Problem: Kettenwurz – Recursive Definition

The recursive sequence is defined as:

• $x_1 = \sqrt{2}$,
• $x_{n+1} = \sqrt{2 + x_n}$.

We aim to:

1. Prove that the sequence $\{x_n\}$ converges to a limit.
2. Calculate the limit.

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Step 1: Existence of the Limit

a) Monotonicity

We first show that $\{x_n\}$ is monotonically increasing:

• Base case ($n = 1$): $x_2 = \sqrt{2 + \sqrt{2}} > \sqrt{2} = x_1$.
• Assume $x_n > x_{n-1}$. Then: $x_{n+1} = \sqrt{2 + x_n} > \sqrt{2 + x_{n-1}} = x_n.$ Hence, $x_{n+1} > x_n$. By induction, the sequence is monotonically increasing.

b) Boundedness

We show that $x_n$ is bounded above by 2:

• Base case ($n = 1$): $x_1 = \sqrt{2} < 2$.
• Assume $x_n < 2$. Then: $x_{n+1} = \sqrt{2 + x_n} \leq \sqrt{2 + 2} = \sqrt{4} = 2.$ Hence, $x_{n+1} < 2$. By induction, $x_n \leq 2$.

Since $\{x_n\}$ is monotonically increasing and bounded above, it converges to a limit $L$.

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Step 2: Calculating the Limit

Let $L$ be the limit of $\{x_n\}$. Taking the limit on both sides of the recursive relation $x_{n+1} = \sqrt{2 + x_n}$, we have: $L = \sqrt{2 + L}.$ Squaring both sides: $L^2 = 2 + L.$ Rearranging: $L^2 - L - 2 = 0.$ Factoring: $(L - 2)(L + 1) = 0.$ Thus: L = 2 \quad \text{(since L > 0)}.

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Final Answer

The sequence $\{x_n\}$ converges, and the limit is: $\boxed{L = 2}.$

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Let me know if you would like further clarification or examples.
Here are five related questions you might find useful:

1. What are the properties of convergent sequences in general?
2. How does one prove monotonicity for more complex sequences?
3. Can this method be extended to other recursively defined sequences?
4. What happens if the initial value $x_1$ is changed? Does the sequence still converge?
5. How can we use numerical methods to approximate the limit of sequences?

Tip: For recursive sequences, always check monotonicity and boundedness to ensure convergence.