/*! 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 25 Show that any convergent sequenc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 25

Show that any convergent sequence is a Cauchy sequence.

Short Answer

Expert verified
Given a convergent sequence \( (a_n) \) with limit L, for any \( \epsilon > 0 \), there exists a natural number N such that for all \( n \geq N \), \( |a_n - L| < \frac{\epsilon}{2} \). Then, for arbitrary indices \( m, n \geq N \), applying the triangle inequality yields \( |a_m - a_n| \leq |a_m - L| + |a_n - L| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \). Thus, for all \( \epsilon > 0 \), there exists a natural number N such that for all \( m, n \geq N \), \( |a_m - a_n| < \epsilon \), proving that any convergent sequence is a Cauchy sequence.

Step by step solution

01

Understand Definitions

First, we should recall the definitions of convergent and Cauchy sequences. - A sequence \( (a_n) \) is said to be convergent if there exists a limit L such that for any \( \epsilon > 0 \), there exists a natural number N such that for all \( n \geq N \), \( |a_n - L| < \epsilon \). - A sequence \( (a_n) \) is said to be a Cauchy sequence if for any \( \epsilon > 0 \), there exists a natural number N such that for all \( m, n \geq N \), \( |a_m - a_n| < \epsilon \).
02

Consider a Convergent Sequence

Let \( (a_n) \) be a convergent sequence with limit L. We want to demonstrate that this sequence must also be a Cauchy sequence. To do this, we need to show that for any given real number \( \epsilon > 0 \), there exists a natural number N such that for all \( m, n \geq N \), \( |a_m - a_n| < \epsilon \).
03

Leverage the Definition of Convergence

Since \( (a_n) \) is a convergent sequence with limit L, for any \( \epsilon > 0 \), there exists a natural number N such that for all \( n \geq N \), \( |a_n - L| < \frac{\epsilon}{2} \). The choice of \( \frac{\epsilon}{2} \) will be useful when we apply the triangle inequality.
04

Apply the Triangle Inequality

Now, consider arbitrary indices \( m, n \geq N \). By the triangle inequality, we have: \( |a_m - a_n| \leq |a_m - L| + |a_n - L| \) Since \( m, n \geq N \), it follows from the definition of convergence that \( |a_m - L| < \frac{\epsilon}{2} \) and \( |a_n - L| < \frac{\epsilon}{2} \).
05

Combine the Inequalities

Substituting these inequalities back into the triangle inequality, we get: \( |a_m - a_n| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \) So, for all \( m, n \geq N \), \( |a_m - a_n| < \epsilon \).
06

Conclude the Proof

Since for any \( \epsilon > 0 \), we have found a natural number N such that for all \( m, n \geq N \), \( |a_m - a_n| < \epsilon \), we can conclude that the convergent sequence \( (a_n) \) is also a Cauchy sequence. QED.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cauchy Sequences
Understanding the core concept of Cauchy sequences is important, as it provides insights into the fundamental properties of sequences in mathematical analysis. A Cauchy sequence is one where, as you progress further into the sequence, the terms become closer and closer to each other. This means:
  • For any small positive number \( \epsilon \), no matter how small, you can find a point in the sequence beyond which all terms are within distance \( \epsilon \) of each other.
  • This implies an internal consistency within the sequence, allowing it to "stay together" as it progresses.
This behavior is pivotal because it sets groundwork for understanding convergence in more abstract terms, where the actual limit isn't important, but rather the behavior of the sequence itself. The definition of a Cauchy sequence focuses on this closeness of succession, without the need to reference an external limit.
Sequence Convergence
Sequence convergence refers to the tendency of a sequence to approach a specific value, called the limit, as the sequence progresses. This concept can be visualized as follows:
  • You have a sequence \( (a_n) \) and a single value \( L \).
  • As \( n \) (the position in the sequence) becomes large, the terms \( a_n \) get arbitrarily close to \( L \).
  • Formally, for any \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n \geq N \), \( |a_n - L| < \epsilon \).
This means that beyond a certain point, no term in the sequence is farther from \( L \) than the small number \( \epsilon \). Convergent sequences are foundational in calculus and analysis, providing a way to ensure that a sequence "settles down" to a particular value, enabling us to use this value in further analysis and calculations.
Triangle Inequality
The triangle inequality is a crucial tool in proving the relationship between convergent sequences and Cauchy sequences. This inequality states that for any real numbers or vectors:
  • The sum of the lengths of any two sides of a triangle is greater than or equal to the length of the third side.
  • In sequence terms, for any sequence elements like \( a_m \), \( a_n \), and a common limit \( L \), it implies:
\[|a_m - a_n| \leq |a_m - L| + |a_n - L|\]This form of the triangle inequality helps in showing that if each term of the sequence is close to a limit \( L \), they are also close to each other. When applied to convergent sequences:
  • If \( a_n \) is convergent, \( \exists N \) such that \( |a_n - L| < \frac{\epsilon}{2} \) for all \( n \geq N \).
  • By triangle inequality, both terms \( |a_m - L| \) and \( |a_n - L| \) being less than \( \frac{\epsilon}{2} \) implies \( |a_m - a_n| < \epsilon \), satisfying the Cauchy condition.
Therefore, the triangle inequality serves as a bridge, connecting the ideas of convergence to the property of being a Cauchy sequence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a) Show that if \(\boldsymbol{x}=\left(x_{n}\right) \in \ell^{p}\), then \(\boldsymbol{x} \in \ell^{q}\) for all \(q>p\). b) Find a sequence \(\boldsymbol{x}=\left(x_{n}\right)\) that is in \(\ell^{p}\) for \(p>1\), but is not in \(\ell^{1}\).

Let \(\mathcal{P}\) denote the metric space of all polynomials over \(\mathbb{C}\), with metric $$ d(p, q)=\sup _{x \in[a, b]}|p(x)-q(x)| $$ Is \(\mathcal{P}\) complete?

Show that the intersection of any collection of closed sets in a metric space is closed.

Let \((M, d)\) be a metric space. The diameter of a nonempty subset \(S \subseteq M\) is $$ \delta(S)=\sup _{x, y \in S} d(x, y) $$ A set \(S\) is bounded if \(\delta(S)<\infty\). a) Prove that \(S\) is bounded if and only if there is some \(x \in M\) and \(r \in \mathbb{R}\) for which \(S \subseteq B(x, r)\). b) Prove that \(\delta(S)=0\) if and only if \(S\) consists of a single point. c) Prove that \(S \subseteq T\) implies \(\delta(S) \leq \delta(T)\). d) If \(S\) and \(T\) are bounded, show that \(S \cup T\) is also bounded.

Let \(S \subseteq \ell^{\infty}\) be the subspace of all binary sequences (sequences of 0 's and 1 's). Describe the metric on \(S\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.