Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(n = 1\) in the formula: \[ a_1 = \frac{\sqrt{3(1)^2 + 2}}{2(1) + 1} = \frac{\sqrt{3 + 2}}{2 + 1} = \frac{\sqrt{5}}{3}. \] Thus, the first term is \(\frac{\sqrt{5}}{3}\).

Step 2: Find the Second Term

To find the second term \(a_2\), substitute \(n = 2\) in the formula: \[ a_2 = \frac{\sqrt{3(2)^2 + 2}}{2(2) + 1} = \frac{\sqrt{12 + 2}}{4 + 1} = \frac{\sqrt{14}}{5}. \] Thus, the second term is \(\frac{\sqrt{14}}{5}\).

Step 3: Find the Third Term

To find the third term \(a_3\), substitute \(n = 3\) in the formula: \[ a_3 = \frac{\sqrt{3(3)^2 + 2}}{2(3) + 1} = \frac{\sqrt{27 + 2}}{6 + 1} = \frac{\sqrt{29}}{7}. \] Thus, the third term is \(\frac{\sqrt{29}}{7}\).

Step 4: Find the Fourth Term

To find the fourth term \(a_4\), substitute \(n = 4\) in the formula: \[ a_4 = \frac{\sqrt{3(4)^2 + 2}}{2(4) + 1} = \frac{\sqrt{48 + 2}}{8 + 1} = \frac{\sqrt{50}}{9} = \frac{5\sqrt{2}}{9}. \] Thus, the fourth term is \(\frac{5\sqrt{2}}{9}\).

Step 5: Find the Fifth Term

To find the fifth term \(a_5\), substitute \(n = 5\) in the formula: \[ a_5 = \frac{\sqrt{3(5)^2 + 2}}{2(5) + 1} = \frac{\sqrt{75 + 2}}{10 + 1} = \frac{\sqrt{77}}{11}. \] Thus, the fifth term is \(\frac{\sqrt{77}}{11}\).

Step 6: Determine Convergence or Divergence

Examine the behavior of \(a_n\) as \(n\) approaches infinity: \[ a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}. \] Dividing the numerator and denominator by \(n\), we get: \[ a_n = \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}}. \] As \(n\) approaches infinity, \(\frac{2}{n^2}\) and \(\frac{1}{n}\) both approach 0, leading to: \[ a_n \approx \frac{\sqrt{3}}{2}. \] Thus, the sequence converges.

Step 7: Find the Limit

Since the sequence converges, the limit is: \[ \lim_{n \to \infty} a_n = \frac{\sqrt{3}}{2}. \] Thus, the limit of the sequence as \(n\) approaches infinity is \(\frac{\sqrt{3}}{2}\).

Key Concepts

Explicit Formula

An explicit formula provides a clear mathematical expression that helps easily find any term in a sequence without needing the terms before it. In sequences, this is very practical as it saves time and reduces error chances. You simply plug in the value of the term number you're interested in, and out comes your result.
For the sequence given in our exercise, the explicit formula is:

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

This formula is a recipe that helps calculate any term \(a_n\) in the sequence by substituting \(n\) with the term number you want, like \(n = 1\) for the first term, \(n = 2\) for the second term, etc. Whether you're searching for the first or the hundredth term, you use the same straightforward steps. This concise way of defining a sequence helps in understanding the structure and pattern without needing prior terms.

Limit of a Sequence

The limit of a sequence is what values the terms of the sequence approach as the number of terms goes up to infinity. It is an essential concept in calculus and plays a crucial role in understanding the behavior of sequences over the long term. In our given sequence, as \(n\) becomes very large, the sequence approaches a particular number—a process known as finding the limit.

In this sequence, as \(n\) approaches infinity, the sequence simplifies as follows:

• \( a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1} \)
• Divide both numerator and denominator by \(n\): \( \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}} \)
• As \(n\) goes to infinity, both \(\frac{2}{n^2}\) and \(\frac{1}{n}\) approach 0.
• Thus the limit becomes \(\lim_{n \to \infty} a_n = \frac{\sqrt{3}}{2} \).

Knowing the limit helps predict the behavior of sequences, especially when \(n\) becomes very large.

Convergence and Divergence

Understanding whether a sequence converges or diverges helps in determining the long-term behavior of the sequence. Sequences that converge approach a specific value, known as the limit, as \(n\) increases indefinitely.

• A sequence is convergent if its terms get arbitrarily close to a certain number as \(n\) becomes large.
• If the sequence terms do not approach any specific value, it is termed divergent.

In our particular sequence:

• As \(n\) increases, the sequence converges towards \(\frac{\sqrt{3}}{2}\).
• This means it behaves predictably as \(n\) grows, settling around the value \(\frac{\sqrt{3}}{2}\).

Recognizing convergence or divergence is crucial in calculus sequences, ensuring the mathematical analysis does not hit unforeseen boundaries.

Calculus Sequences

Calculus provides tools to analyze sequences, especially when examining their limits or behavioral tendencies. Calculus sequences involve the study of functions defined in terms of a sequence, often requiring manipulation of algebraic expressions to evaluate convergence or divergence.

• The process of finding limits aids in simplifying and understanding these sequences.
• Calculus uses concepts like derivatives and integrals to extend the study beyond mere sequences to series and beyond.

In our exercise:

• We used algebraic simplification to understand the behavior of \( a_n \) as \( n \rightarrow \infty \).
• The same principles apply when extending to more complex functions.

The insights gained from calculus help predict and explain how different sequences behave, especially as they grow very large.