Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n \sqrt{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely because the series of absolute values \( \sum \frac{1}{n^{3/2}} \) converges.

Step 1: Define Absolute Convergence

A series \( \sum a_n \) converges absolutely if the series \( \sum |a_n| \) converges. Here, we are given \( a_n = (-1)^n \frac{1}{n\sqrt{n}} \). We need to consider the series of absolute values, \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{1}{n\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} \).

Step 2: Simplify the Series of Absolute Values

We need to consider the positive series \( \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). This is a p-series with \( p = 3/2 \).

Step 3: Determine Convergence of p-Series

A p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Here, \( p = 3/2 \), which is greater than 1. Therefore, the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges.

Step 4: Conclude Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} \) converges, the original series \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{n\sqrt{n}} \) converges absolutely.

Key Concepts

p-Series Test

The **p-Series Test** is a fundamental tool in determining the convergence of infinity series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). This test is crucial when analyzing series that include fractions with powers of \( n \) in the denominator. Here's how it works.

• For a p-series to converge, the power \( p \) must be greater than 1.
• If \( p \leq 1 \), the series will diverge.

In the given problem, we encountered the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). This is a classic p-series with \( p = \frac{3}{2} \). Since \( \frac{3}{2} > 1 \), the series converges. This means it reaches a finite limit as \( n \) approaches infinity.

Understanding the p-Series Test helps in rapidly determining whether certain infinite series exhibit convergence, which is pivotal in mathematical analysis and applications.

Convergence of Series

**Convergence of Series** is a versatile principle in mathematics that allows us to predict the behavior of sequence sums extending to infinity. A series \( \sum a_n \) is convergent if the sequence of partial sums \( S_k = a_1 + a_2 + \ldots + a_k \) approaches a definite value as \( k \) goes to infinity.

When discussing **absolute convergence**, an important refinement of series convergence, you're assessing whether \( \sum |a_n| \) converges.
The given exercise asks us to determine absolute convergence. For this, we first took the absolute values, \( a_n = ( -1 )^n \frac{1}{n\sqrt{n}} \), giving \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \).

• The absolute series \( \sum |a_n| \) was shown to converge using the p-Series Test.
• This implies the original series converges absolutely.

Convergence of Series is essential in calculus, especially in contexts where infinite series serve as functions or approximations in analysis.

Alternating Series

An **Alternating Series** is characterized by terms that are sequentially positive and negative. A simple example is \( (-1)^n \frac{1}{n} \), where the series alternates signs based on whether \( n \) is odd or even.

To check the convergence of alternating series, the **Alternating Series Test** is often employed. However, in terms of absolute convergence, we primarily focus on the absolute values of the terms.

In our problem, the original series was \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{n\sqrt{n}} \). This had:

• Close inspection to assess convergence using absolute values.
• The sign alternation suggested an alternating series, but by converting terms to absolute values helped determine convergence via the p-series.

Understanding alternating series helps differentiate between conditional and absolute convergence, providing a comprehensive understanding of how an infinite series behaves.