Problem 20 - Single Variable Calculus

Question

Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges or diverges. If it converges, find its limit. \(a_{n}=(-1)^{n} \frac{n+2}{3 n+1}\)

Step-by-Step Solution

Answer

The sequence \(a_{n} =(-1)^n \frac{n+2}{3n+1}\) diverges, as the limit of the sequence does not exist due to the oscillating behavior of \((-1)^n\) as n approaches infinity.

Step 1: Write down the given sequence

: We have the sequence \(a_{n} =(-1)^n \frac{n+2}{3n+1}\).

Step 2: Find the limit of the sequence as n approaches infinity

: We need to find the limit of \(a_n\) as n approaches infinity. \[\lim_{n \to \infty} a_n = \lim_{n \to \infty} (-1)^n \frac{n+2}{3n+1}\]

Step 3: Check if the limit exists and is finite

: To check if the limit exists, we will first look at the term \(\frac{n+2}{3n+1}\) as n approaches infinity: \[\lim_{n \to \infty} \frac{n+2}{3n+1} = \frac{1}{3}\] Since the limit of \(\frac{n+2}{3n+1}\) exists and is finite, we now need to consider the term \((-1)^n\) for the overall limit of the sequence. Notice that \((-1)^n\) oscillates between -1 and 1 as n approaches infinity. This means that the sequence \(\left\\{a_{n}\right\\}\) does not approach a constant value because it keeps oscillating between different values, thus: \[\lim_{n \to \infty} a_n = \lim_{n \to \infty} (-1)^n \frac{n+2}{3n+1} \text{ does not exist}\] Since the limit of the sequence does not exist, the sequence \(\left\\{a_{n}\right\\}\) diverges.

Key Concepts

Sequence Divergence

Sequences are collections of numbers arranged in a particular order. These numbers follow a specific rule. When discussing sequences, the concept of convergence and divergence is pivotal. A sequence diverges if it does not settle at a particular constant value as the number of terms increases. In simple terms, divergence means that the sequence either keeps growing without any bound or it keeps oscillating without approaching a single number.
For example, in the original exercise, the sequence \{a_{n}\} defined as \(a_{n}=(-1)^{n} \frac{n+2}{3n+1}\) was investigated for convergence or divergence. The rule for \(a_{n}\) creates values that do not stabilize at any specific point. This behavior tells us that the sequence diverges. It's essential to understand that sequences representing divergent behaviors do not have limits because they don't zero-in on a single value as more terms are added.

Limit of a Sequence

The limit of a sequence is the value that the sequence approaches as more terms are added. In mathematical terms, we are interested in what happens when the term number, usually noted as \(n\), approaches infinity. A convergent sequence has a limit, meaning it approaches a particular number as \(n\) grows very large.
We calculate this by evaluating the expression that defines the sequence as \(n\) goes to infinity. For instance, the part of the sequence \(a_{n}=(-1)^n \frac{n+2}{3n+1}\) without \((-1)^n\) simplifies to \(\frac{n+2}{3n+1}\). As \(n\) becomes very large, this fraction approaches \(\frac{1}{3}\). This value is the limit, for this specific expression part; however, a limit does not exist for the entire sequence because of its oscillating behavior. If any component causes the sequence not to settle at a point, like the alternation factor \((-1)^n\), the whole sequence lacks a limit.

Oscillating Sequence

An oscillating sequence is one that alternates around different values without settling into a pattern that ends in a single number. In the original exercise, the sequence \{a_{n}\} has an oscillating characteristic due to the factor \((-1)^n\). This part of the sequence makes the value alternate between negative and positive at every step as \(n\) increases.

• This alternation means the sequence switches from being positive to negative continuously.
• Because of this alternation, it's impossible for the sequence to stabilize at a single definitive value.
• This oscillation behavior prevents the sequence from having a limit, thereby making it diverge.

Recognizing this oscillating nature is crucial when assessing whether a sequence converges or diverges. If oscillation is present and prevents a single value limit, the sequence will diverge, like \{a_{n}\} in the original content.