Problem 28 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{n+(-1)^{n}}{n} $$

Step-by-Step Solution

Answer

The sequence converges to 1.

Step 1: Understand the Sequence

The sequence given is \( a_n = \frac{n + (-1)^n}{n} \). This means that we have a fraction where the numerator is \( n + (-1)^n \) and the denominator is \( n \). We need to determine if this sequence converges and find the limit if it does.

Step 2: Simplify the Sequence

Simplify the expression by dividing each term in the numerator by the denominator: \[ a_n = \frac{n}{n} + \frac{(-1)^n}{n} = 1 + \frac{(-1)^n}{n} \].

Step 3: Analyze the Convergence

For large \( n \), observe the term \( \frac{(-1)^n}{n} \). The term \( (-1)^n \) will alternate between -1 and 1, but since it is divided by \( n \), this fraction becomes smaller as \( n \) increases. Hence, \( \frac{(-1)^n}{n} \to 0 \) as \( n \to \infty \).

Step 4: Conclude the Convergence

Given that \( \frac{(-1)^n}{n} \to 0 \) as \( n \to \infty \) and \( 1 \) is constant, the sequence \( a_n = 1 + \frac{(-1)^n}{n} \) converges to \( 1 \).

Step 5: Verify with Limit Definition

To verify, use the definition of the limit of a sequence. Calculate \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 + \frac{(-1)^n}{n}\right) = 1 + \lim_{n \to \infty} \frac{(-1)^n}{n} = 1 + 0 = 1 \). Therefore, the sequence converges to 1.

Key Concepts

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus and analysis. It describes the value that the terms of a sequence approach as the index goes to infinity. In simpler terms, if you have an infinite sequence of numbers, the limit tells you what single number these terms are getting closer and closer to.

When we refer to the limit of a sequence \( \{a_n\} \), we often denote it as \( \lim_{n \to \infty} a_n \). For a sequence to have a limit, its terms must settle towards a single finite number as \( n \) becomes very large.

In the given problem, the sequence is \( a_n = 1 + \frac{(-1)^n}{n} \). By observing as \( n \to \infty \), the term \( \frac{(-1)^n}{n} \) diminishes to zero. Thus, the limit of the sequence is the constant term, which is 1. In mathematical terms:

• \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( 1 + \frac{(-1)^n}{n} \right) = 1 \).

Understanding limits helps us determine how sequences behave in the long run, which is crucial for analyzing their convergence or divergence.

Alternating Sequences

An alternating sequence is one where the terms switch between positive and negative values. This happens because each term in the sequence changes sign based on whether the index is odd or even.

In our case, this is represented by \( (-1)^n \). For odd values of \( n \), \((-1)^n\) equals -1, and for even \( n \), it equals +1. Consequently, the sequence alternates between adding \( \frac{1}{n} \) and subtracting \( \frac{1}{n} \) from 1.

• At odd \( n \), \( a_n = 1 - \frac{1}{n} \).
• At even \( n \), \( a_n = 1 + \frac{1}{n} \).

Despite these alternating terms, as the sequence progresses towards infinity, the alternating component \( \frac{(-1)^n}{n} \) grows smaller and smaller, driving the entire sequence towards its limit.

Understanding alternating sequences helps identify how oscillations within sequences behave as they head towards certain values or continue without settling on one specific number.

Convergence and Divergence Analysis

Analyzing whether a sequence converges or diverges is key to understanding its long-term behavior. A sequence converges if its terms get indefinitely close to a specific number, known as the limit. Conversely, a sequence diverges if it fails to approach any definite limit as \( n \to \infty \).

For the sequence in question, \( a_n = 1 + \frac{(-1)^n}{n} \), we observed that as \( n \) increases, the term \( \frac{(-1)^n}{n} \) approaches 0. Hence, regardless of whether \( n \) is odd or even, the sequence draws closer to 1.

Convergence was established through the limit evaluation:

• \( \frac{(-1)^n}{n} \to 0 \)
• Thus, \( 1 + \frac{(-1)^n}{n} \to 1 \)

These conditions confirm convergence because the sequence consistently nears 1, the proposed limit. Divergence would have been suggested if the sequence continued oscillating or growing without commitment to any value.

In summary, convergence and divergence analysis are essential tools for predicting how sequences evolve, highlighting when they stabilize to a limit or spiral without hover.