Problem 34 - Thomas Calculus

Question

Which of the series in Exercises \(15 - 48\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { n ^ { 2 } + 2 n + 1 } $$

Step-by-Step Solution

Answer

The series converges absolutely, as tested by the Alternating Series and p-series tests.

Step 1: Identify the Series Type

The series given is \( \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { n ^ { 2 } + 2 n + 1 } \). This is an alternating series as indicated by the factor \(( - 1 ) ^ { n - 1 }\). We can apply the Alternating Series Test to check for convergence.

Step 2: Simplify the Denominator

Notice that the denominator \(n^2 + 2n + 1\) can be rewritten as \((n+1)^2\), turning our series into \(\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } } { (n+1)^2 }\). This simplification will help in testing convergence.

Step 3: Apply the Alternating Series Test

According to the Alternating Series Test, a series \(\sum (-1)^{n-1} a_n\) converges if the sequence \(a_n\) is decreasing and \(\lim_{n \to \infty} a_n = 0\). Here, \(a_n = \frac{1}{(n+1)^2}\). Clearly, \(a_n\) is positive, decreasing, and \(\lim_{n \to \infty} a_n = 0\), thus the series converges.

Step 4: Test for Absolute Convergence

To check for absolute convergence, consider the absolute values, resulting in the series \(\sum_{n=1}^{\infty} \frac{1}{(n+1)^2}\). This is similar to the p-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which converges as \(p > 1\). Hence, the original series converges absolutely.

Key Concepts

Absolute Convergence

In mathematical analysis, absolute convergence is an essential concept. It helps determine whether a given series converges regardless of the signs of its terms. A series \( \sum a_n \) is said to converge absolutely if the series of absolute values \( \sum |a_n| \) also converges.

When testing for absolute convergence, convert any alternating or mixed-sign series into its absolute form by removing the negative signs. Analyze this new series to understand its behavior better.

• If \( \sum |a_n| \) converges, then \( \sum a_n \) converges absolutely.
• If \( \sum |a_n| \) diverges, the series may still converge, but only conditionally.

In our exercise, the series \( \sum \frac{(-1)^{n-1}}{(n+1)^2} \) becomes \( \sum \frac{1}{(n+1)^2} \) when considering the absolute terms. By recognizing this series as a p-series with \( p = 2 \), we confirm it converges absolutely, since \( p > 1 \), aligning with characteristics of a converging series.

P-Series

A p-series is a specific type of series that takes the general form \( \sum \frac{1}{n^p} \). Understanding this type of series allows you to quickly determine its convergence based on the power \( p \) alone. The behavior of a p-series depends explicitly on the value of \( p \):

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

The term "p-series" arises because the exponent \( p \) plays a crucial role in determining convergence.

For the series in our exercise, after simplifying the denominator, the series became \( \sum \frac{1}{(n+1)^2} \). This is easily comparable with a p-series where \( p = 2 \). Since \( 2 > 1 \), it confirms that this series converges. Understanding and recognizing p-series helps to evaluate series quickly and effectively in mathematical analysis.

Alternating Series Test

The Alternating Series Test is a useful method for determining the convergence of series whose terms alternate in sign. This type of series generally appears in the form \( \sum (-1)^{n-1} a_n \). The test states that an alternating series converges if it meets two conditions:

• The absolute value of the sequence \( a_n \) is decreasing: \( a_{n+1} \leq a_n \).
• The limit of the sequence as \( n \) approaches infinity is zero: \( \lim_{n \to \infty} a_n = 0 \).

Using these criteria to check the exercise, we identified that \( a_n = \frac{1}{(n+1)^2} \). The sequence is positive, decreasing, and its limit as \( n \to \infty \) equals zero. Hence, the original alternating series \( \sum \frac{(-1)^{n-1}}{(n+1)^2} \) converges, satisfying the above conditions of the Alternating Series Test.

Using the test, mathematicians and students can efficiently determine convergence without evaluating entire sums, simplifying complex problems into manageable assessments.