/*! 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 86 (a) Prove that if \(\lim _{n \ri... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 86

(a) Prove that if \(\lim _{n \rightarrow \infty} s_{n}=s\) and \(sN\) then \(s_{n}M\) then \(r

Short Answer

Expert verified
For the first statement, we can choose \(N\) to be that number \(N_1\) for which if \(n > N\) then \(s_{n} < t\). For the second statement, we can pick \(M\) as that number \(M_1\) for which if \(n > M\) then \(r < s_{n}\). These statements are true due to the formal definition of limits of sequences and the ability to choose a specific, suitable epsilon value for establishing the necessary inequalities.

Step by step solution

01

Proof of first statement

Assume \(\lim _{n \rightarrow \infty} s_{n}=s\) and \(s 0\), there exists a number \(N_1\) such that if \(n > N_1\) then \(|s_{n}-s| < \varepsilon\). Let's choose \(\varepsilon = t - s\) which is greater than 0 because \(s N_1\) then \(|s_{n}-s|< t-s\). This means \(s_{n} < t\). Thus, we can conclude, taking N to be \(N_1\), that if \(n > N\) then \(s_{n} < t\)
02

Proof of second statement

Assume \(\lim _{n \rightarrow \infty} s_{n}=s\) and \(r 0\), there exists a number \(M_1\) such that for \(n>M_1\) then \(|s_{n}-s|<\varepsilon\). Again let's choose \(\varepsilon = s - r\) which is greater than 0 because \(r M_1\) then \(|s_{n}-s|M\), then \(r<s_{n}\).

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.

Limit of a Sequence
Understanding the limit of a sequence is crucial in calculus and real analysis as it's the foundation for defining continuity, derivatives, and integrals. When we say that a sequence \(s_n\) approaches the limit \(s\) as \(n\) tends to infinity, symbolically written as \(\lim _{n \rightarrow \infty} s_{n}=s\), we are describing a situation where the terms of the sequence get arbitrarily close to \(s\) as \(n\) increases.

To give a sense of how this concept applies in your textbook exercise, consider this: when \(\lim _{n \rightarrow \infty} s_{n}=s\) and you know \(s
Real Analysis
Real analysis is a branch of mathematics that deals with real numbers and real-valued functions. This field focuses heavily on the rigorous presentation of the limit concept under various contexts such as sequences, series, and functions. Within real analysis, the limit behavior of sequences is explored in depth using structural rules and restrictions to analyze convergence and continuity.

The textbook exercise highlights a fundamental aspect of real analysis: showing that every convergent sequence stabilizes within any chosen arbitrary margin of error when considering sufficiently large terms. This stability is a cornerstone in many other theorems and proofs within real analysis and offers a methodology to demonstrate bounds and the existence of limits even when they cannot be calculated directly.
Epsilon-N Proof
The epsilon-N proof is a formal method used to prove statements about limits in calculus and analysis. It is characterized by two quantities: \(\varepsilon\) and \(N\). The \(\varepsilon\) (epsilon) represents an arbitrarily small positive value, and \(N\) (N) stands for a threshold index. The idea is that for any chosen \(\varepsilon\), there exists a corresponding \(N\) such that for all indices \(n>N\), the terms of the sequence are within an \(\varepsilon\)-wide band around the limit \(s\).

In the exercise provided, the epsilon-N proof is used twice, once to prove that terms of the sequence \(s_n\) will be less than \(t\) after some point, and again to show they'll be greater than \(r\) beyond another point. This approach is fundamental for establishing the precise meaning of limits and understanding how sequences behave as they approach infinity. It's the go-to method for constructing a rigorous argument to demonstrate the convergence of numerical sequences in both routine homework and complex mathematical theories.
Sequence Inequalities
When dealing with sequences, inequalities are used to compare the terms of a sequence with particular values. These comparisons can set bounds on the behavior of a sequence and are particularly important when considering their limits. In real analysis, establishing sequence inequalities often involves finding a specific index after which the inequality will hold for all subsequent terms.

The textbook exercise asks you to think about inequalities in the context of limits: if you have a limit \(s\) for a sequence and another number \(t\) such that \(s

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

Let \(b\) be a nonzero real number with \(|b|<1\) and let \(\varepsilon>0\). (a) Solve the inequality \(|b|^{n}<\varepsilon\) for \(n\). (b) Use part (a) to prove \(\lim _{n \rightarrow \infty} b^{n}=0\).

Suppose that \(\left(a_{n}\right)_{n=1}^{\infty}\) diverges but not to infinity and that \(\alpha\) is a real number. What conditions on \(\alpha\) will guarantee that: (a) \(\left(\alpha a_{n}\right)_{n=1}^{\infty}\) converges? (b) \(\left(\alpha a_{n}\right)_{n=1}^{\infty}\) diverges?

Let \((c)_{n=1}^{\infty}=(c, c, c, \ldots)\) be a constant sequence. Show that \(\lim _{n \rightarrow \infty} c=c\) In proving the familiar limit theorems, the following will prove to be a very useful tool. Lemma 1. (a) Triangle Inequality Let \(a\) and \(b\) be real numbers. Then $$ |a+b| \leq|a|+|b| $$ (b) Reverse Triangle Inequality Let \(a\) and \(b\) be real numbers. Then $$ |a|-|b| \leq|a-b| $$

Show that if \(|r|>1\) then \(\left(r^{n}\right)_{n=1}^{\infty}\) diverges. Will it diverge to infinity?

(a) Provide a rigorous definition for \(\lim _{n \rightarrow \infty} s_{n} \neq s\). (b) Use your definition to show that for any real number a \(\lim _{n \rightarrow \infty}\left((-1)^{n}\right) \neq\) a.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

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.