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Chapter 1: Problem 22

Prove in detail that a convergent sequence is bounded.

Short Answer

Expert verified
A convergent sequence is bounded by its definition of convergence and finite terms.

Step by step solution

01

Understanding the Definition of Convergence

A sequence \((a_n)\) is said to be convergent if there exists a limit \(L\) such that for every \(\epsilon > 0\), there exists a positive integer \(N\) where for all \(n \geq N\), \(|a_n - L| < \epsilon\). This means the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes large.
02

Defining Bounded Sequence

A sequence \((a_n)\) is bounded if there exists a real number \(M\) such that \(|a_n| \leq M\) for all \(n\). In simpler terms, the terms of the sequence do not exceed this magnitude.
03

Developing an Approach

To prove that a convergent sequence is bounded, we need to demonstrate the existence of such an \(M\) for a convergent sequence \((a_n)\). We will use the definition of convergence to construct our argument.
04

Choosing an Epsilon Value

Let \(\epsilon = 1\), based on the definition of convergence. Since \((a_n)\) converges to \(L\), there exists \(N\) such that for all \(n \geq N\), \(|a_n - L| < 1\).
05

Bounding the Terms Beyond N

For \(n \geq N\), we have \(|a_n| - |L| \leq |a_n - L| < 1\). Therefore, \(|a_n| < |L| + 1\).
06

Handling Terms Before N

The remaining terms of the sequence up to \(a_N\) (i.e., \(a_1, a_2, \ldots, a_{N-1}\)) are finite. Let \(K\) be the maximum of \(|a_1|, |a_2|, \ldots, |a_{N-1}|\), and define \(M = \max(K, |L| + 1)\).
07

Conclusion of Boundedness

Since \(|a_n| \leq M\) for all \(n\), as either \(n \geq N\) (where \(|a_n| < |L| + 1 \leq M\)) or \(n < N\) (where \(|a_n| \leq K \leq M\)), the sequence \((a_n)\) is bounded. This completes our proof.

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Key Concepts

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

Bounded Sequence
A bounded sequence is a sequence of numbers where all its terms remain within a certain fixed range. This means that there exists a real number, often denoted as \(M\), such that the absolute value of every term in the sequence is less than or equal to \(M\).

For example, the sequence \((a_n)\) is bounded if for all natural numbers \(n\), \(|a_n| \leq M\). Intuitively, think of a bounded sequence as being "trapped" within a fixed band on the number line. No matter how many terms you compare, none will exceed this limit.
  • The property of being bounded helps in guaranteeing that the sequence is manageable and doesn't "run off" to infinity or negative infinity.
  • This concept is crucial in many areas of analysis, providing a foundation for more complex theorems.
Recognizing that a convergent sequence must also be bounded is an important insight in understanding sequence behavior.
Limit of a Sequence
The limit of a sequence is the value that the terms of the sequence approach as the index (usually \(n\)) goes to infinity. If a sequence \((a_n)\) has a limit \(L\), we say that the sequence converges to \(L\). This is mathematically represented as \( \lim_{n \to \infty} a_n = L \).

When approaching the concept of a limit, it's important to understand these key points:
  • The sequence \(a_n\) gets "arbitrarily close" to \(L\) as \(n\) increases. This means that no matter how small a difference \(\epsilon\) you choose, you can find a point in the sequence after which all terms are within \(\epsilon\) of \(L\).
  • The concept of limit is a backbone in calculus and analysis, offering a way to precisely discuss the behavior of sequences as their terms grow large.
  • While limits provide insight into the behavior at infinity, they also allow us to infer properties like boundedness across the entire sequence.
Understanding limits is crucial for moving on to more advanced topics, particularly in analyses involving continuity, integrals, and series.
Mathematical Proof
In mathematics, a proof is a logical argument that establishes the truth of a given statement. Proofs are the foundation of mathematics, ensuring that our conclusions are based on firm reasoning.

When proving that a convergent sequence is bounded, one must employ a systematic approach:
  • Identify the assumptions: For convergent sequences, the assumption is that it approaches some limit \(L\) as described.
  • Apply definitions: Make use of formal definitions like those of convergence and boundedness to construct the argument.
  • Found arguments on these properties: Use precise logical steps that stem from these definitions to arrive at your conclusion.
Proofs often involve a combination of direct calculation and logical deduction, reflecting the rigorous nature of mathematical thinking. Eleshing a proof not only confirms a statement's truth but often provides a deeper understanding of why the statement holds.
Epsilon-Delta Definition
The epsilon-delta definition is a formal way to define the concept of a limit within calculus, specifically with sequences and functions. It provides a precise framework to characterize how the terms of a sequence come closer to a limit.

For a sequence \((a_n)\) to converge to a limit \(L\), the epsilon-delta definition states:
  • For any small positive number \(\epsilon\), however tiny, there exists a corresponding integer \(N\).
  • For all integers \(n\) that are greater than or equal to \(N\), the terms of the sequence satisfy \(|a_n - L| < \epsilon\).

This definition may feel abstract at first, but it enforces a precise understanding of "closeness" and "approaching a limit." By relying on this foundational framework:
  • We develop a rigorous approach to demonstrate the boundedness of convergent sequences.
  • It clarifies how bounds are maintained even as we examine sequences extending infinitely.
Grasping this definition is imperative to navigating advanced mathematical concepts that build upon it.

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Most popular questions from this chapter

Prove: If \(x_{n}=\mathcal{O}\left(\alpha_{n}\right)\) and \(y_{n}=\mathcal{O}\left(\alpha_{n}\right)\), then \(x_{n}+y_{n}=\mathcal{O}\left(\alpha_{n}\right)\).

Find the Taylor series for \(f(x)=e^{x}\) about the point \(c=3\). Then simplify the series and show how it could have been obtained directly from the series for \(f\) about \(c=0 .\)

Determine whether the following function is continuous, once differentiable, or twice differentiable: $$ f(x)=\left\\{\begin{array}{ll} x^{3}+x-1 & \text { if } x \leq 0 \\ x^{3}-x-1 & \text { if } x>0 \end{array}\right. $$

For the pair \(\left(x_{n}, \alpha_{n}\right)\), is it true that \(x_{n}=\mathcal{O}\left(\alpha_{n}\right)\) as \(n \rightarrow \infty\) ? a. \(x_{n}=5 n^{2}+9 n^{3}+1, \quad \alpha_{n}=n^{2}\) b. \(x_{n}=5 n^{2}+9 n^{3}+1, \quad \alpha_{n}=1\) c. \(x_{n}=\sqrt{n+3}, \quad \alpha_{n}=1\) d. \(x_{n}=5 n^{2}+9 n^{3}+1, \quad \alpha_{n}=n^{3}\) e. \(x_{n}=\sqrt{n+3}, \quad \alpha_{n}=1 / n\)

Criticize this reasoning: The function \(f\) defined by $$ f(x)=\left\\{\begin{array}{ll} x^{3}+x & \text { if } x \leq 0 \\ x^{3}-x & \text { if } x \geq 0 \end{array}\right. $$ has the properties $$ \begin{array}{l} \lim _{x \rightarrow 0^{+}} f^{\prime \prime}(x)=\lim _{x \rightarrow 0^{+}} 6 x=0 \\ \lim _{x \rightarrow 0^{-}} f^{\prime \prime}(x)=\lim _{x \rightarrow 0^{-}} 6 x=0 \end{array} $$ Therefore, \(f^{\prime \prime}\) is continuous.
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