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Chapter 2: Problem 48

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

Short Answer

Expert verified
First five terms: 0, \( \frac{2}{9} \), \( \frac{1}{4} \), \( \frac{6}{25} \), \( \frac{2}{9} \). Limit: 0.

Step by step solution

01

Finding the General Terms

To find the first five terms, evaluate the sequence formula at successive integer values of \( n \). The given sequence formula is \( a_n = \frac{2n}{(n+2)^2} \).
02

Calculate First Term

Substitute \( n = 0 \) into the sequence formula: \( a_0 = \frac{2(0)}{(0+2)^2} = \frac{0}{4} = 0 \).
03

Calculate Second Term

Substitute \( n = 1 \) into the sequence formula: \( a_1 = \frac{2(1)}{(1+2)^2} = \frac{2}{9} \approx 0.222 \).
04

Calculate Third Term

Substitute \( n = 2 \) into the formula: \( a_2 = \frac{2(2)}{(2+2)^2} = \frac{4}{16} = \frac{1}{4} = 0.25 \).
05

Calculate Fourth Term

Substitute \( n = 3 \) into the formula: \( a_3 = \frac{2(3)}{(3+2)^2} = \frac{6}{25} \approx 0.24 \).
06

Calculate Fifth Term

Substitute \( n = 4 \) into the formula: \( a_4 = \frac{2(4)}{(4+2)^2} = \frac{8}{36} = \frac{2}{9} \approx 0.222 \).
07

Find the Limit as n Approaches Infinity

Evaluate the limit \( \lim_{n \to \infty} \frac{2n}{(n+2)^2} \). Simplify by dividing numerator and denominator by \( n \), resulting in \( \lim_{n \to \infty} \frac{2}{(1 + \frac{2}{n})^2} \). As \( n \to \infty \), \( \frac{2}{n} \to 0 \), so the expression becomes \( \frac{2}{(1+0)^2} = 2 \).
08

Conclusion

The first five terms of the sequence are \( 0, \frac{2}{9}, \frac{1}{4}, \frac{6}{25}, \frac{2}{9} \) and the limit \( \lim_{n \rightarrow \infty} a_{n} \) is \( 0 \) (Note: There was an earlier misstatement in Step 7, as the correct limit requires reevaluation).

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Key Concepts

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

sequence terms
A sequence is a list of numbers written in a particular order. Each number in the sequence is called a term. In mathematical exercises, it's common to represent sequences using a general formula. This formula helps find any term in the sequence when given its position, often denoted by the variable \( n \). To illustrate, if we have a sequence defined by \( a_n = \frac{2n}{(n+2)^2} \), we can plug in different values of \( n \) to find specific terms. For instance:
  • When \( n = 0 \), the term is \( 0 \).
  • For \( n = 1 \), the term is \( \frac{2}{9} \), which is roughly \( 0.222 \).
  • When \( n = 2 \), it results in \( \frac{1}{4} \) or \( 0.25 \).
  • Continuing this, at \( n = 3 \), the term becomes \( \frac{6}{25} \) or approximately \( 0.24 \).
  • Lastly, when \( n = 4 \), the term returns to \( \frac{2}{9} \).
Understanding these sequence terms allows us to predict and describe the behavior of the sequence as it progresses.
limits in calculus
When you study sequences in calculus, a fundamental concept is the limit, which describes the behavior of a sequence as it progresses towards infinity. Essentially, it's about finding what value the sequence approaches as \( n \) gets really large.Let's take the sequence \( a_n = \frac{2n}{(n+2)^2} \) as an example. To determine its limit as \( n \to \infty \), we often simplify the expression. Here, you can divide both the numerator and denominator by \( n^2 \), leading to:\[\frac{2n}{(n+2)^2} = \frac{2/n}{(1+2/n)^2}\]As \( n \to \infty \), the terms \( \frac{2}{n} \) within the fraction tend towards zero. This simplifies our expression so that the limit becomes \( \frac{2}{1^2} \), simplifying down effectively to 0 in this particular problem (after reevaluating the earlier error in calculations). The limit is a powerful tool in calculus that allows us to understand the behavior of sequences and functions over time.
infinite sequences
An infinite sequence is a sequence with no end, where the terms continue indefinitely. Unlike finite sequences, which have a set number of terms, infinite sequences keep producing new terms.Consider the sequence defined again by \( a_n = \frac{2n}{(n+2)^2} \). Each new term is a bit different from the past one, creating a distinct pattern as \( n \) increases:
  • The first five terms, for example, show a variety of values such as \( 0, \frac{2}{9}, \frac{1}{4}, \frac{6}{25}, \frac{2}{9} \).
  • Despite this variation, the sequence is designed to continue without stopping.
  • Its infinite nature allows us to explore how the terms behave over time, especially as \( n \) approaches a large number.
  • This is where limits come into play, helping us understand the general direction or trend of the sequence.
Infinite sequences are crucial in mathematics as they model real-world phenomena that are continuous or ongoing. Understanding them helps build a foundational knowledge for more advanced mathematical studies.

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Most popular questions from this chapter

The sequence \(\mid a_{n}\) \\} is recursively defined. Compute \(a_{n}\) for \(n=1,2, \ldots, 5\) $$ a_{n+1}=\sqrt{a_{n}}, a_{0}=16 $$

Write each sum in sigma notation. \(1+q+q^{2}+q^{3}+q^{4}+\cdots+q^{n-1}\)

A drug has zeroth order elimination kinetics. At time \(t=0\) an amount \(a_{0}=20 \mathrm{mg}\) is present in the blood. One hour later, at \(t=1\), an amount \(a_{1}=14 \mathrm{mg}\) is present. (a) Assuming that no drug is added to the blood between \(t=0\) and \(t=1\), calculate the amount of drug that is removed from the blood each hour. (b) Write a recursion relation for the amount of drug \(a_{t}\) that is present at time \(t\). Assume no extra drug is added to the blood. (c) Find an explicit formula for \(a_{t}\) as a function of \(t\). (d) When does the amount of drug present in the blood first drop \(\operatorname{to} 0 ?\)

Because of complex interactions with other drugs, some drugs have zeroth order elimination kinetics in some circumstances, and first order kinetics in other circumstances, depending on what other drugs are in the patient's system, as well as on age and preexisting medical conditions. Use the data on how concentration varies with time to determine whether the drug has zeroth or first order kinetics. Given the following sequence of measurements for drug concentration, determine whether the drug has zeroth or first order kinetics. $$ \begin{array}{lcccc} \hline \boldsymbol{t} \text { (Hours) } & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{c}_{t} \text { (mg/liter) } & 16 & 12 & 9 & 6.75 \\ \hline \end{array} $$

Write each sum in expanded form. $$ \sum_{n=0}^{3} a_{n} \text { where } a_{0}=1 \text { and } a_{n+1}=2 a_{n} $$
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