/*! 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 43 Write the first five terms of th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 43

Write the first five terms of the sequence \(\left\\{a_{n}\right\\}\), \(n=0,1,2,3, \ldots\), and find \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{(-1)^{n}}{n+1} $$

Short Answer

Expert verified
The first five terms are 1, -1/2, 1/3, -1/4, 1/5 and the limit is 0.

Step by step solution

01

Calculate the First Term

To find the first term of the sequence, substitute \(n = 0\) into the formula. \[a_0 = \frac{(-1)^0}{0+1} = \frac{1}{1} = 1.\]
02

Calculate the Second Term

Substitute \(n = 1\) into the formula to find the second term. \[a_1 = \frac{(-1)^1}{1+1} = \frac{-1}{2}.\]
03

Calculate the Third Term

For the third term, substitute \(n = 2\). \[a_2 = \frac{(-1)^2}{2+1} = \frac{1}{3}.\]
04

Calculate the Fourth Term

To find the fourth term, substitute \(n = 3\). \[a_3 = \frac{(-1)^3}{3+1} = \frac{-1}{4}.\]
05

Calculate the Fifth Term

Finally, for the fifth term, substitute \(n = 4\). \[a_4 = \frac{(-1)^4}{4+1} = \frac{1}{5}.\]
06

Determine the Limit

To find the limit of the sequence as \(n\) approaches infinity, consider the expression \(\frac{(-1)^n}{n+1}\). As \(n\) increases, the denominator \(n+1\) becomes very large, so the fraction \(\frac{1}{n+1}\) approaches 0. Thus, \(\lim_{n \to \infty} a_n = 0.\)

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.

Sequence Terms Calculation
Calculating sequence terms involves using a given formula to find each term by substituting values for the sequence index, often labeled as \( n \). In the exercise provided, the sequence is defined by the formula \( a_n = \frac{(-1)^n}{n+1} \). To find each term:
  • Start by substituting \( n = 0 \) to find the first term: \( a_0 = \frac{1}{1} = 1 \).

  • Substitute \( n = 1 \) for the second term: \( a_1 = \frac{-1}{2} \).

  • Continue this process for \( n = 2,\;3, \) and \( 4 \) to find: \( a_2 = \frac{1}{3}, \; a_3 = \frac{-1}{4}, \) and \( a_4 = \frac{1}{5} \).
This iterative method provides a systematic approach to determine each sequence term. Calculating these terms visibly shows how the alternating nature and decreasing magnitude play out in the sequence.
Infinite Limits
Determining the limit of a sequence as \( n \) approaches infinity is crucial in understanding the sequence's behavior in the long run. For the given sequence \( a_n = \frac{(-1)^n}{n+1} \), observe how the fraction behaves as \( n \) increases indefinitely.
  • The term \( (-1)^n \) makes the sequence alternate between positive and negative.

  • The denominator \( n+1 \) grows ever larger, causing the fraction's absolute value \( \frac{1}{n+1} \) to get closer to 0.
Despite the sequence's oscillations due to the \((-1)^n\) factor, the primary influence on the sequence's limit is the denominator. As \( n \) increases without bound, \( \frac{1}{n+1} \) approaches 0; thus, the overall limit is \( \lim_{n \to \infty} a_n = 0 \). Understanding this infinite limit helps us conclude the sequence trends toward 0 despite its alternation in signs.
Alternating Sequences
Sequences that switch between positive and negative terms are known as alternating sequences. These sequences have specific characteristics that influence their pattern and overall limit behavior.
  • The term \((-1)^n\) is essential in creating the alternating pattern. For even \( n \), \((-1)^n = 1 \), and for odd \( n \), \((-1)^n = -1 \).

  • In the sequence \( a_n = \frac{(-1)^n}{n+1} \), this factor means the sequence flips signs with each consecutive term.
Alternating sequences are interesting because their oscillations between positive and negative values can converge to a specific limit, as seen in this case where the sequence approaches 0. These sequences provide good practice in understanding both their pattern through direct calculation and their longer-term behavior through limit evaluation.

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

Formal Definition of Limits: Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists an \(N\) such that \(\left|a_{n}-a\right|<\epsilon\) whenever \(n>N\). $$ \lim _{n \rightarrow \infty} \frac{1}{n^{2}+1}=0 $$

In Problems 103-110, assume that \(\lim _{n \rightarrow \infty} a_{n}\) exists. Find all fixed ooints of \(\left\\{a_{n}\right\\}\), and use a table or other reasoning to guess which ixed point is the limiting value for the given initial condition. $$ a_{n+1}=\frac{1}{2}\left(a_{n}+5\right), a_{0}=1 $$

Investigate the advantage of dimensionless variables. The time to the most recent common ancestor of a pair of individuals from a randomly mating population depends on the population size. Let \(t\) denote the time, measured in units of generations, to the most recent common ancestor, and let \(T\) be equal to \(N\) generations, where \(N\) is the population size of the randomly mating population. Define \(z=t / T .\) Show that \(z\) is dimensionless and that the value of \(z\) does not change, regardless of whether \(t\) and \(T\) are measured in units of generations or in units of, say, years. (Assume that one generation is equal to \(n\) years.)

Investigate the advantage of dimensionless variables. Suppose a bacterium divides every 20 minutes, which we call the characteristic time scale and denote by \(T\). Let \(t\) be the time elapsed since the beginning of an experiment that involves this bacterium. Define \(z=t / T .\) Find \(z\) if \(t=120\) minutes, and show that \(z\) has the same value if \(t\) and \(T\) are measured in units of hours instead.

Investigate the behavior of the discrete logistic equation $$ x_{t+1}=r x_{t}\left(1-x_{t}\right) $$ Compute \(x_{t}\) for \(t=0,1,2, \ldots, 20\) for the given values of \(r\) and \(x_{0}\), and graph \(x_{t}\) as a function of \(t\). \(r=3.1, x_{0}=0.9\)
See all solutions

Recommended explanations on Biology Textbooks

Microbiology

Read Explanation

Biological Processes

Read Explanation

Biological Organisms

Read Explanation

Ecology

Read Explanation

Plant Biology

Read Explanation

Genetic Information

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.