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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.
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.
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.
In the provided solution, both \( a_n \) and \( b_n \) are examples of \(p\)-series with \( p = 2 \). Since 2 is greater than 1, both series converge. This property is crucial for discussing whether the series together can produce a convergent or divergent result, as seen when comparing the series of their ratios.
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.

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Most popular questions from this chapter

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.

Find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x\) ) for those values of \(x .\) $$\sum_{n=0}^{\infty}(-1)^{n} x^{-2 n}$$

A sequence of rational numbers is described as follows: $$ \frac{1}{1}, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \ldots, \frac{a}{b}, \frac{a+2 b}{a+b}, \ldots $$ Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let \(x_{n}\) and \(y_{n}\) be, respectively, the numerator and the denominator of the \(n\) th fraction \(r_{n}=x_{n} / y_{n}\). a. Verify that \(x_{1}^{2}-2 y_{1}^{2}=-1, x_{2}^{2}-2 y_{2}^{2}=+1\) and, more generally, that if \(a^{2}-2 b^{2}=-1\) or \(+1,\) then $$(a+2 b)^{2}-2(a+b)^{2}=+1 \quad \text { or } \quad-1$$ respectively. b. The fractions \(r_{n}=x_{n} / y_{n}\) approach a limit as \(n\) increases. What is that limit? (Hint: Use part (a) to show that \(r_{5}^{2}-2=\pm\left(1 / y_{n}\right)^{2}\) and that \(y_{n}\) is not less than \(n .\)

Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

Which of the sequences converge, and which diverge? Give reasons for your answers. $$a_{n}=\frac{4^{n+1}+3^{n}}{4^{n}}$$

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