Question

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.Sometimes the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1} b_{n}\) converges to a certain fraction of an absolutely convergent series \(\sum_{n=1}^{\infty} b_{n}\) a faster rate. Given that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}\), find \(S=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\cdots .\) Which of the series \(6 \sum_{n=1}^{\infty} \frac{1}{n^{2}}\) and \(S \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}\) gives a better estimation of \(\pi^{2}\) using 1000 terms?

Short answer

The series \(6 \sum_{n=1}^{1000} \frac{1}{n^2}\) provides a better approximation of \(\pi^2\).

Step-by-step solution

Understanding the Alternating Series Test

The alternating series test states that for a series \( \sum_{n=1}^{\infty} (-1)^{n-1} b_n \) to converge, two conditions must be met: 1) The terms \( b_n \) must be positive, decreasing, and approach zero, i.e., \( b_{n+1} \leq b_n \) and \( \lim_{n \to \infty} b_n = 0 \). We need to check if these conditions are satisfied for the given series.

Check the Hypotheses for Convergence

For the series \( S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} \), we notice that the terms \( b_n = \frac{1}{n^2} \) are positive and decreasing, and \( \lim_{n \to \infty} \frac{1}{n^2} = 0 \). Therefore, both hypotheses of the alternating series test are satisfied, and hence, the series \( S \) converges.

Absolute Convergence

To check absolute convergence, we look at the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Since this is a p-series with \( p = 2 > 1 \), it converges. Therefore, the original alternating series \( S \) converges absolutely.

Calculate S

Since the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \), the alternating series converges to a different value related to this. Research shows that in such cases, the alternating series converges to half the value of the non-alternating one. Thus, \( S = \frac{1}{2} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{12} \).

Evaluate Approximation of \(\pi^2\)

For the first series \( 6 \cdot \sum_{n=1}^{1000} \frac{1}{n^2} \), since we know \( \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \), and we're using 1000 terms, it will provide a close approximation of \( \pi^2 \), though not exact due to truncation error. For the alternating series \( S \), its full convergence to \( \frac{\pi^2}{12} \) means \( S \sum_{n=1}^{1000} \frac{(-1)^{n-1}}{n^2} \) offers convergence at a similar rate since fewer terms are accounted in its result.

Compare Series for Better Approximation

Overall, the series \( 6 \sum_{n=1}^{1000} \frac{1}{n^2} \) provides a better approximation of \( \pi^2 \) with minimal error from the final term, compared to \( S \sum_{n=1}^{1000} \frac{(-1)^{n-1}}{n^2} \) which is less direct in approaching \( \pi^2\) due to its lesser total value tendency without sufficient correction.

Key concepts

Absolute Convergence

Absolute convergence refers to a situation where a series converges even when all of its terms are made positive. This is a stronger form of convergence. When a series \( \sum_{n=1}^{\infty} a_n \) is absolutely convergent, it means that the series \( \sum_{n=1}^{\infty} |a_n| \) also converges. For example, consider the alternating series \( S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} \). To check for absolute convergence, we examine the series with all positive terms, \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). This is a p-series with \( p = 2 \), which we know converges because \( p > 1 \). Consequently, the original series converges absolutely. Absolute convergence is important because it guarantees convergence regardless of any alternating signs, which can make a series converge conditionally, but not absolutely.

P-Series

A p-series is a specific type of infinite series defined as: \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \] where \( p \) is a positive real number. The convergence of a p-series depends on the value of \( p \):

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In the exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series with \( p = 2 \). This means it converges, as \( 2 > 1 \). Understanding the behavior of p-series helps determine convergence of related series, such as those appearing in convergence tests like the alternating series test. Since the p-series converges, it supports the conclusion that the alternating series derived from it converges absolutely.

Convergence Criteria

Convergence criteria are essential tools for determining whether a series sums to a finite value. Different tests and criteria apply based on the series form. A common and useful test is the Alternating Series Test. The Alternating Series Test stipulates that for a series of the form \( \sum_{n=1}^{\infty} (-1)^{n-1} b_n \) to converge:

• The terms \( b_n \) must be positive.
• They must decrease in size \( b_{n+1} \leq b_n \).
• The terms must approach zero \( \lim_{n \to \infty} b_n = 0 \).

In the example series \( S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} \), these criteria are satisfied — the terms \( \frac{1}{n^2} \) are positive, decrease with increasing \( n \), and approach zero. Thus, the series meets the convergence criteria under this test. Correctly applying convergence tests allows for accurate conclusions about a series' behavior, essential for mathematical analysis and problem-solving.