Problem 42 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. Let \(\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}\) be an alternating series, where \(a_{n}>0\). If \(\lim _{n \rightarrow \infty} a_{n}=0\), then \(\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}\) converges.

Step-by-Step Solution

Answer

The statement is true. An alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}\), with \(a_{n}>0\), will converge if \(\lim _{n \rightarrow \infty} a_{n}=0\). This was demonstrated by constructing an example in which the decreasing condition from the Alternating Series Test is not met, but the alternating series still converges.

Step 1: Recap the Alternating Series Test

To analyze the convergence of an alternating series, we can use the Alternating Series Test. The test states that an alternating series converges if the following two conditions are met: 1. The sequence \(\{a_n\}\) is a decreasing sequence, i.e., \(a_{n+1} \le a_n\) for all \(n \ge 1\). 2. The limit of the sequence, \(\lim _{n \rightarrow \infty} a_{n}=0\).

Step 2: Analyze the given statement

We're given an alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}\), with \(a_{n}>0\). It's also given that \(\lim _{n \rightarrow \infty} a_{n}=0\). However, we don't know if the sequence \(\{a_n\}\) is decreasing or not. Therefore, we cannot conclude if the series converges using the Alternating Series Test.

Step 3: Create an example to verify the statement

We'll construct an example of an alternating series that satisfies the given condition, but not necessarily the decreasing condition from the Alternating Series Test. Let \(a_n = \frac{1}{n}\) if \(n\) is odd and \(a_n = \frac{1}{n^2}\) if \(n\) is even. Clearly, \(a_n > 0\) for all \(n\), and \(\lim _{n \rightarrow \infty} a_{n}=0\). Then the alternating series can be written as: $$ \sum_{n=1}^{\infty}(-1)^{n+1} a_{n} = 1 - \frac{1}{4} - \frac{1}{3} + \frac{1}{16} + \frac{1}{5} - \frac{1}{36} - \cdots $$ The sequence \(\{a_n\}\) is not a decreasing sequence, since \(a_3 = \frac{1}{3} > \frac{1}{4} = a_2\). Now we'll use the Comparison Test to check the convergence of the series. Since \(a_n \ge \frac{1}{n^2}\) for all \(n\), we have: $$ \sum_{n=1}^{\infty}(-1)^{n+1} a_{n} \ge \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n^2} $$ The right-hand side is an alternating convergent series (as it satisfies the Alternating Series Test). Therefore, the given series also converges, even though the decreasing condition from the Alternating Series Test is not met.

Step 4: Conclude the statement

In conclusion, the statement is true. An alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}\), with \(a_{n}>0\), will converge if \(\lim _{n \rightarrow \infty} a_{n}=0\). The example we constructed demonstrates that even if the decreasing condition from the Alternating Series Test is not met, the alternating series can still converge.