Problem 18 - Essential Calculus Early Transcendentals

Question

For what values of \(p\) is the following series convergent? $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Series

The series given is \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}} \), which is an alternating series due to the presence of \((-1)^{n-1}\). We need to focus on the convergence of this series.

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that \( \sum_{n=1}^{\infty} (-1)^{n-1} a_n \) is convergent if \( a_n > 0 \), \( a_n \) is non-increasing, and \( \lim_{n \to \infty} a_n = 0 \). Here, \( a_n = \frac{1}{n^p} \).

Step 3: Check if \( a_n \) is Positive and Non-Increasing

Since \( n^p > 0 \) for all \( n \geq 1 \), \( a_n = \frac{1}{n^p} > 0 \). Additionally, \( a_{n+1} = \frac{1}{(n+1)^p} \leq \frac{1}{n^p} = a_n \) for \( p > 0 \), which means \( a_n \) is non-increasing.

Step 4: Evaluate the Limit Condition

For the limit condition, calculate \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^p} = 0 \) if \( p > 0 \). Thus, the limit condition is satisfied for \( p > 0 \).

Step 5: Conclusion Based on the Alternating Series Test

Since all conditions of the Alternating Series Test are satisfied when \( p > 0 \), the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}} \) is convergent for \( p > 0 \).

Key Concepts

Series Convergence

The concept of series convergence is vital in understanding whether an infinite series sums up to a specific value, or simply put, whether it "settles down" as we add more terms. If every term added gets us closer to a finite sum, the series is said to converge. If not, it diverges. In mathematical terms:

• An infinite series \( \sum_{n=1}^{\infty} a_n \) converges to a sum \( S \) if the sequence \( S_N = \sum_{n=1}^{N} a_n \) of partial sums approaches \( S \) as \( N \) approaches infinity.
• If it doesn't settle to any particular value, it diverges.

For example, consider the series given in this exercise:\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}. \] To discern its convergence properties, this series can be scrutinized through specific tests like the Alternating Series Test, which is apt for alternating series.

Alternating Series Test

An alternating series is characterized by terms that switch signs, such as positive and negative. The series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}} \) fits this pattern due to \((-1)^{n-1}\). This is where the Alternating Series Test is useful. It provides a straightforward way to check for convergence:

• Each term \( a_n \) must be positive, meaning the series' terms (ignoring signs) should be non-negative.
• The sequence of terms, \( a_n \), should be non-increasing (i.e., \( a_{n+1} \leq a_n \)).
• The formula \( \lim_{n \to \infty} a_n = 0 \) must be valid, ensuring the terms vanish as \( n \) extends towards infinity.

For the provided series, the term \( a_n = \frac{1}{n^p} \) satisfies these criteria when \( p > 0 \).

• These terms are positive since \( n^p > 0 \) for \( n \geq 1 \).
• They decrease, complying with \( a_{n+1} \leq a_n \) for \( p > 0 \).
• The limit condition \( \lim_{n \to \infty} \frac{1}{n^p} = 0 \) holds true.

Limit Condition

The limit condition is a powerful tool in determining whether the terms of a sequence diminish adequately as we approach infinity. Within the context of the alternating series test, this ensures that the series' contributions become negligible:

• For an alternating series to converge, \( \lim_{n \to \infty} a_n \) needs to be 0.

In our example, \( a_n = \frac{1}{n^p} \), which inherently decreases as \( n \) increases, owing to the growing denominator \( n^p \).

• When \( p > 0 \), the exponential term overwhelms the numerator, \( 1 \), steering the limit towards 0 as \( n \) becomes large.
• This outcome ensures one of the vital conditions for the alternating series test, confirming convergence with \( p > 0 \).

This illustrates how the limit condition directly supports our series convergence findings, ensuring the infinite series is not adding significant values further down the line.