Problem 16 - Essential Calculus

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.) $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} n !(x-4)^{n}}{3^{n}} $$

Step-by-Step Solution

Answer

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

Step 1: 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 \[\lim_{n\rightarrow \infty} \left| \frac{(n+1) |x-4|}{3} \right|\]

Step 2: 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 \(x \in (3, 5)\).

Step 3: 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.

Key Concepts

Ratio Test

The Ratio Test is a method used to determine the convergence of a power series. It involves taking the limit of the ratio of successive terms. For a series \(\sum a_n\), the Ratio Test considers \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).

• If \(L < 1\), the series converges absolutely.
• If \(L > 1\), or if \(L = \infty\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In our exercise, by applying the Ratio Test, we find that \(L = \lim_{n \to \infty} \left| \frac{(n+1)|x-4|}{3} \right| < 1\), leading us to the interval \((3, 5)\). This means the series converges within this range of \(x\) values.

Alternating Series Test

An alternating series is one where the terms alternate in sign, typically represented as \(\sum (-1)^n b_n\). The Alternating Series Test provides a set of conditions under which these series converge.

• The first condition is that the terms \(b_n\) must be decreasing: \(b_{n+1} \leq b_n\) for all \(n\).
• The second condition requires that the limit of the terms as \(n\) approaches infinity is zero: \( \lim_{n \to \infty} b_n = 0\).

In our specific series, when \(x = 3\) and \(x = 5\), the series both reduce to an alternating series. Both these conditions are satisfied because the magnitude of terms decreases and approaches zero. Thus, the series converges at these endpoints.

Interval of Convergence

The interval of convergence is the set of all \(x\) values for which a power series converges. It outlines where on the number line the function given by the series is valid.To determine this interval, we use tests like the Ratio Test to find an interval where the series converges, ignoring the endpoints. For our series, the Ratio Test indicated convergence for \(x\) values between 3 and 5, forming the interval \((3, 5)\).After determining this interval, it is crucial to separately check the convergence at the endpoints. This ensures that the solution includes accurate and complete information about convergence.

Convergence at Endpoints

Convergence at endpoints must be evaluated separately because the Ratio Test cannot provide definitive results for these points. At the endpoints, other tests, such as the Alternating Series Test, help verify if the series converges.In the solution, we tested both endpoints:

• For \(x = 3\), the series is an alternating one. It converges as explained with the Alternating Series Test, thanks to its decreasing terms which approach zero.
• For \(x = 5\), the series is identical in form to \(x = 3\), thus also converging.

Verifying convergence at endpoints augments the precision of the interval of convergence, resulting in a final interval of \([3, 5]\), inclusive of both endpoints.