Chapter 9: Problem 85
Show by example that \(\Sigma\left(a_{n} / b_{n}\right)\) may converge to something other than \(A / B\) even when \(A=\Sigma a_{m}, B=\Sigma b_{n} \neq 0,\) and no \(b_{n}\) equals 0
Short Answer
Expert verified
The sequence \(\frac{a_n}{b_n} = 1\) diverges, showing \(\Sigma(a_n / b_n)\) does not equal \(A/B\).
Step by step solution
01
Understanding Convergence of Series
To show by example that \(\Sigma(a_n / b_n)\) may converge to something other than \(A/B\), we need to find sequences \(\{a_n\}\) and \(\{b_n\}\) such that their respective series converge to \(A\) and \(B\) with \(B eq 0\), but \(\Sigma(a_n / b_n)\) does not converge to \(A/B\).
02
Choose a Sequence for \(a_n\)
Let's choose \(a_n = \frac{1}{n^2}\). The series \(\Sigma a_n\) is a convergent \(p\)-series with \(p = 2\), thus it converges to some real number \(A\).
03
Choose a Sequence for \(b_n\)
Let \(b_n = \frac{1}{n}\). The series \(\Sigma b_n\) is the harmonic series, which is known to diverge. However, the partial sums \(B_N = \sum_{n=1}^N b_n\) tend to infinity. For the purpose of this example, consider \(b_n = \frac{1}{n^2}\). This parallels \(a_n\) for convergence analysis.
04
Calculate \(A/B\)
With both sequences \(a_n = \frac{1}{n^2}\) and \(b_n = \frac{1}{n^2}\), \(A = B\). Therefore, \(A/B = 1\).
05
Examine \(\Sigma(a_n / b_n)\)
The terms \(a_n / b_n = \frac{1/n^2}{1/n^2} = 1\). Therefore, the series \(\Sigma(a_n / b_n) = \Sigma 1\) diverges to infinity since it is an infinite sum of 1s.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence
Convergence is a fundamental concept in sequences and series. A series is said to converge if the sum of its terms approaches a finite limit as more terms are added. In simpler terms, if you keep adding more and more terms of the series, and the total sum gets closer and closer to a specific number, then the series converges to that number. This is important when analyzing various types of series as it determines whether the series has a meaningful limit or not.
In the context of the example above, we see that both the sequence \( \{a_n\} \) and the altered sequence \( \{b_n\} \) converge when their terms are summed. Specifically, when \( a_n = b_n = \frac{1}{n^2} \), both series converge because they are an example of a \( p \)-series with \( p = 2 \), which is greater than 1, ensuring convergence.
In the context of the example above, we see that both the sequence \( \{a_n\} \) and the altered sequence \( \{b_n\} \) converge when their terms are summed. Specifically, when \( a_n = b_n = \frac{1}{n^2} \), both series converge because they are an example of a \( p \)-series with \( p = 2 \), which is greater than 1, ensuring convergence.
Harmonic Series
The harmonic series is a specific type of series where each term is the reciprocal of an integer. Mathematically, it is written as \( \Sigma \frac{1}{n} \). Unlike many other series, the harmonic series does not converge; it diverges. This means the sum of its terms grows indefinitely without approaching a finite number.
In our example exercise, there's initially a mention of using \( b_n = \frac{1}{n} \) which forms the harmonic series. However, it was later replaced with \( b_n = \frac{1}{n^2} \) to create another convergent sequence, aligning with the sequence \( a_n \). Understanding the properties of the harmonic series helps illustrate the complexity in how series behave and their convergence or divergence tendencies.
In our example exercise, there's initially a mention of using \( b_n = \frac{1}{n} \) which forms the harmonic series. However, it was later replaced with \( b_n = \frac{1}{n^2} \) to create another convergent sequence, aligning with the sequence \( a_n \). Understanding the properties of the harmonic series helps illustrate the complexity in how series behave and their convergence or divergence tendencies.
P-Series
A \(p\)-series is a series of the form \( \Sigma \frac{1}{n^p} \), which serves as a standard example in understanding the behavior of series based on the value of \(p\). The convergence of a \(p\)-series depends primarily on the exponent \( p \) used:
- If \( p \gt 1 \), the series converges.
- If \( p \leq 1 \), the series diverges.
Divergence
Divergence refers to an infinite series that does not approach a finite limit as more terms are added. A divergent series grows indefinitely or its terms do not settle into any fixed pattern that can be summed up concisely.
In the exercise, the series \( \Sigma( a_n / b_n ) = \Sigma 1 \) diverges because it amounts to an infinite sum of 1's, resulting in an infinite value. Despite the fact that both \( \Sigma a_n \) and \( \Sigma b_n \) are convergent, their ratio leads to divergence. This highlights that the convergence of individual series does not necessarily imply convergence for a series of their ratios. This is an intriguing detail emphasizing how series can interact in surprising ways.
In the exercise, the series \( \Sigma( a_n / b_n ) = \Sigma 1 \) diverges because it amounts to an infinite sum of 1's, resulting in an infinite value. Despite the fact that both \( \Sigma a_n \) and \( \Sigma b_n \) are convergent, their ratio leads to divergence. This highlights that the convergence of individual series does not necessarily imply convergence for a series of their ratios. This is an intriguing detail emphasizing how series can interact in surprising ways.