Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(1, \frac{1}{1-\frac{1}{2}}, \frac{1}{1-\frac{2}{3}}, \frac{1}{1-\frac{3}{4}}, \ldots\)

Step-by-Step Solution

Answer

The sequence diverges because \(a_n = n+1\) increases indefinitely.

Step 1: Identify the Pattern in the Sequence

Given the sequence elements: \(1, \frac{1}{1-\frac{1}{2}}, \frac{1}{1-\frac{2}{3}}, \frac{1}{1-\frac{3}{4}}, \ldots\), let's find the pattern in the denominators. Observe that the general term in the denominator is of the form \(1 - \frac{n}{n+1}\).

Step 2: Simplify the General Term

Simplify the general term: \(1 - \frac{n}{n+1} = \frac{n+1-n}{n+1} = \frac{1}{n+1}\). Thus, the nth term is \(a_n = \frac{1}{\frac{1}{n+1}} = n+1\).

Step 3: Determine Convergence or Divergence

To check if the sequence converges, determine the limit of \(a_n = n+1\) as \(n\) goes to infinity. As \(n\) increases, \(a_n = n+1\) also increases without bound, indicating that the sequence diverges.

Step 4: Conclusion

Since \(a_n = n+1\) and it tends to infinity as \(n\) increases, the sequence diverges.

Key Concepts

Explicit Formulas for Sequences

An explicit formula for a sequence provides a direct way to calculate any term without recourse to previous ones. To derive such a formula, observe the recurring pattern in the given sequence elements. In the provided example, we have a sequence with terms like \(1, \frac{1}{1-\frac{1}{2}}, \frac{1}{1-\frac{2}{3}}, \frac{1}{1-\frac{3}{4}}, \ldots\).
By identifying the pattern, we can express the denominator as \(1 - \frac{n}{n+1}\).
From here, simplifying, \(1 - \frac{n}{n+1} = \frac{1}{n+1}\).
Consequently, the nth term can be represented as
- \(a_n = \frac{1}{\frac{1}{n+1}} = n+1\).
This formula allows us to easily compute any term in the sequence.

• Pattern recognition: Spotting similarities among sequence terms is essential.
• Simplification: Break down complex fractions to find the explicit formula.
• Representation: Express the sequence succinctly for any integer \(n\).

Infinite Limits

Understanding limits as they approach infinity is crucial in determining whether a sequence converges or diverges. In sequences, convergence refers to approaching a specific number, while divergence means there is no limit as \(n\) increases. Observe \(a_n = n+1\) for our sequence.
As \(n\) becomes very large, \(n+1\) also grows without limit, effectively heading towards infinity.
Since the sequence grows indefinitely, it does not settle on a particular value—that's a clear sign of divergence.

• Concept of Infinity: Visualize infinity as growth without bounds.
• Convergence vs Divergence: A converging sequence nears a specific number; here, however, \(a_n\) diverges.
• Limit Assessment: Calculate \(\lim_{n \to \infty} a_n\); if infinity results, divergence occurs.

Sequence Simplification

Simplification often plays a vital role in recognizing a sequence's behavior. It's about reducing expressions to their simplest form, helping us better understand the properties of a sequence.
In the sequence provided, \(a_n = \frac{1}{1 - \frac{n}{n+1}}\) initially seems complex but simplifies to \(n + 1\).
Here is how:

• Observe and simplify: \(1 - \frac{n}{n+1}=\frac{1}{n+1}\).
• Calculate \(a_n = \frac{1}{\frac{1}{n+1}}\), resulting in \(n+1\).
• Simplified forms offer straightforward methods for computing terms or identifying sequence behavior.

This simplification process emphasizes breaking down complicated terms into easily manageable components, which is particularly useful when more complex patterns arise.