Problem 11 - 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 \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{e^{2 n}}{n^{2}+3 n-1}\)

Step-by-Step Solution

Answer

The first five terms are \(\frac{e^2}{3}, \frac{e^4}{9}, \frac{e^6}{17}, \frac{e^8}{27}, \frac{e^{10}}{39}\) and the sequence diverges.

Step 1: Calculate the First Term

For the first term, substitute \( n = 1 \) into the formula, \( a_1 = \frac{e^{2 \times 1}}{1^2 + 3 \times 1 - 1} = \frac{e^2}{3} \).

Step 2: Calculate the Second Term

Substitute \( n = 2 \) into the formula, \( a_2 = \frac{e^{2 \times 2}}{2^2 + 3 \times 2 - 1} = \frac{e^4}{9} \).

Step 3: Calculate the Third Term

Substitute \( n = 3 \) into the formula, \( a_3 = \frac{e^{2 \times 3}}{3^2 + 3 \times 3 - 1} = \frac{e^6}{17} \).

Step 4: Calculate the Fourth Term

Substitute \( n = 4 \) into the formula, \( a_4 = \frac{e^{2 \times 4}}{4^2 + 3 \times 4 - 1} = \frac{e^8}{27} \).

Step 5: Calculate the Fifth Term

Substitute \( n = 5 \) into the formula, \( a_5 = \frac{e^{2 \times 5}}{5^2 + 3 \times 5 - 1} = \frac{e^{10}}{39} \).

Step 6: Determine Convergence or Divergence

Analyze the behavior of the sequence as \( n \to \infty \). As \( n \to \infty \), the exponential term in the numerator, \( e^{2n} \), grows much faster than the polynomial term in the denominator, \( n^2 + 3n - 1 \). Therefore, \( a_n \to \infty \) and the sequence diverges.

Key Concepts

Convergence and Divergence in Sequences

The behavior of sequences in calculus is determined by whether they converge or diverge. In simple terms, convergence refers to a sequence approaching a specific value as the term number goes to infinity. Conversely, divergence means the sequence does not settle on a particular value and may increase or decrease without bound, or even oscillate indefinitely.

To determine if a sequence converges or diverges, we observe its behavior as the term number, often denoted as \( n \), approaches infinity. For instance, in our given sequence \( a_n = \frac{e^{2n}}{n^2 + 3n - 1} \), the terms include an exponential growth component \( e^{2n} \). Since exponential terms grow faster than polynomial terms in the denominator \( n^2 + 3n - 1 \), this sequence diverges because its terms grow larger without bound.

Understanding these concepts helps in identifying the long-term behavior of sequences, which is crucial in various calculus problems.

Limit of a Sequence

The concept of the limit of a sequence is central in determining sequence convergence. When a sequence converges, it approaches a limit, a stable, finite value as the sequence progresses indefinitely. We use the notation \( \lim_{n \to \infty} a_n = L \) to denote that as \( n \) becomes very large, \( a_n \) gets as close as desired to \( L \).

In our sequence \( a_n = \frac{e^{2n}}{n^2 + 3n - 1} \), the limit analysis shows that it does not converge to a finite value, since the exponential numerator dominates the polynomial denominator as \( n \) increases. Consequently, \( a_n \to \infty \), confirming the divergence we discussed earlier.

• Convergent sequences approach a fixed value.
• Divergent sequences do not have a finite limit.

Understanding limits is fundamental for calculus students, as it forms the basis of other complex analyses, such as continuity and derivatives.

Exponential Growth within Sequences

Exponential growth is a powerful concept where quantities grow rapidly, doubling or increasing by a constant ratio over time. In sequences, terms featuring exponential growth are often represented using the expression \( e^x \), where \( e \) is the base of natural logarithms approximately 2.718.

In the sequence \( a_n = \frac{e^{2n}}{n^2 + 3n - 1} \), the term \( e^{2n} \) in the numerator exemplifies exponential growth. This growth rate overpowers the polynomial terms \( n^2 + 3n - 1 \) as \( n \) increases. Exponential functions rise so quickly that they eventually outweigh other forms of growth, such as linear or polynomial.

• Exponential growth can lead to sequences diverging.
• Such growth is common in population models and financial contexts.

Recognizing exponential growth helps predict sequence behavior and is integral in formulating models that require swift progression or accumulation over time.

Explicit Formula in Sequences

An explicit formula in sequences provides a direct way to calculate any term in the sequence without referencing preceding terms. Given a sequence \( a_n \), an explicit formula allows us to plug in the position \( n \) to find the corresponding term value.

For example, in our sequence \( a_n = \frac{e^{2n}}{n^2 + 3n - 1} \), the explicit formula lets us easily compute \( a_1 \), \( a_2 \), \( a_3 \), and so on, without needing to calculate any previous terms. This method simplifies obtaining specific terms and aids in analyzing sequences for patterns of convergence or divergence.

• Explicit formulas simplify calculating terms in large sequences.
• They help identify the sequence’s general behavior quickly.

Having a grasp on explicit formulas is crucial for students dealing with long sequences and for constructing formulas where only the term’s positions vary.