Problem 19 - 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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\left(1+\frac{2}{n}\right)^{n / 2} $$

Step-by-Step Solution

Answer

The sequence converges to \( e \).

Step 1: Write the First Term

To find the first term of the sequence, substitute \(n = 1\) into the explicit formula: \[a_1 = \left( 1 + \frac{2}{1} \right)^{1/2} = \left( 3 \right)^{1/2} = \sqrt{3}\]

Step 2: Calculate the Second Term

Substitute \(n = 2\) into the formula to find the second term: \[a_2 = \left( 1 + \frac{2}{2} \right)^{2/2} = \left( 2 \right)^1 = 2\]

Step 3: Determine the Third Term

For the third term, substitute \(n = 3\) into the explicit formula: \[a_3 = \left( 1 + \frac{2}{3} \right)^{3/2} = \left( \frac{5}{3} \right)^{3/2}\]

Step 4: Compute the Fourth Term

Substitute \(n = 4\) into the formula for the fourth term: \[a_4 = \left( 1 + \frac{2}{4} \right)^{4/2} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}\]

Step 5: Find the Fifth Term

Find the fifth term by substituting \(n = 5\) into the formula: \[a_5 = \left( 1 + \frac{2}{5} \right)^{5/2} = \left( \frac{7}{5} \right)^{5/2}\]

Step 6: Analyze Convergence

For very large \(n\), the sequence \(a_n = \left(1 + \frac{2}{n}\right)^{n/2}\) approaches a well-known limit: \[\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x\]Therefore, \[\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^{n/2} = e^{2/2} = e\] Thus, the sequence converges to \(e\).

Key Concepts

Understanding the Explicit Formula

An explicit formula is a direct way to define the terms in a sequence. Instead of generating a sequence by referring to previous terms, you use a specific formula to calculate each term directly by inserting a position number (usually denoted by \( n \)). In our exercise, the explicit formula is:
\[ a_n = \left( 1 + \frac{2}{n} \right)^{n/2} \]
This formula determines each term \(a_n\) based on the value of \(n\). By substituting different values of \(n\), we can compute the specific term. For instance, for the first term, substitute \(n = 1\):

• \(a_1 = \left( 1 + \frac{2}{1} \right)^{1/2} = \sqrt{3}\)

Explicit formulas make it easier for students to identify and compute terms directly, especially when dealing with sequences that are complex or follow non-obvious patterns.

Delving into Sequence Limits

Sequence limits help us understand the behavior of sequences as the number of terms grows indefinitely, or when \(n\) approaches infinity. Determining the limit of a sequence tells us whether the sequence converges—that is, if it approaches a particular finite value.
In the exercise, we are given:
\[a_n = \left(1 + \frac{2}{n}\right)^{n/2}\]
To analyze its convergence, we look at the sequence's limits as \(n \to \infty\). This form resembles a renowned limit in calculus:

• \[ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x \]

Here, substituting \(x = 1\), results in:

• \[ \lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^{n/2} = e^{2/2} = e \]

Hence, the sequence indeed converges, and its limit is the mathematical constant \(e\). Understanding sequence limits is crucial in various calculus applications, as they reveal the stability or instability of a sequence over the long term.

The Role of Calculus Concepts

Calculus concepts are fundamental when analyzing sequences, especially for evaluating convergence and limits. Concepts from calculus such as limits, differentiation, and integration provide powerful tools to dissect sequences and conclude their behavior as we extend them toward infinity.
In this exercise, limits are crucial. The limit helps determine if a sequence converges and, if so, to what value. Recognizing the pattern that resembles the exponential limit function, noted as:

• \[ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x \]

is essential to finding the convergence of the sequence.
Beyond mere theoretical interest, these tools are practical in real-life problems, such as evaluating series in physics or economics. Grasping these concepts lays a robust foundation for solving complex mathematical problems and enhances reasoning skills across different scientific domains.