Problem 70 - Calculus: Early Transcendentals

Question

(a) If \( \left \\{ a_n \right\\} \) is convergent, show that \( \displaystyle\lim_{n\to\infty} a_{n+1} = \displaystyle\lim_{n\to\infty} a_n \) (b) A sequence \( \left\\{ a_n \right\\} \) is defined by \( a_1 = 1 \) and \( a_{n + 1} = 1/(1 + a_n) \) for \( n \ge 1. \) Assuming that \( \left\\{ a_n \right\\} \) is convergent, find its limit.

Step-by-Step Solution

Answer

The sequence \( \{ a_n \} \) converges to \( \frac{-1 + \sqrt{5}}{2} \).

Step 1: Understanding Convergence

If a sequence \( \{ a_n \} \) is convergent, it means that there exists a real number \( L \) such that \( \lim_{n\to\infty} a_n = L \). This implies that as \( n \) becomes very large, the terms of the sequence \( a_n \) become arbitrarily close to \( L \).

Step 2: Applying the Convergence Definition to Shifted Terms

Given that \( \{ a_n \} \) converges to \( L \), consider \( a_{n+1} \). Since convergence implies eventual stabilizing behavior, \( a_{n+1} \) should also be arbitrarily close to \( L \) as \( n \) grows. Hence, \( \lim_{n\to\infty} a_{n+1} = L \). Thus: \( \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} a_n \).

Step 3: Analyzing the Sequence Definition

The sequence \( \{ a_n \} \) is defined by the recurrence relation \( a_{n+1} = \frac{1}{1 + a_n} \) with the initial term \( a_1 = 1 \). We want to find the limit \( L \) assuming the sequence converges.

Step 4: Setting the Limit Equation

Assume the sequence converges to some limit \( L \). Therefore, both \( \lim_{n\to\infty} a_n = L \) and \( \lim_{n\to\infty} a_{n+1} = L \). Using the recurrence relation, set up the limit equation: \( L = \frac{1}{1 + L} \).

Step 5: Solving the Limit Equation

Multiplying both sides of the equation \( L = \frac{1}{1 + L} \) by \( 1 + L \), we get \( L(1 + L) = 1 \). Simplifying, this becomes \( L^2 + L - 1 = 0 \).

Step 6: Using the Quadratic Formula

The equation \( L^2 + L - 1 = 0 \) is a quadratic equation. Use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), \( c = -1 \).

Step 7: Computing the Roots

Substitute into the quadratic formula: \( L = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2 \times 1} \). Simplify to find \( L = \frac{-1 \pm \sqrt{5}}{2} \). Since \( a_n \) is positive, choose the positive root: \( L = \frac{-1 + \sqrt{5}}{2} \).

Step 8: Conclusion

The limit of the sequence \( \{ a_n \} \) is \( \frac{-1 + \sqrt{5}}{2} \) under the assumption of convergence.

Key Concepts

Sequence Convergence

Understanding sequence convergence is crucial in calculus. When we say a sequence converges, we mean that it approaches a specific value, called the limit, as the index goes to infinity. Consider a sequence \( \{ a_n \} \) that converges to a limit \( L \). As \( n \) (the index) increases, the terms \( a_n \) get closer and closer to this number \( L \). This implies that the difference between \( a_n \) and \( L \) becomes negligible as \( n \) grows very large, mathematically expressed as \( \lim_{n\to\infty} a_n = L \).

Furthermore, if \( \{a_n\} \) converges, any subsequent term \( a_{n+1} \) will also tend towards \( L \). Thus, \( \lim_{n\to\infty} a_{n+1} = L \). This concept of stability and settling around a central value makes convergence a key part of understanding the behavior of sequences over time.

Recurrence Relations

Recurrence relations represent sequences where each term is formulated based on the preceding terms, thus establishing a fascinating link between the data points. In our exercise, the sequence \( \{ a_n \} \) is defined by a recurrence relation: \( a_{n+1} = \frac{1}{1 + a_n} \). This means each term is derived from its previous term. The initial term is given as \( a_1 = 1 \), and from here, all subsequent terms can be computed iteratively.

To understand this sequence's long-term behavior, we hypothesize that it converges to a limit \( L \). By substituting \( L \) into the recurrence relation like so: \( L = \frac{1}{1 + L} \), we set up a mathematical equation representing the system's steady state. Solving such an equation often gives insights into the sequence's eventual pattern or limit.

Quadratic Formula

The quadratic formula is an indispensable tool for solving quadratic equations, those of the form \( ax^2 + bx + c = 0 \). In the sequence's context, after setting up the limit equation, \( L = \frac{1}{1 + L} \), we manipulated it into a quadratic equation: \( L^2 + L - 1 = 0 \).

Solving this requires using the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \). Plugging these values into the formula provides potential limits: \( L = \frac{-1 \pm \sqrt{5}}{2} \). Since we consider positive terms, we choose the solution that fits this criterion, giving us \( L = \frac{-1 + \sqrt{5}}{2} \). The quadratic formula finely extracts the potential values which offer insights into the behavior of the recurring sequence based on the initial conditions and the recurrence relation.