Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=(-1)^{n} \frac{n}{n+2}\)

Step-by-Step Solution

Answer

The sequence diverges as it oscillates indefinitely.

Step 1: Identify the Explicit Formula

The given formula for the sequence is \(a_{n} = (-1)^{n} \frac{n}{n+2}\). This alternates the sign of the terms because of \((-1)^n\).

Step 2: Compute the First Five Terms

Let's calculate the first five terms by substituting \(n = 1, 2, 3, 4, 5\) in the explicit formula.- For \(n = 1\), \(a_1 = (-1)^1 \frac{1}{1+2} = -\frac{1}{3}\).- For \(n = 2\), \(a_2 = (-1)^2 \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}\).- For \(n = 3\), \(a_3 = (-1)^3 \frac{3}{3+2} = -\frac{3}{5}\).- For \(n = 4\), \(a_4 = (-1)^4 \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}\).- For \(n = 5\), \(a_5 = (-1)^5 \frac{5}{5+2} = -\frac{5}{7}\).The first five terms are: \(-\frac{1}{3}, \frac{1}{2}, -\frac{3}{5}, \frac{2}{3}, -\frac{5}{7}\).

Step 3: Observe the Behavior of the Sequence

To determine whether the sequence converges or diverges, examine the general term \((-1)^{n} \frac{n}{n+2}\) as \(n\) approaches infinity.

Step 4: Evaluate \(\lim_{n \to \infty} a_n\)

As \(n\) becomes very large, the term \(\frac{n}{n+2}\) approaches 1 because it can be approximated as \(\frac{n}{n}\). However, the term \((-1)^n\) causes the sign to alternate between positive and negative, making the sequence oscillate between values close to 1 and -1.

Step 5: Conclude Convergence or Divergence

Since the sequence does not approach a single finite value and instead oscillates indefinitely between values close to 1 and -1, it diverges.

Key Concepts

Convergence and Divergence

When we talk about the convergence or divergence of a sequence, we're essentially assessing its behavior as the index \(n\) approaches infinity. In simple terms:

• A sequence converges if it approaches a particular finite value at infinity.
• A sequence diverges if it does not approach a single finite value.

In the original exercise, the sequence given is an example of divergence. Even though the fraction \( \frac{n}{n+2} \) approaches 1 as \( n \to \infty \), the sequence includes a multiplier \((-1)^n\) which alternates the sign of each term. This alternation prevents the sequence from settling on a single value, causing it to oscillate indefinitely. Thus, it diverges.

Limit of a Sequence

The limit of a sequence, when it exists, is the value that the terms of the sequence get closer to as \(n\) becomes very large. The notation for this is \( \lim_{n \to \infty} a_{n} \). If a sequence converges, this limit will be a real number. However, if a sequence diverges, no such limit exists at a finite value.In the provided sequence \((-1)^n \frac{n}{n+2}\), we attempted to determine the limit. Although the fraction \( \frac{n}{n+2} \) seems to approach 1 as \( n \to \infty \) due to the dominance of \(n\) in both the numerator and denominator, the sequence overall does not have a real limit because of the alternating sign factor \((-1)^n\). This oscillation prevents us from assigning a definitive limit to the sequence.

Alternating Sequences

In mathematics, alternating sequences are those sequences in which the sign of terms alternates between positive and negative. This often results from a factor like \((-1)^n\), where \((-1)\) raised to an odd power gives negative one and to an even power gives positive one. The sequence given in the exercise, \((-1)^n \frac{n}{n+2}\), is a perfect example of this phenomenon because:

• The term \((-1)^n\) causes each consecutive element to flip in sign, inhibiting convergence.

The alternating nature introduces complexity when evaluating convergence. Indeed, such sequences often lead to oscillation between two values, and without additional structure, they cannot settle on a single limit. Overall, this factor typically leads to divergence unless some cancellation or compensatory pattern is present.