/*! 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 53 In Problems \(53-60\), write the... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems \(53-60\), write the first five terms of the sequence \(\left\\{a_{n}\right\\}\) \(n=0,1,2,3, \ldots\), and determine whether \(\lim _{n \rightarrow \infty} a_{n}\) exists. If the limit exists, find it. $$ a_{n}=\frac{n^{2}}{n+1} $$

Short Answer

Expert verified
The sequence terms are 0, \(\frac{1}{2}\), \(\frac{4}{3}\), \(\frac{9}{4}\), and \(\frac{16}{5}\). The limit does not exist as \(n \to \infty\).

Step by step solution

01

Calculate the First Term

To find the first term, substitute \( n = 0 \) into the sequence formula \( a_n \). \[ a_0 = \frac{0^2}{0+1} = 0 \] So, the first term \( a_0 \) is 0.
02

Calculate the Second Term

Next, substitute \( n = 1 \) into the sequence formula. \[ a_1 = \frac{1^2}{1+1} = \frac{1}{2} \] Thus, the second term \( a_1 \) is \( \frac{1}{2} \).
03

Calculate the Third Term

Substitute \( n = 2 \) into the sequence formula. \[ a_2 = \frac{2^2}{2+1} = \frac{4}{3} \] Therefore, the third term \( a_2 \) is \( \frac{4}{3} \).
04

Calculate the Fourth Term

Substitute \( n = 3 \) into the sequence formula. \[ a_3 = \frac{3^2}{3+1} = \frac{9}{4} \] Thus, the fourth term \( a_3 \) is \( \frac{9}{4} \).
05

Calculate the Fifth Term

Substitute \( n = 4 \) into the sequence formula. \[ a_4 = \frac{4^2}{4+1} = \frac{16}{5} \] Therefore, the fifth term \( a_4 \) is \( \frac{16}{5} \).
06

Determine the Limit as \( n \to \infty \)

To find the limit of \( a_n \) as \( n \to \infty \), consider the dominant terms in the numerator and denominator. The expression becomes: \[ \lim_{n \to \infty} \frac{n^2}{n+1} \] Divide both the numerator and the denominator by \( n \): \[ \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \] As \( n \to \infty \), \( \frac{1}{n} \to 0 \). Therefore, the limit is \( \lim_{n \to \infty} \frac{n}{1} = n \). Thus, the limit \( e \infty \) does not exist.

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Key Concepts

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

Term Calculation
Understanding how to calculate the terms of a sequence is crucial when working with sequences and limits. In our example, we begin by evaluating the first few values of the sequence \( \left\{a_{n}\right\} \), defined by the formula \( a_n = \frac{n^2}{n+1} \). Let's see how it works for each of the first five terms:
  • **First Term**: Insert \( n = 0 \) into the formula to get \( a_0 = \frac{0^2}{0+1} = 0 \). The first term is simply 0.

  • **Second Term**: Insert \( n = 1 \) to find \( a_1 = \frac{1^2}{1+1} = \frac{1}{2} \). This gives us \( \frac{1}{2} \) for the second term.

  • **Third Term**: Calculate \( a_2 \) by using \( n = 2 \). So, \( a_2 = \frac{2^2}{2+1} = \frac{4}{3} \), giving \( \frac{4}{3} \) as the third term.

  • **Fourth Term**: For \( n = 3 \), \( a_3 = \frac{3^2}{3+1} = \frac{9}{4} \), which is \( \frac{9}{4} \).

  • **Fifth Term**: Finally, use \( n = 4 \) to find \( a_4 = \frac{4^2}{4+1} = \frac{16}{5} \), resulting in \( \frac{16}{5} \).
Each term is calculated by plugging in a different \( n \) value into the given formula, showing how sequences serve to generalize patterns.
Infinite Sequence
Infinite sequences are essentially a list of numbers in a specific order that goes on forever. An infinite sequence is represented as \( \{a_0, a_1, a_2, \ldots \} \), where each term can be calculated using a formula. In the case of our sequence, \( a_n = \frac{n^2}{n+1} \), each term in the sequence is derived from a function of \( n \).

Working with infinite sequences involves understanding how the sequence behaves as it progresses towards infinity. Rather than merely looking at a small segment, infinite sequences help illustrate long-term behavior in mathematical modeling. By analyzing an infinite sequence, students can explore concepts that lay the groundwork for calculus and real analysis studies.
Limit Analysis
When analyzing limits of sequences, we seek to find the value that the elements of a sequence approach as \( n \) becomes very large. Here's how it works for our specific example of \( a_n = \frac{n^2}{n+1} \).
  • We start with the formula for \( a_n \) and consider \( \lim_{n \to \infty} \frac{n^2}{n+1} \).

  • By dividing both the numerator and the denominator by \( n \), simplify to \( \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \).

  • As \( n \to \infty \), \( \frac{1}{n} \to 0 \), simplifying further to \( \lim_{n \to \infty} n \).
However, this does not produce a single finite value. It tells us the sequence grows without bound, meaning in this case, the limit does not exist as a finite number. In simpler terms, this sequence will keep increasing endlessly, without settling at any stable point, demonstrating how limits provide insight into infinite growth behavior.

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Most popular questions from this chapter

Model painkillers that are absorbed into the blood from a slow release pill. Ourmathematical model for the amount, \(a_{t}\), of drug in the blood t hours after the pill is taken must include the amount absorbed from the pill each hour. Our model starts with the word equation. $$ \begin{array}{c} a_{t+1}=a_{t}+\begin{array}{l} \text { amount absorbed } \\ \text { from the pill } \end{array}-\begin{array}{l} \text { amount eliminated } \\ \text { from the blood } \end{array} \end{array} $$ Assume the amount absorbed from the pill between time \(t\) and time \(t+1\) is \(20 \cdot(0.2)^{t}\). (a) The drug has first order elimination kinetics. \(40 \%\) of the drug is eliminated from the blood each hour. Write down the recursion relation for \(a_{t+1}\) in terms of \(a_{t}\) (b) Assuming that \(a_{0}=0\), meaning that no drug is present in the blood initially, calculate the amount of drug present at times \(t=1,2, \ldots, 6\) (c) What is the maximum amount of drug present at any time in this interval? At what time is this maximum amount reached? (d) Use a spreadsheet to calculate the amount of drug present in hourly intervals from \(t=0\) up to \(t=24\). (e) Show that, when \(t\) is large, the amount of drug present in the blood decreases approximately exponentially with \(t .\) Hint: Plot the values that you computed for \(a_{t}\) against \(t\) on semilogarithmic axes.

Investigate the behavior of the discrete logistic equation $$ x_{t+1}=R_{0} x_{t}\left(1-x_{t}\right) $$ 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 \(t .\) \(R_{0}=3.8, x_{0}=0.9\)

Investigate the behavior of the discrete logistic equation $$ x_{t+1}=R_{0} x_{t}\left(1-x_{t}\right) $$ 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 \(t .\) \(R_{0}=3.8, x_{0}=0\)

The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=1+2 a_{n}, a_{0}=0 $$

Assume that the population growth is described by the Beverton-Holt recruitment curve with parameters \(R_{0}\) and a. Find the population sizes for \(t=1,2, \ldots, 5\) and find \(\lim _{t \rightarrow \infty} N_{t}\) for the given initial value \(N_{0} .\) \(R_{0}=4, a=1 / 60, N_{0}=2\)

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