We are given the recursive sequence:

$x_1 = 2, \quad x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right)$

We are tasked with showing that $x_n$ converges to $\sqrt{2}$.

Step 1: Prove that the sequence is bounded and monotonic

1. Boundedness:

To show that $x_n$ is bounded, we need to show that there exists some constant $M$ such that $x_n \leq M$ for all $n$. We will prove this by induction.

• Base case (n = 1): We know that $x_1 = 2$, which is clearly greater than $\sqrt{2}$.

• Inductive step: Assume that $x_n \geq \sqrt{2}$ for some $n \geq 1$. Now, let's show that $x_{n+1} \geq \sqrt{2}$.

$x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right)$ For $x_n \geq \sqrt{2}$, observe that:

$x_n + \frac{2}{x_n} \geq \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$

Thus,

$x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right) \geq \frac{1}{2} \times 2\sqrt{2} = \sqrt{2}$

Hence, $x_{n+1} \geq \sqrt{2}$ whenever $x_n \geq \sqrt{2}$. Therefore, by induction, $x_n \geq \sqrt{2}$ for all $n$.

Next, let's show that $x_n$ is bounded above.

• For $x_1 = 2$, we can see that the sequence is decreasing and approaches $\sqrt{2}$. To verify this, we need to show that $x_{n+1} \leq 2$ for all $n$.

$x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right)$ We will show that if $x_n \leq 2$, then $x_{n+1} \leq 2$:

$x_n + \frac{2}{x_n} \leq 2 + \frac{2}{2} = 3$

Thus,

$x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right) \leq \frac{1}{2} \times 3 = 1.5$

This shows that $x_n$ remains bounded between $\sqrt{2}$ and $2$.

2. Monotonicity:

Next, we prove that the sequence is decreasing.

We know that:

$x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right)$

To show that $x_{n+1} \leq x_n$, we want to check if:

$\frac{1}{2} \left( x_n + \frac{2}{x_n} \right) \leq x_n$

Multiplying both sides by 2:

$x_n + \frac{2}{x_n} \leq 2x_n$

$\frac{2}{x_n} \leq x_n$

$2 \leq x_n^2$

This inequality is true if $x_n \geq \sqrt{2}$, which we have already established.

Step 2: Convergence

Now that we know the sequence $x_n$ is bounded and monotonic, it must converge. Let the limit of the sequence be $L$. Taking the limit of both sides of the recursive formula as $n \to \infty$:

$L = \frac{1}{2} \left( L + \frac{2}{L} \right)$

Multiplying both sides by 2:

$2L = L + \frac{2}{L}$

Subtract $L$ from both sides:

$L = \frac{2}{L}$

Multiplying both sides by $L$:

$L^2 = 2$

Thus, $L = \sqrt{2}$ (since $L \geq \sqrt{2}$).

Conclusion

We have shown that the sequence $x_n$ converges to $\sqrt{2}$.

Would you like further details or explanations? Here are some related questions for you to explore:

1. What is the behavior of the sequence for different initial values $x_1$?
2. How can we prove convergence using the Banach fixed-point theorem?
3. What are some other sequences similar to this, such as the Newton-Raphson method for finding square roots?
4. Can we generalize this sequence for finding the square root of any positive number?
5. How would the sequence behave if $x_1$ was less than $\sqrt{2}$?

Tip: If you're working with recursive sequences, always check for boundedness and monotonicity as key steps in proving convergence.