Problem 14 - Calculus Early Transcendentals

Question

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum k !(x-10)^{k}$$

Step-by-Step Solution

Answer

Answer: The power series diverges for all x, and therefore, the interval of convergence is an empty set.

Step 1: Applying the Ratio Test

First, let's apply the Ratio Test to the given power series. The Ratio Test states that the radius of convergence is given by: $$R = \lim_{k \to \infty} \left|\frac{a_k}{a_{k+1}}\right|$$ where \(a_k\) is the general term of the power series. For the given power series, we have \(a_k = k!(x-10)^k\). Then, the next term is \(a_{k+1} = (k+1)!(x-10)^{k+1}\).

Step 2: Compute the limit

Now, let's compute the limit involved in the Ratio Test. $$R = \lim_{k\to\infty} \left|\frac{a_k}{a_{k+1}} \right| = \lim_{k\to\infty} \left|\frac{k!(x-10)^k}{(k+1)!(x-10)^{k+1}}\right|$$ To simplify, we can cancel out \((x-10)^k\) from both the numerator and denominator and expand (k+1)! $$R = \lim_{k\to\infty} \left|\frac{k!}{(k+1)k!}\right|\cdot\left|\frac{1}{x-10}\right|$$ The k! terms cancel each other, and we are left with: $$R=\lim_{k\to\infty} \left|\frac{1}{k+1}\right|\cdot\left|\frac{1}{x-10}\right|$$ Since the limit of \(\frac{1}{k + 1}\) as \(k \to \infty\) is 0, the radius of convergence R is given by: $$R = \frac{1}{|x - 10|} \cdot 0 = 0$$ But this is not possible, the radius must be positive. In this case, the Ratio Test is inconclusive and we must try a different method.

Step 3: Applying the Root Test

Let's try using the Root Test instead. The Root Test states that the radius of convergence is also given by: $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|a_k|}}$$ Therefore, we have: $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|k!(x-10)^k|}}$$

Step 4: Compute the limit using the Root Test

Let's compute the limit involved in the Root Test. $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|k!(x-10)^k|}} = \frac{1}{\lim_{k \to \infty} |(x-10)\sqrt[k]{k!}|}$$ Now, the original power series converges if the limit of the expression inside the absolute value is less than 1. In other words, we are looking for the values of x for which $$|(x-10)\sqrt[k]{k!}| < 1$$ As \(k \to \infty\), according to the Stirling's approximation formula, \(\sqrt[k]{k!} \sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\). Therefore, $$R = \frac{1}{\lim_{k \to \infty} |(x-10) \sqrt{2\pi k}\left(\frac{k}{e}\right)^k|}$$ Since Stirling's approximation grows rapidly with k, the only way this limit can be other than 0 is if \((x-10) = \infty\). Thus, the power series diverges for all x.

Step 5: Interval of Convergence

Since the power series diverges for all x, there is no need to test the endpoints and the interval of convergence is an empty set.