Question

For each of the following sequences, if the divergence test applies, either state that \(\lim _{n \rightarrow \infty} a_{n}\) does not exist or find \(\lim _{n \rightarrow \infty} a_{n} .\) If the divergence test does not apply, state why. \(a_{n}=\frac{(2 n+1)(n-1)}{(n+1)^{2}}\)

Short answer

The sequence converges to 2.

Step-by-step solution

Evaluate the Limit

To determine the convergence or divergence of the sequence \(a_n = \frac{(2n+1)(n-1)}{(n+1)^2}\), we first find \(\lim_{n \to \infty} a_n\). Let's express this limit:\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(2n+1)(n-1)}{(n+1)^2}\]

Simplify the Expression

Simplify \((2n+1)(n-1)\):\[(2n+1)(n-1) = 2n^2 - 2n + n - 1 = 2n^2 - n - 1\]So, the sequence becomes:\[\lim_{n \to \infty} \frac{2n^2 - n - 1}{(n+1)^2}\]

Use Highest Power in Numerator and Denominator

Notice that in both the numerator and denominator, the highest power of \(n\) is \(n^2\). To simplify the limit, divide every term by \(n^2\):\[\lim_{n \to \infty} \frac{2 - \frac{1}{n} - \frac{1}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}\]As \(n\) approaches infinity, \(\frac{1}{n}\), \(\frac{1}{n^2}\) terms tend towards zero.

Calculate the Limit

Substitute \(\frac{1}{n} \to 0\) and \(\frac{1}{n^2} \to 0\):\[\lim_{n \to \infty} \frac{2 - 0 - 0}{1 + 0 + 0} = \frac{2}{1} = 2\]Thus, the limit \(\lim_{n \to \infty} a_n = 2\) exists.

Conclusion on Divergence Test

In the Divergence Test, if the \(\lim_{n \to \infty} a_n eq 0\), the sequence is divergent. However, in this case, since the limit exists and is a finite number (2), the sequence does not apply to the divergence test. The sequence converges to 2.

Key concepts

Limit of a Sequence

A fundamental concept in calculus and mathematical analysis is the limit of a sequence. Let's simplify this idea. Imagine a sequence as a set of numbers getting increasingly closer to a certain value as we progress through the sequence. This "certain value" is called the limit.

If we consider the sequence given in the exercise, \( a_n = \frac{(2n+1)(n-1)}{(n+1)^2} \), its limit as \( n \) goes to infinity is what we're trying to find. By manipulating and simplifying the terms in the sequence, we can better understand the behavior of the sequence as \( n \) becomes very large.

This means that by evaluating the limit \( \lim_{n \to \infty} a_n \), we’re assessing how the sequence behaves when \( n \) grows large. In this case, after simplifying and substituting, we find the limit to be 2. This implies that as \( n \) increases, the values in the sequence get closer and closer to 2.

Convergence and Divergence

The terms convergence and divergence describe the behavior of a sequence in relation to its limit. A sequence is said to converge if its terms approach a specific limit as \( n \) approaches infinity. Conversely, a sequence is divergent if it does not approach any particular value.

In simple terms:

• A convergent sequence settles to a specific number.
• A divergent sequence either approaches infinity or does not settle to a particular number.

In the context of our example, the sequence \( a_n \) converges because the limit \( \lim_{n \to \infty} a_n = 2 \) exists and is finite. Thus, the sequence approaches the value 2 as \( n \) becomes very large. It demonstrates how, through calculating and simplifying limits, we can determine whether a sequence converges or diverges.

Simplifying Expressions

Simplifying expressions is an integral part of working with sequences and limits since it often reveals the underlying pattern of a sequence. In our given sequence \( a_n = \frac{(2n+1)(n-1)}{(n+1)^2} \), simplification is key to accurately determining the sequence's limit.

Let's break down the simplification process:

• First, expand the expression in the numerator: \((2n+1)(n-1)\).
• This results in: \(2n^2 - 2n + n - 1 \), which simplifies to \(2n^2 - n - 1 \).
• Next, notice that both numerator and denominator involve \(n^2\) terms.
• Dividing every term by \(n^2\) simplifies the limit evaluation: \(\frac{2 - \frac{1}{n} - \frac{1}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}\).

As \(n\) increases, expressions like \(\frac{1}{n}\) and \(\frac{1}{n^2}\) approach zero, simplifying our limit further to a clear finite value — in this case, 2. This simplification process makes it much easier to identify the behavior of a sequence at infinity.