/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Explain the fallacy in the follo... [FREE SOLUTION] | 91Ó°ÊÓ

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Explain the fallacy in the following argument. Let \(x=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots .\) It follows that \(2 y=x+y\), which implies that \(x=y .\) On the other hand, $$x-y=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots>0$$ is a sum of positive terms, so \(x>y .\) Thus, we have shown that \(x=y\) and \(x>y\).

Short Answer

Expert verified
Answer: The fallacy in the argument lies in assuming that both series are convergent, and it incorrectly manipulates these series as if they were convergents. The claim that x = y is incorrect because both series are actually divergent. It is true that x > y, as the series for x consists of reciprocals of odd integers which are larger than the series for y containing reciprocals of even integers.

Step by step solution

01

Breaking down the series

First, let's break down the series provided for \(x\) and \(y\): $$x=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$$ $$y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$$ Notice that \(x\) is the sum of the reciprocals of odd positive integers, and \(y\) is the sum of the reciprocals of even positive integers.
02

Analyzing the argument

The argument initially claims that \(2y = x + y\), which implies \(x = y\). Then, the argument calculates the difference between \(x\) and \(y\) and shows that the difference is the sum of positive terms, i.e., \(x > y\). This is a contradiction to the previous claim that \(x = y\). We need to investigate these claims to identify the fallacy.
03

Evaluating 2y = x + y

Let's first evaluate the claim that \(2y = x + y\). Expanding the series for \(x\) and \(y\) and doubling the series of \(y\): $$2y = 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots\right) = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$ Now let's add the series of \(x\) and \(y\): $$x + y = \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right) + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots\right)$$ Shifting the even terms to the series of \(x\) while maintaining the order of the terms: $$x + y = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$$ Comparing the obtained series for \(2y\) and \(x + y\), we can say that \(2y = x + y\).
04

Evaluating x = y

If \(2y = x + y\), then we can conclude that \(x = y\). This means that both series \(x\) and \(y\) must have the same sum, which seems contradictory since \(x\) is the sum of reciprocals of odd numbers and \(y\) is the sum of reciprocals of even numbers. The fallacy lies in the assumption that both series converge to a finite sum. In fact, both series are divergent, which renders this comparison invalid.
05

Analyzing the difference x - y

Despite the fact that the first part of the argument is incorrect, let's analyze the second part, where the difference between the terms of \(x\) and \(y\) is calculated: $$x - y = \left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots$$ Each term in the series of \(x - y\) is positive, which suggests that \(x > y\). This actually makes sense as \(x\) includes larger reciprocals compared to \(y\).
06

Conclusion

The fallacy in the argument lies in assuming that both series are convergent and that the manipulation of divergent series can be treated as if they were convergents. The claim that \(x=y\) is incorrect, and it is true that \(x>y\), as the series for \(x\) consists of reciprocals of odd integers, which are larger than the series for \(y\), containing reciprocals of even integers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series
An infinite series is a sum of an infinite sequence of terms. Unlike a finite sum, which has a limited number of terms, an infinite series keeps adding terms endlessly. The series given in the exercise are examples of infinite series, where each series is constructed by adding the reciprocals of an infinite sequence of either odd or even integers.

For instance, the series for \( x \) is \( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots \), consisting of reciprocals of odd numbers, while the series for \( y \) is \( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots \), involving reciprocals of even numbers. An essential aspect of working with series is understanding their behavior as more terms are added. Are they approaching a specific value (convergence), or are they growing without bound (divergence)?
  • Infinite series can be finite in actual value given they converge.
  • Convergent series approach a specific limit as the number of terms increases indefinitely.
  • Divergent series obviously do not have this property and tend to infinity.
Series Convergence
In mathematics, series convergence is a vital concept that helps to determine whether the infinite series settles on a particular value. A series is convergent if the value of the partial sums approaches a fixed number as the number of terms becomes very large.

However, in the provided exercise, both \( x \) and \( y \) are divergent, meaning their sums do not approach a fixed limit. The fallacy in the problem is rooted in treating these divergent series as if they converge to a finite sum.

For example,
  • The harmonic series \( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \), which includes both series \( x \) and \( y \) as parts, diverges.
  • Divergent series, like the ones in the exercise, continue growing without bounds as more terms are added.
Understanding this difference between convergence and divergence is crucial when analyzing infinite series.
Mathematical Fallacy
A mathematical fallacy is usually an error in an argument that leads to an incorrect conclusion. In the case of the exercise, the fallacy arises from treating divergent series, like \( x \) and \( y \), as if they were convergent.

The argument mistakenly claims \( 2y = x + y \), thus deducing \( x = y \). However, given that both series are divergent, this assumption is invalid. The disparity between \( x \) and \( y \) can be seen when calculating \( x - y \), as this value shows that \( x \) is indeed greater than \( y \).
  • Fallacies may involve incorrect application of mathematical rules or invalid comparisons.
  • In this exercise, assuming both series are convergent leads directly to a misinterpretation of their relationship.
Taking into account the divergent nature of the series unveils the error in reasoning, affirming that \( x > y \) logically follows from the terms' arrangement.

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Most popular questions from this chapter

Consider the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{p}},\) where \(p\) is a real number. a. Use the Integral Test to determine the values of \(p\) for which this series converges. b. Does this series converge faster for \(p=2\) or \(p=3 ?\) Explain.

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Given any infinite series \(\Sigma a_{k},\) let \(N(r)\) be the number of terms of the series that must be summed to guarantee that the remainder is less than \(10^{-r}\), where \(r\) is a positive integer. a. Graph the function \(N(r)\) for the three alternating \(p\) -series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{p}},\) for \(p=1,2,\) and \(3 .\) Compare the three graphs and discuss what they mean about the rates of convergence of the three series. b. Carry out the procedure of part (a) for the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k !}\) and compare the rates of convergence of all four series.

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).

Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using. the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}},\) for \(n=0,1,2,3, \ldots .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}\) where \(p > 0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

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