/*! 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 69 Is it possible for a sequence to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 69

Is it possible for a sequence to converge to two different numbers? If so, give an example. If not, explain why not.

Short Answer

Expert verified
No, it is not possible for a sequence to converge to two different numbers. A sequence can only have one unique limit to which it converges.

Step by step solution

01

Concept of Converging Sequences

A sequence converges if it approaches a particular point, known as a limit, as the term number becomes infinitesimally large. In mathematical terms, for a sequence \(a_n\), it converges to a limit 'l' if for any \( \epsilon > 0 \), there exists an integer 'N' such that \( | a_n - l | < \epsilon \) for all \( n > N \). The limit 'l' denotes the number to which the sequence converges.
02

Addressing multiple limits

Assume it's possible that a sequence could converge to two different limits, say, 'L' and 'M', where \( L \neq M \). By the definition of limit, for a small positive number \( \epsilon \) , for \( n > N1 \), \( | a_n - L | = | a_n - M | < \epsilon \) is simultaneously true. This eventually leads to \( | L - M | = 0 \) which contradicts the original assumption that \( L \neq M \). Hence establishes that a sequence cannot converge to two different limits.
03

Conclusion

Therefore, by contradiction, it has been shown that a sequence cannot have more than one limit. So a sequence cannot converge to two different numbers.

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!

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

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{1} \cos x^{2} d x $$

Finding a Limit In Exercises \(59-62,\) use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{\sin x}{x} $$

Area In Exercises 71 and \(72,\) use a power series to approximate the area of the region. Use a graphing utility to verify the result. $$ \int_{0.5}^{1} \cos \sqrt{x} d x $$

Using a Binomial Series In Exercises \(17-26,\) use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt{1+x^{3}} $$

the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. \(a_{1}=2, a_{n+1}=\frac{2 n+1}{5 n-4} a_{n}\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete 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.