/*! 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 34 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$a_{n}=\frac{1-n^{3}}{70-4 n^{2}}$$

Short Answer

Expert verified
The sequence \( a_n \) diverges.

Step by step solution

01

Analyze the Sequence

The given sequence is \( a_n = \frac{1-n^3}{70-4n^2} \). To determine if it converges or diverges, observe the degrees of the terms in the numerator and the denominator. The dominant term in the numerator is \( -n^3 \), and the dominant term in the denominator is \( -4n^2 \). Since the numerator and the denominator have different degrees, we will look at the highest degree terms to understand the behavior as \( n \to \infty \).
02

Simplify the Dominant Terms

Focus on the dominant terms of the numerator and the denominator: \( a_n \approx \frac{-n^3}{-4n^2} = \frac{-1}{-4}n = \frac{n}{4} \). As \( n \) increases, the term \( \frac{1}{4}n \) increases without bound.
03

Determine Convergence or Divergence

Since \( \frac{n}{4} \to \infty \) as \( n \to \infty \), the sequence \( a_n \) diverges. A sequence must converge to a finite number, but here it does not; it instead tends to infinity.

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 91Ó°ÊÓ!

Key Concepts

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

Limits of Sequences
To understand the behavior of sequences, it's crucial to comprehend the concept of limits. The limit of a sequence describes the value that the terms of the sequence approach as the index, usually denoted as \( n \), goes to infinity. If a sequence approaches a particular finite number, we say that it converges to that limit. The mathematical notation for the limit of the sequence \( \{a_n\} \) converging to \( L \) is \( \lim_{{n \to \infty}} a_n = L \).

For example, consider a sequence like \( a_n = \frac{1}{n} \). As \( n \to \infty \), \( a_n \) approaches 0. Thus, we say this sequence converges to 0. When studying sequences, identifying the dominant term helps determine whether it converges, especially for complex sequences like polynomials.

However, if the terms of the sequence grow larger in magnitude without approaching any fixed number, we say the sequence diverges. This understanding helps to recognize divergent behaviors early by simplifying terms to their respective dominant components.
Divergent Sequences
A divergent sequence is one that does not settle down to any particular value as \( n \) increases. Essentially, it's a sequence that doesn't have a finite limit. When analyzing a sequence, one key indicator of divergence is when the terms grow infinitely large in magnitude. For example, if terms tend towards infinity or negative infinity, the sequence is considered divergent.

Looking at our example, \( a_n = \frac{1-n^3}{70-4n^2} \), we identify the dominant terms \(-n^3\) in the numerator and \(-4n^2\) in the denominator. Simplifying yields \( \frac{-1}{-4}n = \frac{n}{4} \), which grows without bound as \( n \to \infty \). Thus, the sequence diverges to infinity.

Practicing how to identify divergent sequences using degree analysis of terms can greatly aid in quickly determining if a sequence converges or diverges, especially when terms have polynomial components like in our example.
Sequence Analysis
Analyzing sequences involves breaking them down to their core components to determine behavior over time. This particular concept can help in both academic exercises and real-life scenarios where understanding growth trends or settlement values is required.

In sequence analysis, terms with the highest degree, whether polynomial or exponential, often dominate the behavior as \( n \to \infty \). This is crucial when performing sequence analysis on complex sequences. It involves identifying the dominant terms and evaluating their impact on the sequence's convergence or divergence.

For our sequence \( a_n = \frac{1-n^3}{70-4n^2} \), the analysis involves comparing \(-n^3\) against \(-4n^2\). When simplified, the sequence behaves like \( \frac{n}{4} \), indicating divergence. Such systematic evaluation aids in drawing conclusions regarding the limits or divergent nature of sequences.
  • Step 1: Identify and simplify the dominant terms from both the numerator and the denominator.
  • Step 2: Compare the resulting simplified form.
  • Step 3: Determine whether the sequence approaches a finite value (convergence) or grows towards infinity (divergence).
Using these steps, sequence analysis becomes a structured process.

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

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=\sqrt[n]{n}$$

Find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x\) ) for those values of \(x .\) $$\sum_{n=0}^{\infty} 2^{n} x^{n}$$

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

In each of the geometric series, write out the first few terms of the series to find \(a\) and \(r\), and find the sum of the series. Then express the inequality \(|r|<1\) in terms of \(x\) and find the values of \(x\) for which the inequality holds and the series converges. $$\sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n}$$

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\).

See all solutions

Recommended explanations on Math Textbooks

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.