/*! 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 14 Write out the first five terms o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 14

Write out the first five terms of the sequence, determine whether the sequence converges, and if so find its limit. $$ \left\\{\frac{(-1)^{n+1}}{n^{2}}\right\\}_{n=1}^{+\infty} $$

Short Answer

Expert verified
The sequence converges and its limit is 0.

Step by step solution

01

Identify the Sequence Formula

The given sequence is \( a_n = \frac{(-1)^{n+1}}{n^2} \). This formula will be used to calculate the first five terms by substituting different values of \( n \).
02

Calculate the First Term

Substitute \( n = 1 \) into the sequence formula: \( a_1 = \frac{(-1)^{1+1}}{1^2} = \frac{1}{1} = 1 \).
03

Calculate the Second Term

Substitute \( n = 2 \) into the formula: \( a_2 = \frac{(-1)^{2+1}}{2^2} = \frac{-1}{4} = -\frac{1}{4} \).
04

Calculate the Third Term

Substitute \( n = 3 \) into the formula: \( a_3 = \frac{(-1)^{3+1}}{3^2} = \frac{1}{9} \).
05

Calculate the Fourth Term

Substitute \( n = 4 \) into the formula: \( a_4 = \frac{(-1)^{4+1}}{4^2} = \frac{-1}{16} \).
06

Calculate the Fifth Term

Substitute \( n = 5 \) into the formula: \( a_5 = \frac{(-1)^{5+1}}{5^2} = \frac{1}{25} \).
07

List the First Five Terms

The first five terms of the sequence are: \( 1, -\frac{1}{4}, \frac{1}{9}, -\frac{1}{16}, \frac{1}{25} \).
08

Analyze Convergence

As \( n \to \infty \), the terms of the sequence are \( \frac{(-1)^{n+1}}{n^2} \). Since \( \frac{1}{n^2} \to 0 \) as \( n \to \infty \) and \((-1)^{n+1}\) results in alternating positive and negative signs, the absolute value of terms approaches zero, thus the sequence converges.
09

Determine the Limit

The limit of the sequence is 0 because \( \lim_{{n \to \infty}} \frac{1}{n^2} = 0 \) and the alternating sign does not affect the magnitude of convergence to zero.

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.

Alternating Series
An alternating series is a sequence whose terms switch signs between positive and negative as the sequence progresses. In our example, the sequence given by \( a_n = \frac{(-1)^{n+1}}{n^2} \) is a classic alternating series. Here, the factor \((-1)^{n+1}\) dictates the alternating signs:
  • When \( n \) is odd, \((-1)^{n+1}\) will be positive, making the term positive.
  • When \( n \) is even, \((-1)^{n+1}\) will be negative, making the term negative.
This alternating behavior is essential because it can affect convergence properties, often helping the series converge when terms decrease in absolute magnitude. Alternate series like this are crucial in mathematical fields and applications due to their unique convergence properties.
Limit of a Sequence
The concept of the limit of a sequence is fundamental in calculus and analysis. It describes the value that the terms in a sequence approach as the index \( n \) becomes very large. For the sequence \( a_n = \frac{(-1)^{n+1}}{n^2} \), to find its limit, we focus on the magnitude \( \frac{1}{n^2} \). As \( n \) approaches infinity, this term approaches zero:
  • \( \frac{1}{n^2} \to 0 \) as \( n \to \infty \)
  • The factor \((-1)^{n+1}\) only affects the sign and not the size of the terms.
Therefore, even though the terms of the sequence alternate in sign, they get closer to zero. Thus, the limit of this sequence is \( 0 \). Understanding limits is essential as it is the base for many more advanced mathematical concepts like continuity and differentiability.
Sequence Analysis
Sequence analysis involves examining the properties and behavior of a sequence. In this scenario, we analyzed our sequence \( a_n = \frac{(-1)^{n+1}}{n^2} \) by:
  • Calculating the initial terms (e.g., first five: 1, \( -\frac{1}{4} \), \( \frac{1}{9} \), \( -\frac{1}{16} \), \( \frac{1}{25} \)).
  • Describing its behavior as \( n \to \infty \).
  • Determining its convergence to a limit.
Through this process, we confirmed that the sequence is both alternating and convergent, and found its limit to be \( 0 \). Such analysis allows us to better understand mathematical behavior and predict outcomes, making it a crucial tool in any mathematician's toolkit. This thorough examination helps provide deeper insights and understanding beyond simple computation.

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

Determine whether the series converges. $$ \sum_{k=1}^{\infty} \frac{1}{\sqrt[k]{e}} $$

Use Maclaurin series to approximate the integral to three decimal-place accuracy. $$\int_{0}^{1 / 2} \tan ^{-1}\left(2 x^{2}\right) d x$$

(a) Show that the hypotheses of the integral test are satisfied by the series \(\sum_{k=1}^{\infty} 1 /\left(k^{3}+1\right)\) (b) Use a CAS and the integral test to confirm that the series converges. (c) Construct a table of partial sums for \(n=10,20,\) \(30, \ldots, 100,\) showing at least six decimal places. (d) Based on your table, make a conjecture about the sum of the series to three decimal-place accuracy. (e) Use part (b) of Exercise 36 to check your conjecture.

Let \(\sum u_{k}\) be a series and define series \(\sum p_{k}\) and \(\sum q_{k}\) so that $$p_{k}=\left\\{\begin{array}{ll}{u_{k},} & {u_{k}>0} \\ {0,} & {u_{k} \leq 0}\end{array} \quad \text { and } \quad q_{k}=\left\\{\begin{aligned} 0, & u_{k} \geq 0 \\\\-u_{k}, & u_{k}<0 \end{aligned}\right.\right.$$ (a) Show that \(\sum u_{k}\) converges absolutely if and only if \(\sum p_{k}\) and \(\sum q_{k}\) both converge. (b) Show that if one of \(\sum p_{k}\) or \(\sum q_{k}\) converges and theother diverges, then \(\sum u_{k}\) diverges. (c) Show that if \(\sum u_{k}\) converges conditionally, then both $$\sum p_{k} \text { and } \sum q_{k} \text { diverge. }$$

Determine whether the series converges. $$ \sum_{k=1}^{\infty} \operatorname{sech}^{2} k $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Applied 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.