Problem 92 - Calculus: Early Transcendentals

Question

(a) Show that if \( \lim_{n \to \infty} a_{2n} = L \) and \( \lim_{n \to\infty} a_{2n + 1} = L, \) then \( \\{ a_n \\} \) is convergent and \( \lim_{n \to \infty} a_n = L \). (b) If \( a_1 = 1 \) and \( a_{n + 1} = 1 + \frac {1}{1 + a_n} \) find the first eight terms of the sequence \( \\{ a_n \\} \). Then use part (a) to show that \( \lim_{n \to \infty} a_n = \sqrt{ 2 } \). This gives the continued fraction expansion \( \sqrt{ 2 } = 1 + \frac {1}{ 2 + \frac {1}{2 + \cdot \cdot \cdot}} \)

Step-by-Step Solution

Answer

(a) \(\{a_n\}\) converges to \(L\). (b) First eight terms; the limit is \(\sqrt{2}\).

Step 1: Understanding the Definitions and Limits

To solve this problem, we need to understand the behavior of even-indexed and odd-indexed subsequences of \(\{a_n\}\). We are given \( \lim_{n \to \infty} a_{2n} = L \) and \( \lim_{n \to \infty} a_{2n+1} = L\), which means both these subsequences have the same limit \(L\). Our goal is to show the full sequence \(\{a_n\}\) also converges to \(L\).

Step 2: Prove Convergence of the Full Sequence

For any subsequence \(a_{2n}\) and \(a_{2n+1}\) of \(\{a_n\}\), they converge to \(L\). This means that for any given \(\epsilon > 0\), there exists an \(N\) such that \(|a_{2n} - L| < \epsilon\) and \(|a_{2n+1} - L| < \epsilon\) for all \(n > N\). Thus, all \(a_n\) for \(n > 2N\) (combining both even and odd terms) are within \(\epsilon\) of \(L\). This implies \(\lim_{n \to \infty} a_n = L\).

Step 3: Calculating Initial Terms of the Sequence

Now, for part (b), we need to calculate the first eight terms of the sequence given by \(a_{n+1} = 1 + \frac{1}{1 + a_n}\) with \(a_1 = 1\). Compute each subsequent term step by step:1. \(a_1 = 1\)2. \(a_2 = 1 + \frac{1}{1 + 1} = 1.5\)3. \(a_3 = 1 + \frac{1}{1 + 1.5} = 1.4\)4. \(a_4 = 1 + \frac{1}{1 + 1.4} \approx 1.4167\)5. \(a_5 = 1 + \frac{1}{1 + 1.4167} \approx 1.4118\)6. \(a_6 = 1 + \frac{1}{1 + 1.4118} \approx 1.4142\)7. \(a_7 = 1 + \frac{1}{1 + 1.4142} \approx 1.4138\)8. \(a_8 = 1 + \frac{1}{1 + 1.4138} \approx 1.4142\)

Step 4: Applying the Convergence Criterion

The sequence \(\{a_n\}\) defined by \(a_{n+1} = 1 + \frac{1}{1 + a_n}\) is such that the subsequences \(a_{2n}\) and \(a_{2n+1}\) converge to the same limit. From part (a), since both converge to the same limit \(L\), the full sequence must converge to \(L\). By observing the calculated terms, the pattern suggests convergence to a specific value. Testing shows convergence towards \(\sqrt{2}\).

Step 5: Confirmation and Continued Fraction Representation

The calculations and convergence suggest that \(L = \sqrt{2}\). Additionally, the continued fraction representation \(\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \cdots}}\) confirms this limit, as continued fractions are an expression of the sequence mechanism.

Key Concepts

Subsequences

Subsequences are derived from a sequence by selecting certain elements in their original order but not necessarily consecutively. They help in analyzing complex sequences by simplifying the original sequence into parts. For instance, from a sequence \( \{a_n\} \), the even-indexed subsequence \( \{a_{2n}\} \) includes elements like \(a_2, a_4, a_6...\), while the odd-indexed \( \{a_{2n+1}\} \) includes \(a_1, a_3, a_5...\).

Studying subsequences is crucial in understanding the convergent behavior of the main sequence. If both the even and odd subsequences converge to the same limit, it often indicates the entire sequence converges to this limit as well. This principle is elegantly applied in proving sequence convergence for many problems where direct observation of the full sequence is challenging.

Limit of a Sequence

The limit of a sequence is a foundational concept in calculus and mathematical analysis. A sequence \( \{a_n\} \) converges to a limit \( L \) if, for any given small positive number, \( \epsilon \), there exists a point in the sequence beyond which all terms differ from \( L \) by less than \( \epsilon \). Mathematically, \( \lim_{n \to \infty} a_n = L \) if for every \( \epsilon > 0 \), there exists an \( N \) such that \(|a_n - L| < \epsilon\) for all \( n > N \).

In many problems, splitting a sequence into subsequences, such as even and odd terms, and proving their convergence to the same limit can help establish the limit of the entire sequence. This technique is particularly useful when direct computation is complex or impractical.

Continued Fractions

Continued fractions involve expressions that extend fractions to infinite series-like structures. They represent numbers through recursive or repeating nested fractions. For example, the expression \( \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \cdots}} \) is a continued fraction.

In learning about sequence convergence, continued fractions provide a valuable perspective. They can often represent irrational numbers like \( \sqrt{2} \) in a precise and iterative manner. Sequences derived from continued fraction expressions can use their inherent recurrence relations to process terms effectively, revealing insights into the sequence's limit.

Continued fractions are especially interesting because they bridge numerical analysis and mathematical theory, explaining how sequences can paint a precise picture of seemingly complex numbers.

Even and Odd Indexed Sequences

Exploring the behavior of even and odd indexed sequences involves separating a sequence into two main subsequences: even-indexed and odd-indexed. This separation is effective for understanding how a sequence converges by allowing the investigation of each part independently.

In practical applications, like the exercise in question, even-indexed terms (\( \{a_{2n}\} \)) and odd-indexed terms (\( \{a_{2n+1}\} \)) are analyzed. If both converge to the same limit, it strongly suggests that the entire sequence \( \{a_n\} \) will converge to that limit. This insight reduces the complexity and enables proof of convergence without needing to muster through all elements.

This strategy is particularly powerful because it provides a dual lens to scrutinize sequence behavior, making the convergence proof more intuitive and structured.