Problem 12 - 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 \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{e^{2 n}}{4^{n}}\)

Step-by-Step Solution

Answer

The sequence converges, and \( \lim_{n \to \infty} a_n = 0 \).

Step 1: Identify the Explicit Formula

The given explicit formula for the sequence is: \[ a_n = \frac{e^{2n}}{4^n} \] where \( n \) represents the term number in the sequence.

Step 2: Calculate the First Five Terms

To find the first five terms, substitute \( n = 1, 2, 3, 4, 5 \) into the formula:For \( n = 1 \):\( a_1 = \frac{e^{2 imes 1}}{4^1} = \frac{e^2}{4} \)For \( n = 2 \):\( a_2 = \frac{e^{2 imes 2}}{4^2} = \frac{e^4}{16} \)For \( n = 3 \):\( a_3 = \frac{e^{2 imes 3}}{4^3} = \frac{e^6}{64} \)For \( n = 4 \):\( a_4 = \frac{e^{2 imes 4}}{4^4} = \frac{e^8}{256} \)For \( n = 5 \):\( a_5 = \frac{e^{2 imes 5}}{4^5} = \frac{e^{10}}{1024} \)So, the first five terms are: \( \frac{e^2}{4}, \frac{e^4}{16}, \frac{e^6}{64}, \frac{e^8}{256}, \frac{e^{10}}{1024} \).

Step 3: Determine Sequence Behavior

The sequence behavior as \( n \) approaches infinity can be determined by analyzing the growth of \( e^{2n} \) relative to \( 4^n \). Since \( e^{2n} \) grows exponentially, but at a rate less than \( 4^n \) (since \( e^2 < 4 \)), the denominator grows faster, implying that the sequence values get smaller as \( n \) increases.

Step 4: Evaluate the Limit as \( n \) Approaches Infinity

Determine if the sequence converges by evaluating:\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{e^{2n}}{4^n} = \lim_{n \to \infty} \left(\frac{e^2}{4}\right)^n \]Since \( \frac{e^2}{4} < 1 \), this term approaches zero as \( n \to \infty \). Thus, the sequence converges to 0.

Key Concepts

Sequence Terms

In a sequence, each element is called a **term**, and understanding these terms can help illustrate the behavior of the whole sequence. When you have an explicit formula like \( a_n = \frac{e^{2n}}{4^n} \), it essentially gives you a direct way to find any term without needing to know the others beforehand. This gives you the flexibility to calculate any individual value of a sequence. Here, you simply plug in values for \( n \) to get the terms. For the first five terms, if you substitute \( n \) with 1, 2, 3, 4, and 5 respectively, you get:

• For \( n = 1 \): \( a_1 = \frac{e^2}{4} \)
• For \( n = 2 \): \( a_2 = \frac{e^4}{16} \)
• For \( n = 3 \): \( a_3 = \frac{e^6}{64} \)
• For \( n = 4 \): \( a_4 = \frac{e^8}{256} \)
• For \( n = 5 \): \( a_5 = \frac{e^{10}}{1024} \)

These terms show how the sequence progresses and can suggest patterns or trends. Notice how each term is formed by a number in the numerator growing exponentially with 2, while the number in the denominator grows exponentially with 4.

Explicit Formula

The **explicit formula** is a mathematical expression that directly connects each term in a sequence with its position. This contrasts with a recursive formula, which requires calculating each term based on previous ones. For our sequence, the explicit formula is \( a_n = \frac{e^{2n}}{4^n} \).
Within this formula, the numerator \( e^{2n} \) shows how the number \( e \), the base of natural logarithms, is raised to an increasingly higher power. In parallel, the denominator, \( 4^n \), increases faster since its base is larger than the effective base \( e^2 \) of the numerator under repeated multiplication.
This formula provides a clear structure to predict the growth or decay pattern of terms, a vital aspect for analyzing sequence behavior. Since it lays out the growth operation explicitly, it's handy for both generating terms and identifying long-term trends, like convergence.

Limit Evaluation

**Limit evaluation** is crucial for understanding what happens to a sequence as its terms progress towards infinity. In mathematical terms, you're often looking to find out what the terms "converge" to. If the terms approach a specific value, the sequence is said to converge; if not, it diverges.
For the sequence \( a_n = \frac{e^{2n}}{4^n} \), evaluating the limit as \( n \) approaches infinity involves checking the behavior of the fraction \( \left(\frac{e^2}{4}\right)^n \).
Here, since \( \frac{e^2}{4} < 1 \), raising it to increasingly higher powers results in increasingly smaller values, eventually approaching zero. This behavior indicates that the sequence converges, with the limit evaluated as zero. Understanding this limit involves understanding the relationship between the growth rates of the numerator and denominator in the explicit formula; when the denominator grows faster, the term itself shrinks, leading to convergence to zero. This concept emphasizes the importance of comparing growth rates when assessing long-term behavior in sequences.