Problem 3 - 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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{4 n^{2}+2}{n^{2}+3 n-1} $$

Step-by-Step Solution

Answer

The first five terms are 2, 2, \(\approx 2.235\), \(\approx 2.444\), \(\approx 2.615\). The sequence converges to 4 as \(n\) approaches infinity.

Step 1: Write the First Five Terms of the Sequence

To find the first five terms of the sequence \( \{a_n\} \), substitute \(n = 1, 2, 3, 4, 5\) into the given explicit formula \(a_n = \frac{4n^2 + 2}{n^2 + 3n - 1}\):- For \(n = 1\), \(a_1 = \frac{4(1)^2 + 2}{(1)^2 + 3(1) - 1} = \frac{6}{3} = 2\).- For \(n = 2\), \(a_2 = \frac{4(2)^2 + 2}{(2)^2 + 3(2) - 1} = \frac{18}{9} = 2\).- For \(n = 3\), \(a_3 = \frac{4(3)^2 + 2}{(3)^2 + 3(3) - 1} = \frac{38}{17} \approx 2.235\).- For \(n = 4\), \(a_4 = \frac{4(4)^2 + 2}{(4)^2 + 3(4) - 1} = \frac{66}{27} \approx 2.444\).- For \(n = 5\), \(a_5 = \frac{4(5)^2 + 2}{(5)^2 + 3(5) - 1} = \frac{102}{39} \approx 2.615\).

Step 2: Attempt direct substitution

Try substituting the value the variable approaches directly into the expression.

Step 3: Handle indeterminate forms

If direct substitution gives an indeterminate form, apply L'Hopital's Rule, algebraic manipulation, or other techniques.

Step 4: Evaluate the limit

Compute the final value of the limit.

Step 5: State the conclusion

Express the final answer.

Step 6: Conclude with the answer

The first five terms are 2, 2, \(\approx 2.235\), \(\approx 2.444\), \(\approx 2.615\). The sequence converges to 4 as \(n\) approaches infinity.

Key Concepts

Sequence Terms

When dealing with sequences in mathematics, a sequence is simply an ordered list of numbers, each number being called a "term". In this exercise, we are given an explicit formula for the sequence terms: \[ a_n = \frac{4n^2 + 2}{n^2 + 3n - 1} \]By using this formula, we can generate the terms of the sequence by plugging in different values of \( n \). For example, to find the first five terms of this particular sequence, we substitute the values \( n = 1, 2, 3, 4, \) and \( 5 \) into the formula, calculating each term one by one.

• For \( n = 1 \), the term is \( a_1 = 2 \).
• For \( n = 2 \), the term is also \( a_2 = 2 \).
• For \( n = 3 \), the term is approximately \( a_3 \approx 2.235 \).
• For \( n = 4 \), the term is approximately \( a_4 \approx 2.444 \).
• For \( n = 5 \), the term is approximately \( a_5 \approx 2.615 \).

This process illustrates how sequence terms can vary or remain constant depending on the structure of the sequence formula.

Limits in Calculus

In calculus, understanding the behavior of a sequence as \( n \) approaches infinity is essential. This involves the concept of a "limit". The limit of a sequence \( \{a_n\} \) is a value that the terms of the sequence approach as \( n \) becomes very large.Limits are crucial for determining whether sequences converge or diverge. For a sequence to converge, it means that the terms are getting closer to a specific number, known as the limit. In our example, we look for \[ \lim_{n \to \infty} a_n \]To find this, we apply calculus to analyze the structure of the formula, sometimes simplifying it to recognize its dominant behavior as \( n \) grows. Practically, if the highest power terms in the numerator and the denominator dominate, we can simplify the expression:\[ \lim_{n \to \infty} \frac{4n^2 + 2}{n^2 + 3n - 1} = \frac{4}{1} = 4 \]Thus, the limit here is 4, indicating that the sequence converges to the value 4 as \( n \) increases.

Divergence of Sequences

A sequence is said to diverge if the terms do not approach any single value as \( n \) becomes very large. Instead of getting closer to one number, they might increase or decrease without bound, or oscillate between different values without settling down.In our exercise, we determined convergence by identifying that the sequence approaches a limit. However, if we discovered that the terms did not settle towards any finite number, we would declare the sequence as divergent.

• If the expression within the sequence grows infinitely or decreases infinitely, the sequence diverges.
• In some cases, if terms fluctuate erratically, it can indicate divergence.
• Divergence suggests the absence of a specific pattern of stabilization as \( n \) increases.

It's important to analyze the sequence characteristics, and the limits technique often helps in verifying if a sequence diverges or converges. For sequences that appear erratic or grow indefinitely, careful application of limit calculation will reveal this behavior.