Problem 32 - Calculus: Early Transcendentals

Question

For what value of \( p \) is each series convergent? \( \displaystyle \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n^p} \)

Step-by-Step Solution

Answer

The series converges for \( p > 0 \).

Step 1: Understanding the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} \). It is an alternating series because of the \((-1)^{n-1}\) term, which makes the sign of each term alternate between positive and negative.

Step 2: Identifying Convergence Criteria

For an alternating series \( \sum (-1)^{n-1} a_n \), where \( a_n = \frac{1}{n^p} \), to converge, two conditions must be satisfied: \( a_n \to 0 \) as \( n \to \infty \), and the sequence \( a_n \) must be non-increasing (i.e., \( a_{n+1} \leq a_n \)).

Step 3: Checking the Limit Condition

The first condition is \( a_n \to 0 \). Since \( a_n = \frac{1}{n^p} \), we have \( \lim_{n \to \infty} \frac{1}{n^p} = 0 \) provided \( p > 0 \). Hence \( a_n \to 0 \) for \( p > 0 \).

Step 4: Ensuring the Non-Increasing Condition

The second condition requires checking if the sequence \( a_n \) is non-increasing. Since \( a_n = \frac{1}{n^p} \), it's clear that \( a_{n+1} = \frac{1}{(n+1)^p} \leq \frac{1}{n^p} = a_n \) for all \( n \), when \( p > 0 \). Thus, \( a_n \) is non-increasing for \( p > 0 \).

Step 5: Conclusion on the Value of p

Both conditions are satisfied when \( p > 0 \). Therefore, the given alternating series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} \) is convergent for \( p > 0 \).

Key Concepts

Convergence Criteria

In mathematics, understanding whether a series converges or diverges is crucial. To determine this, specific criteria must be satisfied. For an alternating series like \( \sum (-1)^{n-1} a_n \), we focus on two primary criteria for convergence:

• The terms \( a_n \) must approach zero as \( n \to \infty \). This ensures that the series doesn't spiral out of control in magnitude.
• The sequence \( a_n \) should be non-increasing, meaning each term should be at most equal to the previous one.

These criteria help us check whether the infinite sum will settle at a particular value instead of diverging or alternating erratically.

Series Convergence

Series convergence refers to the behavior of the sum of the terms in an infinite series. For a series to converge, as you add more terms, the total sum should approach a specific finite number.

In the context of the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} \), we are dealing with an alternating series. When working with such series, it is important to check if adding an infinite number of terms will lead to a stable sum.

The convergence of this series largely depends on the value of \( p \). Specifically, the series converges if \( p > 0 \). This means that for the sum to eventually stabilize, \( p \) must be greater than zero, as shown in our original step-by-step solution.

Alternating Series Test

The alternating series test is a useful tool in determining the convergence of an alternating series. When applied to a series such as \( \sum (-1)^{n-1} a_n \), we focus on ensuring that it satisfies two conditions:

• First, each term in the series \( a_n \) approaches zero as \( n \) becomes very large.
• Second, the entire sequence should be non-increasing. This means each term should be no larger than the one before it.

If both conditions are met, the alternating series test confirms that the series converges. For the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} \), these conditions are satisfied when \( p > 0 \), ensuring series convergence.

Non-Increasing Sequence

A non-increasing sequence is a sequence where each term is less than or equal to the one before it. Understanding this property is essential when analyzing series for convergence.

For the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p} \), we look at the terms \( a_n = \frac{1}{n^p} \). To determine if this sequence is non-increasing, we check:

• Compare \( a_{n+1} = \frac{1}{(n+1)^p} \) with \( a_n = \frac{1}{n^p} \).
• If \( a_{n+1} \leq a_n \), then the sequence is non-increasing.

Given that \( \frac{1}{(n+1)^p} \leq \frac{1}{n^p} \) holds true for all \( n \) when \( p > 0 \), the sequence is indeed non-increasing. Hence, the criteria for the alternating series test are met, aiding in proving series convergence.