3: Rewrite \((k+1)!\) in the denominator

Rewrite \((k+1)!\) as \((k+1)k!\): $$\lim_{k\to\infty} \frac{(x-1)k!}{(k+1)k!}$$ Then cancel out the common factors of \(k!\): $$\lim_{k\to\infty} \frac{x-1}{k+1}$$

Step 4: Find the limit

As \(k\) approaches infinity, the denominator goes to infinity while the numerator \((x-1)\) remains constant. This means that the limit is zero: $$\lim_{k\to\infty} \frac{x-1}{k+1} = 0$$ Since the limit is 0, the Ratio Test tells us that the power series converges for all \(x\). Hence, the radius of convergence is \(\infty\).

Step 5: Test the endpoints

Since the radius of convergence is infinity, the interval of convergence is \((-\infty, \infty)\). The power series converges for all values of \(x\).