Problem 15 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n} $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n} \), which is an alternating series because of the factor \((-1)^{n+1}\).

Step 2: Alternating Series Test

For an alternating series \( \sum (-1)^{n}a_n \) to converge, two conditions must be met: 1. \( a_{n} > 0 \) for all \( n \) and decreases monotonically.2. \( \lim_{n \to \infty}a_n = 0 \).Here, \( a_n = (0.1)^n \), which is positive and decreases as \( n \to \infty \), and \( \lim_{n \to \infty}(0.1)^n = 0 \). Thus, the series converges by the Alternating Series Test.

Step 3: Test for Absolute Convergence

To determine absolute convergence, we consider its absolute series: \( \sum_{n=1}^{\infty}(0.1)^{n} \), a geometric series with common ratio \( r = 0.1 \). A geometric series converges if \( |r| < 1 \). Since \( |0.1| < 1 \), the series \( \sum_{n=1}^{\infty}(0.1)^{n} \) converges absolutely.

Step 4: Conclusion

The original alternating series converges absolutely because the absolute series \( \sum_{n=1}^{\infty}(0.1)^{n} \) converges.

Key Concepts

Alternating Series

An alternating series is recognized by its terms alternating between positive and negative signs. This happens due to the presence of a factor like \((-1)^{n+1}\), which switches the sign of each term as \(n\) increases. In the case of our series \(\sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n}\), the negative and positive signs alternate because of the \((-1)^{n+1}\) factor.

For an alternating series to converge, two key conditions must be satisfied:

• The sequence of its terms \(a_n\) must be positive for all \(n\) and decrease monotonically (i.e., each term is less than or equal to the previous term).
• The limit of the sequence \(a_n\) as \(n\) approaches infinity must be zero. This is expressed as \(\lim_{n \to \infty} a_n = 0\).

Given our series, we identify \(a_n = (0.1)^n\), which is strictly positive and decreases as \(n\) increases. Furthermore, since \((0.1)^n\) approaches zero as \(n\) goes to infinity, both conditions for convergence are met.

Thus, by the Alternating Series Test, our series converges.

Absolute Convergence

A series is said to converge absolutely if the series formed by taking the absolute value of each term also converges. In simpler terms, if you ignore the negative signs in an alternating series and it still converges, it converges absolutely.

To check for absolute convergence of the series \(\sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n}\), we look at the absolute series \(\sum_{n=1}^{\infty}(0.1)^{n}\). This means we remove the alternating sign factor and check the convergence of the resulting series.

In our scenario, this absolute series is a geometric series, which leads to straightforward convergence checking.

Geometric Series

A geometric series is a series where each term is a constant multiple (called the common ratio \(r\)) of the previous term. The standard form for a geometric series is \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term and \(r\) is the common ratio.

Our absolute series \(\sum_{n=1}^{\infty}(0.1)^{n}\) fits this form with \(a = 0.1\) and the common ratio \(r = 0.1\). To determine whether the geometric series converges, we simply check if the absolute value of the common ratio \(|r|\) is less than 1.

In this instance, \(|0.1| = 0.1 < 1\), so our geometric series converges.

Because the absolute series converges, we conclude that the original alternating series converges absolutely. This simple rule helps us resolve many such problems effectively.