/*! 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 36 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. $$a_{n}=(-1)^{n}\left(1-\frac{1}{n}\right)$$

Short Answer

Expert verified
The sequence \(a_n = (-1)^n \left(1 - \frac{1}{n}\right)\) diverges.

Step by step solution

01

Understand the sequence

The sequence given is \(a_n = (-1)^{n} \left(1 - \frac{1}{n}\right)\). This is an alternating sequence because of the factor \((-1)^n\), which causes the signs of the terms to alternate between positive and negative.
02

Analyze the behavior of the terms

Examine the term \(1 - \frac{1}{n}\). As \(n\) becomes very large, the fraction \(\frac{1}{n}\) approaches zero, thus \(1 - \frac{1}{n}\) approaches 1. However, the factor \((-1)^n\) will continue to cause the terms to alternate signs.
03

Determine if the sequence converges or diverges

A sequence converges if its terms approach a single finite number as \(n\) approaches infinity. In this sequence, as \(n\) becomes larger, while the magnitude of \(1 - \frac{1}{n}\) approaches 1, the sign alternates continuously due to \((-1)^n\). As such, the sequence does not approach a single value. Thus, the sequence diverges.

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.

Alternating Sequences
An alternating sequence is a type of sequence in which consecutive terms switch between positive and negative values. This alternation is often a result of the term \((-1)^n\) or \((-1)^{n+1}\). In our given sequence \(a_n = (-1)^n \left(1-\frac{1}{n}\right)\), the presence of \((-1)^n\) means that each consecutive term has the opposite sign of the previous term.
This pattern can create quite an interesting behavior because, even as the magnitude of each term becomes very predictable, the sign keeps changing.

For example, consider the terms if \(n = 1, 2, 3,\)... you'll have \(a_1 = \left(1-1\right) = 0 \), \(a_2 = -\left(1-\frac{1}{2}\right) = -0.5 \), \(a_3 = \left(1-\frac{1}{3}\right) = 0.666...\)
Notice how the terms keep alternating between positive and negative as the sequence progresses.

Alternating sequences are important because they often require different methods to analyze their convergence. While non-alternating sequences may smoothly approach a limit, alternating sequences can appear scattered due to their sign changes, complicating observation of any limiting behavior.
Limits of Sequences
The concept of the limit of a sequence is foundational in understanding how sequences behave as they progress. A sequence converges to a limit if as \(n\) becomes very large, the terms of the sequence \(a_n\) get arbitrarily close to some fixed number, known as the limit.
To determine a sequence's limit, we examine the behavior of its terms as \(n\) tends to infinity. In typical cases, simplifying the sequence to find a predictable pattern or expression as \(n\) increases is key.

For instance, looking at the sequence \(a_n = \left(1 - \frac{1}{n}\right)\), as \(n\) goes to infinity, \frac{1}{n}\ tends toward zero, making the sequence behave like \(1 - 0 = 1\). However, due to the alternating nature of the sequence, \(-1)^n\) is present, flipping the sign of the terms and complicating convergence.

Even when the magnitude appears to narrow in on a single value, if the sign is inconsistent, like in alternating sequences, reaching a single number or limit may not be possible.The careful observation and understanding of limits allow mathematicians to determine the bounding behaviors of sequences.
Divergent Sequences
A sequence diverges when its terms do not approach any single, finite number as \(n\) becomes increasingly large. Instead, the terms might grow without bound, oscillate, or otherwise fail to hone in on a particular value.
A classic hallmark of a divergent sequence is a pattern that lacks directionality when considering the entire set of terms over time.

Take our sequence \(a_n = (-1)^n (1-\frac{1}{n})\). While the magnitude of the terms \(1-\frac{1}{n}\) moves towards \(1\), the alternating factor \((-1)^n\) prevents the sequence from settling into a single value.
Therefore, while the terms have a predictable size, their continually swapping signs means no single finite value is approached as \(n\) increases.

Recognizing that such a sequence does not converge helps categorize it correctly and determines how it might be treated mathematically or in applications. Divergence in sequences is quite common when encountering complex or compound expressions, especially those involving periodic sign changes.

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

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. a. \(x_{0}=1, \quad x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}\) b. \(x_{0}=1, \quad x_{n+1}=x_{n}-\frac{\tan x_{n}-1}{\sec ^{2} x_{n}}\) c. \(x_{0}=1, \quad x_{n+1}=x_{n}-1\)

Show that \(\Sigma_{n=1}^{\infty} 2^{(n-1)} / n !\) diverges. Recall from the Laws of Exponents that \(2^{(n)}=\left(2^{n}\right)^{n}\)

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. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

Which of the sequences converge, and which diverge? Give reasons for your answers. $$a_{n}=\frac{n+1}{n}$$

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.