Problem 58 - Essential Calculus Early Transcendentals

Question

(a) A sequence \(\left\\{a_{n}\right\\}\) is defined recursively by the equation \(a_{n}=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)\) for \(n \geqslant 3,\) where \(a_{1}\) and \(a_{2}\) can be any real numbers. Experiment with various values of \(a_{1}\) and \(a_{2}\) and use your calculator to guess the limit of the sequence. (b) Find \(\lim _{n \rightarrow \infty} a_{n}\) in terms of \(a_{1}\) and \(a_{2}\) by expressing \(a_{n+1}-a_{n}\) in terms of \(a_{2}-a_{1}\) and summing a series.

Step-by-Step Solution

Answer

The limit is \( \frac{a_1 + a_2}{2} \).

Step 1: Understand the Recursive Formula

The sequence \( \{a_{n}\} \) is defined by a recursive relationship \( a_{n} = \frac{1}{2}(a_{n-1} + a_{n-2}) \) for \( n \geq 3 \). This means each term is the average of the two preceding terms. The initial terms \(a_1\) and \(a_2\) are given as any real numbers.

Step 2: Experiment and Calculate

Try different values for \( a_1 \) and \( a_2 \) using a calculator to find subsequent terms of the sequence. Let’s take an example: \( a_1 = 1 \) and \( a_2 = 3 \). Calculate \( a_3 = \frac{1}{2}(3 + 1) = 2 \), \( a_4 = \frac{1}{2}(2 + 3) = 2.5 \), and continue this for more terms to observe the trend.

Step 3: Observe Pattern and Convergence

Notice the sequence \( a_n \) seems to converge to a value. For varying inputs of \( a_1 \) and \( a_2 \), calculate several terms to see that differences between successive terms decrease, suggesting convergence.

Step 4: Find Limit in Terms of Initial Conditions

Suppose \( L \) is the limit of the sequence. Then, as \( n \to \infty \), both \( a_{n-1} \to L \) and \( a_{n-2} \to L \), giving \( L = \frac{1}{2}(L + L) \). Solving gives \( L = L \), showing consistency but not dependency on specific initial values. Use another approach: calculate \( a_{n+1} - a_{n} = \frac{1}{2}(a_{n} - a_{n-2}), \) to form a sum of differences telescoping to 0.

Step 5: Express Limit Explicitly

Using summation and telescoping properties, represent \( a_{n} \) in terms of the sum of initial differences. As \( n \to \infty \), \( a_n \) converges to \( a_2 - a_1 \). Therefore, \( \lim_{n \to \infty} a_n = \frac{a_1 + a_2}{2} \).

Key Concepts

Sequence Convergence

In mathematics, sequence convergence is a fundamental concept that describes how a sequence of numbers approaches a specific value as the terms progress infinitely. For a sequence to converge, the terms must get arbitrarily close to a certain number called the limit as you go further and further into the sequence. The closer the terms get to this number, the more we say the sequence is converging.To explore convergence, consider the given sequence defined by the recursive formula: \( a_{n} = \frac{1}{2}(a_{n-1} + a_{n-2}) \) for \( n \geq 3 \). The initial terms \( a_1 \) and \( a_2 \) are any real numbers. By experimenting with these starting points, you'll notice that over time the sequence stabilizes, suggesting it converges.When we say a sequence converges, it means after several terms, the sequence looks like it has settled towards a particular value. Although it never actually reaches this "limit" completely, it gets closer with each additional term.

Limit of a Sequence

The limit of a sequence is the value that the terms of a sequence "tend" to as the number of terms becomes infinitely large. For our example, we can find the limit by considering the behavior of \(a_n\) as \(n \to \infty\).If we assume that \(L\) is the limit of the sequence, it implies that \(a_{n}, a_{n-1}, a_{n-2} \rightarrow L\) as \(n \rightarrow \infty\). By substituting into the recursive formula, we have:\[ L = \frac{1}{2}(L + L) \]Solving this gives the trivial equation \(L = L\), which doesn't provide information about \(a_1 \) and \(a_2\). Therefore, we use the differences between terms and telescoping properties showing that the sequence actually tends towards the average of the initial terms. Thus, we can conclude:\[ \lim_{n \to \infty} a_n = \frac{a_1 + a_2}{2} \]

Telescoping Series

A telescoping series is a type of series where many terms cancel each other out within the sum, leading to a simplified expression. This concept is particularly useful for finding limits of recursively defined sequences like the one in our original problem.In this exercise, the sequence is modeled by a recursion relation. To discover its behavior, you calculate \(a_{n+1} - a_n\), resulting in \(\frac{1}{2}(a_n - a_{n-2})\), which is a form that reveals a telescoping effect when summed over successive terms.In a telescoping series, as you consider more and more terms, intermediate values cancel out, often leaving a simple relationship or just boundary terms. For the recursive sequence, when summed, this technique ultimately shows that the influence of initial term differences diminishes, settling the sequence towards the average value of the starting terms. Hence, telescoping reveals:\[ \lim_{n \to \infty} a_n = \frac{a_1 + a_2}{2} \] Understanding this process helps simplify the analysis, making it easier to determine the limit without needing daunting computations.