Define (xn)n≥1 by x1 := 1 and xn+1 := (3xn + 3)/4 for n ≥ 1. Prove thatlimn→∞ xn = 3.(Hint: First show that xn ≤ 3 for all n ≥ 1.

To prove that $\lim_{n \to \infty} x_n = 3$, we will follow the steps outlined:

1. Break Down the Problem

We need to show two parts:

1. That $x_n \leq 3$ for all $n \geq 1$.
2. That $\lim_{n \to \infty} x_n = 3$.

2. Relevant Concepts

We start from the recursive definition given:

• $x_1 = 1$
• $x_{n+1} = \frac{3x_n + 3}{4}$

3. Analysis and Detail

Step 1: Prove that $x_n \leq 3$ for all $n \geq 1$.

• Base Case: For $n = 1$, $x_1 = 1 \leq 3$
• Inductive Step: Assume $x_n \leq 3$. We need to show that $x_{n+1} \leq 3$: $x_{n+1} = \frac{3x_n + 3}{4}$ Since $x_n \leq 3$, $x_{n+1} \leq \frac{3 \cdot 3 + 3}{4} = \frac{12}{4} = 3$ Thus, by induction, $x_n \leq 3$ for all $n \geq 1$.

Step 2: Show that $x_n$ is increasing and bounded above.

• Show $x_n$ is increasing: We need to show that $x_{n+1} \geq x_n$: $x_{n+1} - x_n = \frac{3x_n + 3}{4} - x_n = \frac{3x_n + 3 - 4x_n}{4} = \frac{-x_n + 3}{4}$ Since we have shown that $x_n \leq 3$, it follows that $-x_n + 3 \geq 0$, thus $x_{n+1} - x_n \geq 0$. Therefore, $x_n$ is increasing.

Since $(x_n)$ is increasing and bounded above by 3, by the Monotone Convergence Theorem, $\lim_{n \to \infty} x_n$ exists.

4. Verify and Summarize

Let $L = \lim_{n \to \infty} x_n$. Taking the limit on both sides of the recursive definition we have: $L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty}\left(\frac{3x_n + 3}{4}\right) = \frac{3L + 3}{4}$ Multiplying by 4, we get: $4L = 3L + 3 \implies L = 3$

Final Answer

Thus, $\lim_{n \to \infty} x_n = 3$.