/*! 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 97 In Exercises \(97-100,\) determi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 97

In Exercises \(97-100,\) determine if the sequence is nondecreasing and if it is bounded from above. $$ a_{n}=\frac{3 n+1}{n+1} $$

Short Answer

Expert verified
The sequence is nondecreasing and bounded from above by 3.

Step by step solution

01

Write the sequence formula

The sequence given is \(a_n = \frac{3n + 1}{n + 1}\). Our task is to determine if this sequence is nondecreasing and if it is bounded from above.
02

Check if the sequence is nondecreasing

To check if a sequence \(a_n\) is nondecreasing, we must ensure that \(a_{n+1} \geq a_n\) for all \(n\). Calculate \(a_{n+1}\): \[a_{n+1} = \frac{3(n+1) + 1}{(n+1) + 1} = \frac{3n + 4}{n + 2}.\] We need to compare \(a_{n+1}\) and \(a_n\): \[\frac{3n+4}{n+2} \geq \frac{3n+1}{n+1}.\] Clear the fractions by cross-multiplying: \[(3n+4)(n+1) \geq (3n+1)(n+2).\] Expanding both sides gives: \[3n^2 + 7n + 4 \geq 3n^2 + 7n + 2.\] Simplifying gives: \[4 \geq 2,\] which is true, confirming the sequence is nondecreasing.
03

Determine if the sequence is bounded from above

To check if the sequence \(a_n\) is bounded from above, find the limit as \(n\) approaches infinity. The degree of the numerator and denominator both being 1, examine \(\lim_{n \to \infty} \frac{3n + 1}{n + 1} = \frac{3}{1} = 3\). This shows the sequence approaches 3, thereby bounding the sequence above by 3.
04

Conclusion

The sequence \(a_n = \frac{3n+1}{n+1}\) is nondecreasing, as shown by \(a_{n+1} \geq a_n\). Additionally, it is bounded from above by 3, as the limit as \(n\) approaches infinity shows the series tending towards 3.

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.

Nondecreasing Sequences
A nondecreasing sequence is one where each term is not less than the previous term. In mathematical terms, this means for any term \( a_n \), the next term \( a_{n+1} \) is either equal to or greater than \( a_n \). For the sequence \( a_n = \frac{3n + 1}{n + 1} \), we confirmed that it is indeed nondecreasing.
To determine this, we compared \( a_{n+1} \) and \( a_n \). After simplification, it showed:
  • The inequality \( a_{n+1} \geq a_n \).
  • This holds for all \( n \), meaning the entire sequence maintains the nondecreasing condition.

The inclusion of terms like \( 3n + 4 \) and \( n + 2 \) should not confuse the essential understanding. It's all about confirming one term doesn鈥檛 dip below the one before it.
Bounded Sequences
A sequence is termed 'bounded from above' if there exists a number such that no term in the sequence exceeds that number. For the sequence \( a_n = \frac{3n + 1}{n + 1} \), we sought to find if there was an upper boundary. The crucial method here is analyzing the behavior of the sequence as \( n \) becomes very large.

What's fascinating about this sequence is that both its numerator and denominator tend towards infinity, yet at the same rate. This relationship culminates in:
  • The limit as \( n \to \infty \) is \( \lim_{n \to \infty} \frac{3n + 1}{n + 1} = 3 \).
  • Which means the sequence settles around and does not exceed 3 after sufficiently large \( n \).
Thus, the sequence is bounded from above at 3, framing its growth within observed limits.
Limits of Sequences
Understanding the concept of limits in sequences is key in analyzing their long-term behavior. Limits tell us the value that the terms of a sequence approach as \( n \) becomes very large. For the sequence \( a_n = \frac{3n + 1}{n + 1} \), we determined that its limit, as \( n \to \infty \), is 3.

In sequences, the approach to the limit is vital because:
  • A sequence approaching a finite limit \( L \) implies stability over time.
  • The elements of the sequence become arbitrarily close to \( L \) as \( n \) increases.
By determining limits, we gain profound insights into whether sequences converge (approach a finite number) or diverge (grow without bounds). In our example, the sequence converges to 3, giving a neat picture of its "end behavior."

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

Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x) .\) Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) , $$ \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) $$

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1\)

Newton's method The following sequences come from the recursion formula for Newton's method, $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ Do the sequences converge? If so, to what value? In each case, begin by identifying the function \(f\) that generates the sequence. $$ \begin{array}{l}{\text { a. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}} \\\ {\text { b. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{\tan x_{n}-1}{\sec ^{2} x_{n}}} \\ {\text { c. } x_{0}=1, \quad x_{n+1}=x_{n}-1}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

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.