Problem 6 - 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 \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{\sqrt{3 n^{2}+2}}{2 n+1}\)

Step-by-Step Solution

Answer

The sequence converges to \(\frac{\sqrt{3}}{2}\).

Step 1: Find the first term

To find the first term \(a_1\), substitute \(n = 1\) into the formula \(a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}\). \(a_1 = \frac{\sqrt{3(1)^2 + 2}}{2(1) + 1} = \frac{\sqrt{5}}{3}\).

Step 2: Find the second term

Substitute \(n = 2\) into the formula to find \(a_2\).\(a_2 = \frac{\sqrt{3(2)^2 + 2}}{2(2) + 1} = \frac{\sqrt{14}}{5}\).

Step 3: Find the third term

Substitute \(n = 3\) into the formula to find \(a_3\).\(a_3 = \frac{\sqrt{3(3)^2 + 2}}{2(3) + 1} = \frac{\sqrt{29}}{7}\).

Step 4: Find the fourth term

Substitute \(n = 4\) into the formula to find \(a_4\).\(a_4 = \frac{\sqrt{3(4)^2 + 2}}{2(4) + 1} = \frac{\sqrt{50}}{9}\).

Step 5: Find the fifth term

Substitute \(n = 5\) into the formula to find \(a_5\).\(a_5 = \frac{\sqrt{3(5)^2 + 2}}{2(5) + 1} = \frac{\sqrt{77}}{11}\).

Step 6: Analyze convergence

To determine if the sequence converges, consider the limit as \(n\) approaches infinity of the sequence \(a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}\).

Step 7: Simplify the expression inside the limit

Rewrite \(a_n\) by dividing the numerator and the denominator by \(n\):\[ a_n = \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}} \]

Step 8: Calculate the limit

As \(n \to \infty\), \(\frac{2}{n^2} \to 0\) and \(\frac{1}{n} \to 0\), transforming the expression into:\[ \lim_{{n \to \infty}} a_n = \frac{\sqrt{3 + 0}}{2 + 0} = \frac{\sqrt{3}}{2} \]

Step 9: Conclude the convergence and limit

Since the limit exists and equals \(\frac{\sqrt{3}}{2}\), the sequence converges to this value.

Key Concepts

Explicit Sequence Formula

Understanding an explicit sequence formula is crucial when dealing with sequences in mathematics. An explicit formula allows you to find the terms in a sequence directly from their position number, which is denoted by \(n\).
For example, in our exercise, the explicit sequence formula given is:

• \(a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}\)

With this formula, you can directly calculate any term in the sequence by substituting \(n\) with the desired term number. For instance, to find the first term \(a_1\), just set \(n = 1\) and solve. This approach is not only efficient but also provides a clear understanding of how the sequence progresses as \(n\) changes.
Each term is derived from a simple plug-and-calculate method, making explicit formulas extremely helpful for generating terms without needing previous ones. This method simplifies identifying patterns or behaviours within the sequence, and it's the starting point for deeper analyses, such as convergence.

Convergence Analysis

Convergence analysis examines whether a sequence settles on a permanent value as \(n\) becomes very large, or indefinitely progresses without such a resolution. To determine this, we need to analyze the behavior of the sequence \(a_n\) using its explicit formula.
The key idea is that a sequence converges if its terms approach a single defined limit as \(n\) increases towards infinity. For our exercise, we began with the sequence:

• \(a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}\)

During analysis, we simplified the sequence by dividing both numerator and denominator by \(n\), resulting in the expression:

• \(a_n = \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}}\)

As \(n\) tends towards infinity, the terms \(\frac{2}{n^2}\) and \(\frac{1}{n}\) vanish, simplifying the limit and making it much easier to evaluate. Hence, understanding convergence through simplification is essential in identifying the behavior of complex sequences.

Limit of a Sequence

The concept of the limit of a sequence is fundamental to understanding convergence.
The limit specifies the value that the terms of a sequence approach as \(n\) becomes infinitely large. In rigorous terms, if \(\lim_{{n \to \infty}} a_n = L\), then the sequence \(a_n\) converges and approaches the limit \(L\). In our case:

• The sequence \(a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}\) simplifies at the limit to \(\frac{\sqrt{3}}{2}\).

By calculating this limit, we've shown that the values of \(a_n\) get arbitrarily close to \(\frac{\sqrt{3}}{2}\) for very large \(n\), confirming the sequence converges to this value.
Limits are not only helpful for proving convergence but also play a crucial role in various areas of calculus and real analysis. They help provide a deeper understanding of the behavior and characteristics of sequences and functions as they are subjected to boundary conditions, like infinity.