/*! 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 36 Find the limit of the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit of the following sequences or determine that the limit does not exist. Verify your result with a graphing utility. $$a_{n}=\frac{(-1)^{n} n}{n+1}$$

Short Answer

Expert verified
Answer: The limit does not exist.

Step by step solution

01

Rewrite the sequence with a convenient form

Since \((-1)^n\) takes the value \(1\) for even n and \(-1\) for odd n, we can write the sequence as follows: $$a_n=(-1)^n\cdot\frac{n}{n+1}$$ This separates the alternating aspect from the rest of the function.
02

Analyze the behavior of the function as \(n\to\infty\)

Let's analyze the behavior of the function \(\frac{n}{n+1}\) as \(n\to\infty\). We write it as: $$\lim_{n\to\infty}\frac{n}{n+1}$$ We notice that the denominator and the numerator grows at the same pace, so we can divide both of them by \(n\). The expression becomes: $$\lim_{n\to\infty}\frac{1}{1+\frac{1}{n}}$$
03

Evaluate the limit

Now, we can evaluate the limit by plugging infinity into the denominator and simplifying: $$\lim_{n\to\infty}\frac{1}{1+\frac{1}{n}} = \frac{1}{1+0}=1$$
04

Where the limit does not exist

Since we found the limit for the function \(\frac{n}{n+1}\), we can see that this part converges to 1 as \(n\to\infty\). However, the sequence also contains the factor \((-1)^n\), which oscillates between 1 and -1 for even and odd values of n. The original sequence can be written as: $$a_n=(-1)^n\cdot\frac{n}{n+1}$$ As \(n\to\infty\), \((-1)^n\) will continue to alternate between positive and negative values, causing the sequence to oscillate between positive and negative values of \(\frac{n}{n+1}\). Since the sequence oscillates between positive and negative values in this manner, it does not converge to a limit as \(n\to\infty\). Therefore, we can conclude that the limit does not exist.
05

Verify with a graphing utility

Using a graphing utility such as Desmos or Grapher, plot the function \(a_n=(-1)^n\cdot\frac{n}{n+1}\) with n being a discrete variable (e.g., from 1 to 100). You will notice that the sequence oscillates between positive and negative values and never converges to a single limit, confirming our result that the limit does not exist.

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Key Concepts

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

Convergence and Divergence of Sequences
Understanding the behavior of sequences is crucial in mathematics, especially when dealing with infinite processes. A sequence is essentially a list of numbers generated by a specific rule. When exploring sequences, determining whether they converge or diverge as they progress towards infinity is of particular interest.

Convergence of a sequence occurs when the terms of the sequence approach a specific number, known as the limit, as the sequence progresses. In mathematical terms, a sequence \( a_n \) converges to a limit \( L \) if, for every positive number \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n \geq N \), the distance between \( a_n \) and \( L \) is less than \( \epsilon \). In simpler terms, the terms get as close as we want to the limit, and stay that way.

Divergence, on the other hand, occurs when the terms of a sequence do not approach a single value. This can happen in several ways, such as the sequence terms getting indefinitely large, or if the sequence oscillates without settling down to a limit. The sequence given in the exercise, \(\frac{(-1)^{n} n}{n+1}\), diverges because it oscillates and does not approach a specific number as \(n\) increases.
Limits at Infinity
The concept of limits at infinity is a cornerstone in calculus, describing the behavior of functions or sequences as the variable approaches infinity. A limit at infinity of a function or a sequence answers the question: 'What value does the function or the sequence approach as the variable gets indefinitely large?'

For a sequence, the notation \(\lim_{n\to\infty}a_n = L\) indicates that as \(n\) becomes very large, the sequence \(a_n\) approaches the value \(L\). If such a finite limit exists, we say that the sequence converges to \(L\). Conversely, if there is no such value, or the terms grow without bound, the sequence is said to diverge. When dealing with fractions as in our exercise, we often divide the terms by the highest power present to simplify the limit computation. However, it's important to note that the presence of an oscillating factor, like \((-1)^n\), can cause divergence despite the fraction alone appearing to converge.
Oscillating Sequences
An oscillating sequence is one that switches back and forth between values in a pattern that prohibits settling down to a single number. This behavior defines a divergent sequence because it fails to approach a specific limit. A classic example of an oscillating sequence is the one defined by \((-1)^n\), which alternates between 1 and -1 indefinitely.

In the case of the sequence from our exercise, \(\frac{(-1)^{n} n}{n+1}\), the oscillating factor \((-1)^n\) plays a crucial role. As \(n\) increases, this term causes the sequence to jump between positive and negative values. Even though the fractional part \(\frac{n}{n+1}\) approaches 1, the entire sequence doesn't settle at any point due to the alternating signs introduced by \((-1)^n\), thus confirming divergence through oscillation.
Graphing Sequences
Graphing sequences is a valuable tool for visualizing their behavior, especially when it comes to understanding convergence or divergence. To graph a sequence, we plot the terms \(a_n\) with respect to their position \(n\) in the sequence on the Cartesian plane. This is different from graphing continuous functions, as sequences are inherently discrete; that is, they have values only at integer points.

When graphing the sequence \(\frac{(-1)^{n} n}{n+1}\), we should expect to see points alternating above and below the horizontal axis, indicative of an oscillating sequence. The distance from the points to the axis will appear to decrease as \(n\) increases—due to the fractional part \(\frac{n}{n+1}\) approaching 1—but the alternation will not cease. A graphing utility can provide this visual representation and support the analytical conclusion that the sequence does not converge.

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Most popular questions from this chapter

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

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\(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. Population growth When a biologist begins a study, a colony of prairie dogs has a population of \(250 .\) Regular measurements reveal that each month the prairie dog population increases by \(3 \%\) Let \(p_{n}\) be the population (rounded to whole numbers) at the end of the \(n\) th month, where the initial population is \(p_{0}=250\).

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