Problem 69 - Single Variable Calculus

Question

Consider the sequence \(\left\\{a_{n}\right\\}\) defined by \(a_{1}=\sqrt{2}\) and \(a_{n}=\sqrt{2+a_{n-1}}\) for \(n \geq 2\). Assuming that the sequence converges, find its limit. Note: Using the principle of mathematical induction, it can be shown that \(\left\\{a_{n}\right\\}\) is increasing and bounded by 2 and, hence, by Theorem 6 is convergent.

Step-by-Step Solution

Answer

The limit of the sequence \(\left\{a_{n}\right\}\) defined by \(a_{1}=\sqrt{2}\) and \(a_{n}=\sqrt{2+a_{n-1}}\) for \(n \geq 2\) is \(L = 1 + \sqrt{3} \approx 2.732\).

Step 1: Establish the Limit of the Sequence

Let's denote the limit by L. That is, if the sequence \(\left\{a_{n}\right\}\) converges, we will have \(L=\lim_{n \rightarrow \infty} a_{n}\).

Step 2: Use the Recursive Formula

According to the recursive formula given, if the sequence converges, then the equation \(L = \sqrt{2 + L}\) must hold.

Step 3: Solve for the Limit

We will now solve this equation for L. If we square both sides of the equation, we find \(L^{2} = 2 + L\), which simplifies to \(L^{2} - L - 2 = 0\).

Step 4: Using quadratic formula

The quadratic formula is used to solve this quadratic equation. The roots of the equation are given by \(L=\frac{1\pm \sqrt{1+8}}{2}\), which simplifies to \(L=1\pm \sqrt{3}\).

Step 5: Discard the Negative Solution

The limits for sequences of real numbers are real numbers and the given sequence is increasing and bounded above by 2, so we only consider the positive root, \(L = 1 + \sqrt{3}\).

Step 6: Confirm the Limit

It must be checked that the limit \(L = 1 + \sqrt{3}\) doesn't exceed 2 in order to be within the sequence's bound. Indeed, \(1 + \sqrt{3} \approx 2.732\), hence, \(L = 1 + \sqrt{3}\) is the limit of the given sequence.

Key Concepts

Recursive Sequences

A recursive sequence is a sequence in which each term is defined based on the terms that come before it. Our given sequence, \(\{a_{n}\}\), is a classic example. It starts with \(a_1 = \sqrt{2}\), and each subsequent term is defined by the rule \(a_n = \sqrt{2 + a_{n-1}}\).

Key points to understand recursive sequences include:

• Initial Value: You need a starting point, like \(a_1 = \sqrt{2}\).
• Recursive Formula: This describes how to generate the terms, as shown by \(a_n = \sqrt{2 + a_{n-1}}\).
• Behavior Over Time: It’s essential to consider how the sequence behaves as it progresses, which often involves examining its limit or whether it converges.

Recursive sequences can have interesting properties, such as convergence. They allow us to build complex sequences from simple rules.

Mathematical Induction

Mathematical induction is a powerful method used to prove that a statement holds for all natural numbers. In our sequence problem, induction helps us show that the sequence \(\{a_{n}\}\) is both increasing and bounded by 2.

The induction process involves two main steps:

• Base Case: Verify the statement for the first term. For example, showing \(a_1 = \sqrt{2}\) is less than 2.
• Inductive Step: Assume the statement is true for some \(n = k\), then prove it for \(n = k+1\). This would typically involve demonstrating that \(a_{k+1} = \sqrt{2 + a_k}\) is greater than \(a_k\) but still less than 2.

Using induction, we ensure the truth of the statement across the entire sequence, aiding us in confirming the sequence’s convergence.

Quadratic Formula

The quadratic formula is a vital tool for solving equations of the form \(ax^2 + bx + c = 0\). In our sequence, after setting the limit equation \(L = \sqrt{2 + L}\), and squaring both sides, we derive \(L^2 - L - 2 = 0\).

The quadratic formula is:

• \[L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For our equation, \(a = 1\), \(b = -1\), and \(c = -2\), so:

• \[L = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2}\]
• This simplifies to \(L = \frac{1 \pm 3}{2}\), giving us two potential roots: \(L = 2\) and \(L = -1\).

Considering the sequence is increasing and must be less than or equal to 2, we only accept the appropriate positive root.

Limits of Sequences

The limit of a sequence is the value that the sequence approaches as the number of terms goes to infinity. In mathematical terms, for sequence \(\{a_n\}\), if \(\lim_{n \to \infty} a_n = L\), then \(L\) is its limit.

In our specific problem:

• The sequence is given by a recursive formula and is showed to be increasing and bounded. Hence it converges according to a theorem in sequence theory.
• The recursive nature leads us to solve \(L = \sqrt{2 + L}\) for \(L\), resulting in two solutions: \(L = 2\) and \(L = -1\).

By analyzing the sequence's properties and constraints, we identify that \(L = 2\) satisfies the problem's conditions. This sophisticated balancing act between limits, sequence behavior, and boundary conditions emphasizes the elegance of limits. It's key in understanding the stability and end-value of sequences.