Question

In Exercises \(7-28,\) find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} n !(x-4)^{n}}{3^{n}} $$

Short answer

The interval of convergence of the given power series is \(x \in [3, 5]\).

Step-by-step solution

Apply the Ratio Test

In order to use the Ratio Test, calculate \(a_{n+1}/a_n\), then the absolute value of that, and finally take the limit as \(n \rightarrow \infty\). For this specific power series, this gives: \[\lim_{n\rightarrow \infty} \left| \frac{(-1)^{n+1} (n+1) !(x-4)^{n+1}}{3^{n+1}} / \frac{(-1)^n n !(x-4)^n}{3^n} \right|\] Simplifying this gives \[\lim_{n\rightarrow \infty} \left| \frac{(n+1) |x-4|}{3} \right|\]

Determine the Limit and Convergence Range

If the power series is to converge, this limit must be less than 1. Therefore, set \[\frac{(n+1) |x-4|}{3} < 1\] and solve for \(x\). This gives the inequality \(3 - \frac{3}{n+1} < x < 3 + \frac{3}{n+1}\). As the limit goes to infinity, the extra terms go to 0, so the range of convergence is \(x \in (3, 5)\).

Test the Endpoints

The Ratio Test is inconclusive for the endpoints, so these need to now be tested using other techniques. For \(x=3\), the power series simplifies to: \[\sum_{n=0}^{\infty} \frac{(-1)^{n} n !}{3^{n}}\] This is an Alternating Series, and it satisfies the conditions of the Alternating Series Test as its terms decrease in magnitude and go to zero, so it converges. For \(x=5\), the power series simplifies to: \[\sum_{n=0}^{\infty} \frac{(-1)^{n} n !}{3^{n}}\] This is the same series and thus also converges.