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Chapter 9: Problem 48

Consider the following sequences. a. Find the first four terms of the sequence. b. Based on part (a) and the figure, determine a plausible limit of the sequence. $$a_{n}=\frac{n^{2}}{n^{2}-1} ; n=2,3,4, \dots$$

Short Answer

Expert verified
Answer: The first four terms of the sequence are $$a_2 =\frac{4}{3}, a_3 = \frac{9}{8}, a_4 = \frac{16}{15}, a_5 = \frac{25}{24}$$. The plausible limit of the sequence is 1.

Step by step solution

01

Find the first four terms of the sequence.

Use the formula for the sequence, $$a_n = \frac{n^2}{n^2 - 1}$$. For the first term, let n = 2: $$a_2 = \frac{2^2}{2^2 - 1} = \frac{4}{3}$$ For the second term, let n = 3: $$a_3 = \frac{3^2}{3^2 - 1} = \frac{9}{8}$$ For the third term, let n = 4: $$a_4 = \frac{4^2}{4^2 - 1} = \frac{16}{15}$$ For the fourth term, let n = 5: $$a_5 = \frac{5^2}{5^2 - 1} = \frac{25}{24}$$ So the first four terms of the sequence are $$a_2 =\frac{4}{3}, a_3 = \frac{9}{8}, a_4 = \frac{16}{15}, a_5 = \frac{25}{24}$$.
02

Determine a plausible limit of the sequence.

We notice that the terms of the sequence have the form ${ \frac{(n+1)(n-1) + 1}{(n+1)(n-1)}}$. Furthermore, we see that each denominator is just one less than the numerator, and as n grows larger, the difference between the numerator and denominator becomes less significant. Thus, it is plausible to conclude that the limit of the sequence is 1, since the ratio of the terms approaches 1 for larger n: $$\lim_{n\to\infty} \frac{n^2}{n^2 - 1} = 1$$ In conclusion, the first four terms of the sequence are: $$a_2 =\frac{4}{3}, a_3 = \frac{9}{8}, a_4 = \frac{16}{15}, a_5 = \frac{25}{24}$$, and the plausible limit of the sequence is 1.

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Key Concepts

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

sequences and series
A sequence is a list of numbers arranged in a specific order. Each number in a sequence is called a term. In mathematics, sequences are a fundamental concept as they can describe an ordered collection of objects, such as the natural numbers or even more complicated formulations. Sequences can be finite, with a certain number of terms, or infinite, continuing indefinitely. In the original exercise, we look at the sequence given by the formula \( a_n = \frac{n^2}{n^2 - 1} \). To find the terms of a sequence, substitute integer values of \( n \) into the sequence鈥檚 formula.
  • The first four terms found using \( n = 2, 3, 4, \) and \( 5 \) are \( \frac{4}{3}, \frac{9}{8}, \frac{16}{15}, \frac{25}{24} \).
A series, on the other hand, is what you get when you add up the terms of a sequence. While the exercise focuses on the sequence itself, understanding series is essential for topics like calculus, where you might need to add an infinite number of terms. Each series can be described as either convergent, meaning it approaches a specific value as more terms are added, or divergent, where it does not have a finite limiting value.
convergence
Convergence is a key idea when studying sequences and series. A sequence converges if its terms approach a specific value, known as the limit, as \( n \) becomes very large. In simple terms, if the terms of the sequence settle down to a particular number as you move further and further down the sequence, it's said to "converge."
In the original exercise, the terms of the sequence \( a_n = \frac{n^2}{n^2 - 1} \) approach the value 1 as \( n \) increases.
  • For small values of \( n \), the numerator and denominator differ slightly, but as \( n \) grows, the impact of subtracting 1 from \( n^2 \) becomes negligible.
  • Thus, the terms get closer and closer to 1, indicating that the sequence converges to 1.
Understanding convergence is crucial not only in sequences but also in series. With series, just like sequences, convergence means that as you add more terms, the total sum approaches a specific number. This concept is foundational in many areas of mathematics and essential for understanding the behavior of functions and their approximations.
limit of a sequence
The limit of a sequence is the value that the terms of the sequence get closer to as the sequence progresses towards infinity. Calculating the limit involves determining the long-term behavior of the sequence, and it helps in predicting what the terms approach.
In our example, the sequence \( a_n = \frac{n^2}{n^2 - 1} \) has a limit of 1.
  • As \( n \) grows very large, the effect of that -1 in the denominator vanishes relative to \( n^2 \), leading the fraction \( \frac{n^2}{n^2 - 1} \) closer to \( \frac{n^2}{n^2} = 1 \).
To find the limit, you often look for patterns or behaviors as \( n \) becomes large. Advanced methods such as L'H么pital's Rule or manipulating expressions algebraically can also be used to determine limits.
Knowing the limit of a sequence is valuable in fields like calculus and analysis, where understanding the limiting behavior of functions and sums can provide deep insights into continuity, differentiability, and integrals.

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Most popular questions from this chapter

In Section 3, we established that the geometric series \(\Sigma r^{k}\) converges provided \(|r|<1\). Notice that if \(-1

a. Consider the number 0.555555...., which can be viewed as the series \(5 \sum_{k=1}^{\infty} 10^{-k} .\) Evaluate the geometric series to obtain a rational value of \(0.555555 \ldots\) b. Consider the number \(0.54545454 \ldots,\) which can be represented by the series \(54 \sum_{k=1}^{\infty} 10^{-2 k} .\) Evaluate the geometric series to obtain a rational value of the number. c. Now generalize parts (a) and (b). Suppose you are given a number with a decimal expansion that repeats in cycles of length \(p,\) say, \(n_{1}, n_{2} \ldots \ldots, n_{p},\) where \(n_{1}, \ldots, n_{p}\) are integers between 0 and \(9 .\) Explain how to use geometric series to obtain a rational form of the number. d. Try the method of part (c) on the number \(0.123456789123456789 \ldots\) e. Prove that \(0 . \overline{9}=1\)

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month, while 80 fish are harvested each month. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. a. Write out the first five terms of the sequence \(\left\\{F_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{F_{n}\right\\}\). c. Does the fish population decrease or increase in the long run? d. Determine whether the fish population decreases or increases in the long run if the initial population is 5500 fish. e. Determine the initial fish population \(F_{0}\) below which the population decreases.

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0}\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G\). a. Show that \(a_{n}>b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

The expression where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of where \(a\) and \(b\) are positive real numbers.
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