Let 0 < b1 < a1 and define the sequences (an) and (bn) by an+1 = (an+bn)/2 and bn+1 = √anbn
for all n ∈ N. Show that both (an) and (bn) converge

To show that the sequences \( (a_n) \) and \( (b_n) \) converge, we can proceed with the following steps:

### 1. Monotonicity of the Sequences

#### Sequence \( (a_n) \):

The sequence \( (a_n) \) is defined by the recursive formula:
\[
a_{n+1} = \frac{a_n + b_n}{2}
\]
Since \( b_n \) is less than \( a_n \) (i.e., \( b_n < a_n \)), the arithmetic mean \( a_{n+1} \) will be less than or equal to \( a_n \). Thus:
\[
a_{n+1} = \frac{a_n + b_n}{2} < \frac{a_n + a_n}{2} = a_n
\]
So, \( (a_n) \) is a decreasing sequence.

#### Sequence \( (b_n) \):

The sequence \( (b_n) \) is defined by:
\[
b_{n+1} = \sqrt{a_n b_n}
\]
Since the geometric mean is less than or equal to the arithmetic mean, we have:
\[
\sqrt{a_n b_n} \leq \frac{a_n + b_n}{2} = a_{n+1}
\]
Given that \( a_n > b_n \), it follows that \( b_{n+1} \geq b_n \). Thus, \( (b_n) \) is an increasing sequence.

### 2. Boundedness of the Sequences

#### Sequence \( (a_n) \):

The sequence \( (a_n) \) is decreasing and \( b_1 > 0 \). Since \( b_n \) is increasing and \( b_n \leq a_n \), it follows that \( a_n \) is bounded below by \( b_1 \). Therefore, \( (a_n) \) is bounded.

#### Sequence \( (b_n) \):

The sequence \( (b_n) \) is increasing and \( b_n < a_n \). Since \( (a_n) \) is bounded, so is \( (b_n) \).

### 3. Convergence of the Sequences

Both sequences \( (a_n) \) and \( (b_n) \) are monotonic and bounded. Therefore, they must converge.

Let \( L_1 \) and \( L_2 \) be the limits of the sequences \( (a_n) \) and \( (b_n) \), respectively:
\[
\lim_{n \to \infty} a_n = L_1 \quad \text{and} \quad \lim_{n \to \infty} b_n = L_2
\]
Taking the limit in the recursive definitions:
\[
L_1 = \frac{L_1 + L_2}{2}
\]
\[
L_2 = \sqrt{L_1 L_2}
\]

### 4. Equating the Limits
From \( L_1 = \frac{L_1 + L_2}{2} \), we get:
\[
2L_1 = L_1 + L_2 \quad \Rightarrow \quad L_1 = L_2
\]
So \( L_1 = L_2 = L \).

From \( L_2 = \sqrt{L_1 L_2} \), we also have:
\[
L = \sqrt{L \cdot L} \quad \Rightarrow \quad L = L
\]
Thus, both sequences converge to the same limit \( L \).

Therefore, the sequences \( (a_n) \) and \( (b_n) \) converge to the same limit \( L \).

---

Would you like further details or have any questions?

Here are some related questions to think about:
1. How does the convergence rate of \( (a_n) \) compare to \( (b_n) \)?
2. Can you determine an explicit formula for the limit \( L \) in terms of \( a_1 \) and \( b_1 \)?
3. What happens if \( b_1 > a_1 \)? Will the sequences still converge?
4. Is there a generalization of this process for sequences defined with different means?
5. How does this problem relate to the arithmetic-geometric mean inequality?

**Tip:** Understanding monotonicity and boundedness is crucial when determining the convergence of sequences.

Explore the convergence of sequences (an) and (bn) defined by recursive formulas. Understand the monotonicity and boundedness of these sequences, leading to their convergence. Includes detailed steps and proofs.

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