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To solve the given problem, we have sequences \( \{a_n\} \) and \( \{b_n\} \) defined recursively as:

- Initial values: \( a_1 = 2, b_1 = 1 \)
- For \( n \geq 1 \): 
  \[
  a_{n+1} = \frac{a_n + b_n}{2}, \quad b_{n+1} = \frac{2a_n b_n}{a_n + b_n}
  \]

We need to show three things:

1. \( b_n \leq a_n \) for all \( n \).
2. \( b_{n+1} \geq b_n \) for all \( n \).
3. Both sequences \( \{a_n\} \) and \( \{b_n\} \) converge to the same limit, \( \sqrt{2} \).

### **Proof:**

#### **(i) Show that \( b_n \leq a_n \) for all \( n \)**:
We start by induction.

- **Base case**: For \( n = 1 \), we have \( a_1 = 2 \) and \( b_1 = 1 \). Clearly, \( b_1 \leq a_1 \).

- **Inductive step**: Suppose \( b_n \leq a_n \) for some \( n \). Now, we need to show that \( b_{n+1} \leq a_{n+1} \).

From the recursive definitions:

\[
a_{n+1} = \frac{a_n + b_n}{2}, \quad b_{n+1} = \frac{2a_n b_n}{a_n + b_n}
\]

Since \( b_n \leq a_n \), \( a_n + b_n \geq 2b_n \). This implies:

\[
\frac{a_n + b_n}{2} \geq b_n
\]

Also, for \( b_{n+1} \):

\[
b_{n+1} = \frac{2a_n b_n}{a_n + b_n} \leq \frac{2a_n b_n}{2b_n} = a_n \quad \Rightarrow \quad b_{n+1} \leq a_{n+1}
\]

Thus, by induction, \( b_n \leq a_n \) for all \( n \).

#### **(ii) Show that \( b_{n+1} \geq b_n \) for all \( n \)**:
We need to show that \( b_{n+1} - b_n \geq 0 \).

\[
b_{n+1} = \frac{2a_n b_n}{a_n + b_n}
\]

Consider the difference:

\[
b_{n+1} - b_n = \frac{2a_n b_n}{a_n + b_n} - b_n = \frac{2a_n b_n - b_n(a_n + b_n)}{a_n + b_n}
\]
\[
b_{n+1} - b_n = \frac{2a_n b_n - a_n b_n - b_n^2}{a_n + b_n} = \frac{a_n b_n - b_n^2}{a_n + b_n}
\]
\[
b_{n+1} - b_n = \frac{b_n(a_n - b_n)}{a_n + b_n}
\]

Since \( a_n \geq b_n \), it follows that \( a_n - b_n \geq 0 \), and hence \( b_{n+1} \geq b_n \).

#### **(iii) Show that both sequences \( \{a_n\} \) and \( \{b_n\} \) converge to the same limit \( \sqrt{2} \)**:
Let \( L \) be the common limit of \( \{a_n\} \) and \( \{b_n\} \).

- As \( n \to \infty \), \( a_n \approx a_{n+1} \approx L \) and \( b_n \approx b_{n+1} \approx L \).

From the recursive definitions:

\[
L = \frac{L + L}{2} = L \quad \text{and} \quad L = \frac{2L^2}{L + L} = \frac{2L^2}{2L} = \sqrt{2}
\]

Thus, both sequences converge to \( \sqrt{2} \).

---

If you'd like a more detailed breakdown of any part or have questions, let me know!

### **Relative Questions:**

1. What other methods can be used to analyze the convergence of recursive sequences?
2. Can we generalize this approach for other initial values of \( a_1 \) and \( b_1 \)?
3. How does the arithmetic mean-geometric mean inequality relate to these sequences?
4. What happens if we modify the recurrence relation for \( b_{n+1} \)?
5. How can this convergence be visualized graphically?

### **Tip**:
When analyzing recursive sequences, always start by establishing bounds and then investigate monotonicity and convergence properties.

Let a₁ = 2, b₁ = 1 and for n ≥ 1, aₙ₊₁ = (aₙ + bₙ)/2, bₙ₊₁ = (2aₙbₙ)/(aₙ + bₙ). Show that (i) bₙ ≤ aₙ for all n, (ii) bₙ₊₁ ≥ bₙ for all n, (iii) the sequences {aₙ} and {bₙ} converge to the same limit √2.

This problem involves recursively defined sequences {aₙ} and {bₙ}, where aₙ₊₁ = (aₙ + bₙ)/2 and bₙ₊₁ = (2aₙbₙ)/(aₙ + bₙ). The task is to prove that (i) bₙ ≤ aₙ for all n, (ii) bₙ₊₁ ≥ bₙ for all n, and (iii) both sequences converge to √2. Explore concepts such as inequalities, monotonicity, and limits.

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