Problem 119 Which of the sequences converge,... [FREE SOLUTION]

Chapter 9: Problem 119

Which of the sequences converge, and which diverge? Give reasons for your answers. $$a_{n}=\left((-1)^{n}+1\right)\left(\frac{n+1}{n}\right)$$

Short Answer

Expert verified

The sequence $a_n$ diverges because it oscillates between 0 and approximately 2.

Step by step solution

Understand the Components of the Sequence

The sequence given is $ a_{n} = ((-1)^n + 1) \times \left(\frac{n+1}{n}\right) $. Notice the sequence can be split into two parts: $(-1)^n + 1$ and $\frac{n+1}{n}$.

Analyze the Oscillating Part $(-1)^n + 1$

The term $(-1)^n$ oscillates between -1 and 1. Thus, $(-1)^n + 1$ oscillates between 0 (when $n$ is odd) and 2 (when $n$ is even). This means $(-1)^n + 1$ is not converging, as it does not approach a single value.

Simplify the Fractional Part $\frac{n+1}{n}$

Simplify $\frac{n+1}{n}$ to $1 + \frac{1}{n}$. As $n$ approaches infinity, $\frac{1}{n}$ approaches 0, hence this part converges to 1.

Consider the Product of the Two Parts

Given $a_n = ((-1)^n + 1) \times (1 + \frac{1}{n})$, evaluate this for even and odd $n$. For even $n$, it becomes $2 \times (1 + \frac{1}{n})$ which approaches 2 as $n$ grows. For odd $n$, it becomes $0 \times (1 + \frac{1}{n}) = 0$. Hence, since $a_n$ oscillates between these two values, it does not converge.

Conclude about the Convergence or Divergence

Since the full sequence $a_n$ does not settle down to a single value regardless of whether $n$ is even or odd, the sequence does not converge.

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.

Oscillating Sequences

Oscillating sequences are a unique category of sequences that does not converge to a single, fixed limit. Instead, these sequences alternate or switch between different values as the term number increases. A classic example of an oscillating sequence is one involving powers of negative one:

For instance, $(-1)^n$ alternates between -1 and 1.
As seen in our original exercise, the term $(-1)^n + 1$ oscillates between 0 and 2. For even numbers, this expression results in 2; for odd numbers, it turns into 0.
These changes mean the sequence's terms do not approach a single value. Thus, they produce oscillations without converging.

Oscillating behavior is essential in determining whether a sequence converges or diverges, as it disrupts the convergence by continuously moving between different values.

Sequence Divergence

A sequence is considered divergent if it does not settle towards a single finite limit as you proceed to infinity. This can occur in several ways:

If the sequence's terms grow indefinitely large (positively or negatively), it diverges.
Sequences that oscillate without settling into a predictable pattern, as is our exercise's case with terms moving between 0 and 2, are also divergent.
Mathematically speaking, a divergent sequence will not meet the formal definition of convergence, where given any small tolerance level $\epsilon$, there exists a point beyond which all subsequent terms of the sequence keep within $\epsilon$ distance from the limit.

In practice, identifying divergent sequences is crucial to understanding their behavior and anticipating that their values will not stabilize.

Limits of Sequences

Understanding limits is key to analyzing sequences. The limit of a sequence is a value that the terms of the sequence get closer to as the index becomes very large. Let's see what this means:

For convergent sequences, there is a specific number that the terms approach as the sequence progresses.
In the expression $\frac{n+1}{n}$, the limit as $n$ tends to infinity is 1. We observe this because $\frac{1}{n}$ approaches 0 as $n$ becomes very large.
However, for our exercise's sequence $a_n = ((-1)^n + 1) \times \left(1 + \frac{1}{n}\right)$, the limit does not exist because the sequence's behavior oscillates.For even $n$, terms approach 2, and for odd $n$, terms are 0.

When evaluating limits, it is this consistency towards a single value that differentiates converging sequences from divergent or oscillating sequences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Which of the sequences converge, and which diverge? Give reasons for your answers. $$a_{n}=\left((-1)^{n}+1\right)\left(\frac{n+1}{n}\right)$$

Short Answer

Step by step solution

Understand the Components of the Sequence

Analyze the Oscillating Part \((-1)^n + 1\)

Simplify the Fractional Part \(\frac{n+1}{n}\)

Consider the Product of the Two Parts

Conclude about the Convergence or Divergence

Key Concepts

Oscillating Sequences

Sequence Divergence

Limits of Sequences

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Logic and Functions

Discrete Mathematics

Mechanics Maths

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.