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Chapter 4: Problem 45

Complete the details of the following argument: If \(\sum_{n=1}^{\infty} \frac{1}{n}\) converges to a finite sum \(s\), then \(\frac{1}{2} s=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots\) and \(s-\frac{1}{2} s=1+\frac{1}{3}+\frac{1}{5}+\cdots\). Why does this lead to a contradiction?

Short Answer

Expert verified
The harmonic series diverges, so assuming it converges to a finite sum contradicts known results.

Step by step solution

01

Understanding the Original Series

The series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is the harmonic series. It is known that the harmonic series diverges, meaning it does not sum to a finite number. However, the hypothesis assumes it converges to some finite sum \( s \).
02

Formula for Half of the Series

If we take every second term from the original series, we get \( \sum_{n=1}^{\infty} \frac{1}{2n} = \sum_{n=1}^{\infty} \frac{1}{2} \cdot \frac{1}{n} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{2} s \).
03

Alternating Terms in the Series

The problem suggests that the remaining terms, after taking every second term, are \( s - \frac{1}{2}s = \frac{1}{2} s \). This assumes that \( 1 + \frac{1}{3} + \frac{1}{5} + \dots = \frac{1}{2} s \).
04

Verifying the Split into Two Series

If \( s = \frac{1}{2} s + \frac{1}{2} s \), this implies \( s = s \), which is trivially true. However, we need to be cautious as the harmonic series diverges, and \( s \) does not exist as a finite sum.
05

Identifying the Contradiction

Assuming that \( s \) is finite leads to the arrangement \( s = \frac{1}{2}s + \frac{1}{2}s \), suggesting \( s = s \), but each piece was assumed finite. The convergence of the harmonic series was falsely assumed. Since the harmonic series diverges, this assumption leads to a logical contradiction.

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Key Concepts

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

Convergence
When we talk about convergence in the context of series, we're looking at whether the sum of an infinite series approaches a finite limit. If it does, we say the series converges. But how can you tell if a series converges? This usually involves using specific tests or known series types. For example, consider the statement: "- If the series \[\sum_{n=1}^{\infty} \frac{1}{n} \]converges to a finite sum, say \( s \), then some conclusions can be drawn based on this assumption. However, not all series behave this way.In fact, a key series type is the **harmonic series**, which is fundamental in understanding convergence. Unlike some other series, it tends to infinity, never settling on a single finite number. This is why some statements about finite convergence lead to contradictions, especially for the harmonic series. So, for a series to truly converge, it must add up to a steady, manageable number over its endless terms. We need to identify that the harmonic series specifically does not behave like convergent series due to its divergence.
Divergence
Divergence is the opposite phenomenon to convergence, where the terms of a series do not approach a fixed sum, instead spreading wide and far without settling. The series \[\sum_{n=1}^{\infty} \frac{1}{n}\] is a classic example of divergence. It looks like it might calm down and add up to something specific, given that each fraction term gets smaller and smaller. However, it doesn't. As more terms are added, the sum grows without bound.Here's why the harmonic series diverges:- The terms decrease very slowly, and while they do get closer to zero, they're not fast enough.- Even the smaller and smaller pieces keep contributing significantly to the overall sum.This nature of divergence creates curiosities and complications in mathematical proofs, making it an important concept to recognize. In the original exercise, the assumption of convergence for the harmonic series challenges reality, thus leading to an eventual contradiction.
Infinite Series
An infinite series is a summation of an endless sequence of numbers, often expressed as:\[\sum_{n=1}^{\infty} a_n,\] where \(a_n\) represents the sequence's individual terms. Understanding infinite series is the key to many advanced topics in mathematics, from calculus to complex analysis.There are notable features of infinite series:- If it converges, it will sum to a specific value even with infinitely many terms.- If it diverges, like our harmonic series example, it fails to sum to a finite number.In examining infinite series, the harmonic series teaches a valuable lesson: not all infinite series have neat answers. This is why they are tested rigorously.- For convergence checks, techniques like the comparison test, ratio test, and others help us assess whether a series converges or diverges.With infinite series, always approach them with an understanding of these possible outcomes: convergence may lead to powerful and beautiful conclusions, while divergence can surprise and remind us of the limits of infinite processes.

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Most popular questions from this chapter

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula \(x_{n+1}=b x_{n}\), where \(x_{n}\) is the population of houseflies at generation \(n\), and \(b\) is the average number of offspring per housefly who survive to the next generation. Assume a starting population \(x_{0}\). Find an expression for \(S_{n}=\sum_{i=0}^{n} x_{i}\) in terms of \(b\) and \(x_{0}\). What does it physically represent?

In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{k}=2^{k} /\left(\begin{array}{l} 3 k \\ k \end{array}\right) $$

In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{k}=2^{\sin (1 / k)} $$

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{\pi^{\ln n}}{(\ln n)^{n}} .\) Show that \(\frac{a_{2 n}}{a_{n}} \rightarrow 0\) as \(n \rightarrow \infty\).

A version of von Bertalanffy growth can be used to estimate the age of an individual in a homogeneous species from its length if the annual increase in year \(n+1\) satisfies \(a_{n+1}=k\left(S-S_{n}\right)\), with \(S_{n}\) as the length at year \(n, S\) as a limiting length, and \(k\) as a relative growth constant. If \(S_{1}=3, S=9\), and \(k=1 / 2\), numerically estimate the smallest value of \(n\) such that \(S_{n} \geq 8\). Note that \(S_{n+1}=S_{n}+a_{n+1} .\) Find the corresponding \(n\) when \(k=1 / 4\)
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