Problem 55 - Calculus Early Transcendentals

Question

Find the radius of convergence of \(\sum \frac{k ! x^{k}}{k^{k}}\)

Step-by-Step Solution

Answer

Answer: The radius of convergence for the power series is \(e\).

Step 1: Identify the sequence terms

In this power series \(\sum \frac{k ! x^{k}}{k^{k}}\), we can identify the sequence terms \(a_k = \frac{k ! x^{k}}{k^{k}}\).

Step 2: Apply the Ratio Test

To use the Ratio Test, we must first find the ratio of consecutive terms \(\frac{a_{k+1}}{a_k}\). Note that \(a_{k+1} = \frac{(k+1) ! x^{k+1}}{(k+1)^{k+1}}\). Divide \(a_{k+1}\) by \(a_k\): \(\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1) ! x^{k+1}}{(k+1)^{k+1}}}{\frac{k ! x^{k}}{k^{k}}} = \frac{(k+1) ! x^{k+1} k^{k}}{k ! x^{k} (k+1)^{k+1}}\)

Step 3: Simplify the Ratio Expression

Now we simplify the expression: \(\frac{a_{k+1}}{a_k} = \frac{(k+1) k ! x^{k} k^{k} x}{k ! (k+1)^{k} (k+1) x^{k}}\) Since we have \((k+1) k ! = (k+1) !\), this expression simplifies to: \(\frac{a_{k+1}}{a_k} = \frac{x k^{k}}{(k+1)^{k}}\)

Step 4: Find the Limit

Now, we find the limit as \(k \to \infty\): \(L = \lim_{k \to \infty} \frac{x k^{k}}{(k+1)^{k}} = x \lim_{k \to \infty} \frac{k^{k}}{(k+1)^{k}}\) Consider the expression inside the limit: \(\lim_{k \to \infty} \frac{k^{k}}{(k+1)^{k}} = \lim_{k \to \infty} \left(\frac{k}{k+1}\right)^{k}\) This is a well-known limit that approaches \(\frac{1}{e}\). Therefore: \(L = x \cdot \frac{1}{e}\)

Step 5: Determine Convergence

To ensure convergence, \(L\) must be less than 1: \(x \cdot \frac{1}{e} < 1\) Solving for \(x\), we have: \(|x| < e\)

Step 6: Find the Radius of Convergence

Now we can find the radius of convergence \(R\): \(R = e\) So, the radius of convergence for the power series \(\sum \frac{k ! x^{k}}{k^{k}}\) is \(e\).