Problem 44 - Essential Calculus Early Transcendentals

Question

Give an example of a pair of series \(\Sigma a_{n}\) and \(\Sigma b_{n}\) with posi- tive terms where \(\lim _{n \rightarrow \infty}\left(a_{n} / b_{n}\right)=0\) and \(\Sigma b_{n}\) diverges, but \(\sum a_{n}\) converges. [Compare with Exercise \(42 . ]\)

Step-by-Step Solution

Answer

The series \( \sum a_n = \sum \frac{1}{n^2} \) and \( \sum b_n = \sum \frac{1}{n} \) meet the criteria.

Step 1: Understand the Problem

The task is to find two series \( \sum a_n \) and \( \sum b_n \) with positive terms such that \( \lim _{n \rightarrow \infty}(a_n / b_n) = 0 \). In addition, the series \( \sum b_n \) must diverge, while the series \( \sum a_n \) must converge.

Step 2: Select the Divergent Series

A common example of a divergent series is the harmonic series \( \sum \frac{1}{n} \). We can choose this series as \( \sum b_n \) where \( b_n = \frac{1}{n} \). It is known to diverge.

Step 3: Find a Convergent Series

For the series \( \sum a_n \), let's choose an example like \( a_n = \frac{1}{n^2} \). The series \( \sum \frac{1}{n^2} \) is known to converge because it is a p-series with \( p = 2 > 1 \).

Step 4: Verify the Limit Condition

We need to check that \( \lim _{n \rightarrow \infty}(a_n / b_n) = 0 \). Here, \( a_n = \frac{1}{n^2} \) and \( b_n = \frac{1}{n} \), so \( \frac{a_n}{b_n} = \frac{1/n^2}{1/n} = \frac{1}{n} \). Thus, \( \lim _{n \rightarrow \infty} \frac{a_n}{b_n} = \lim _{n \rightarrow \infty} \frac{1}{n} = 0 \). This condition is satisfied.

Key Concepts

p-series

A p-series is a type of series characterized by the form \[ \sum \frac{1}{n^p}, \]where \( p \) is a constant.
The behavior of the series, whether it converges or diverges, is determined by the value of \( p \):

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

For example, the series \( \sum \frac{1}{n^2} \) converges because \( p = 2 \) which is greater than 1. This is a classic example of a convergent p-series. If you think about it, as \( n \) gets very large, the terms \( \frac{1}{n^2} \) become so small that their sum approaches a finite limit, rather than growing infinitely large. Understanding the role of "\( p \)" is critical to mastering convergence and divergence of p-series.

harmonic series

The harmonic series is a specific example of a p-series where \( p = 1 \). The series is given by:\[ \sum \frac{1}{n}. \]It is a well-known fact that the harmonic series diverges. Even though the terms \( \frac{1}{n} \) get smaller and smaller, they do not decrease fast enough for the series to converge.
In simpler terms, although each additional term adds less to the total, the sum grows without bound as you continue to add more terms. This occurs due to the relatively slow decline of the series terms. The harmonic series can be surprising because its divergence is not apparent at first glance. However, when compared to other series with faster declining terms, the difference in convergence behavior is clear.

limit comparison test

The limit comparison test is a useful tool to determine whether a series converges or diverges by comparing it with another series whose behavior is known. Here’s a quick rundown of how it works:

• Consider two series \( \sum a_n \) and \( \sum b_n \) with positive terms.
• If \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) and \( 0 < c < \infty \), both series either converge or diverge together.
• If \( c = 0 \) and \( \sum b_n \) diverges, then \( \sum a_n \) converges.

In the given exercise, \( \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \) where \( a_n = \frac{1}{n^2} \) and \( b_n = \frac{1}{n} \).
Since the harmonic series \( \sum b_n \) diverges and the limit is 0, \( \sum a_n \) converges per the limit comparison test. This example perfectly illustrates how the test is applied, showing how convergent behavior can be established in complex scenarios.This tool is powerful in comparing series and can make seemingly complicated problems manageable with the right setup.