Xn=√2+(Xn-1) show that Xn is convergent and find its limit

Given the recursive sequence defined by \( X_n = \sqrt{2 + X_{n-1}} \) for \( n \geq 1 \), with some initial value \( X_0 \), we are to show that the sequence \( \{X_n\} \) is convergent and find its limit.

### Step 1: Show that the sequence is bounded.

Let's assume that the sequence \( \{X_n\} \) converges to a limit \( L \). Then, taking the limit on both sides of the recursive equation, we have:

\[
L = \lim_{n \to \infty} X_n = \lim_{n \to \infty} \sqrt{2 + X_{n-1}} = \sqrt{2 + \lim_{n \to \infty} X_{n-1}} = \sqrt{2 + L}
\]

Thus, we obtain the equation:

\[
L = \sqrt{2 + L}
\]

### Step 2: Solve for the limit \( L \).

To find the value of \( L \), we square both sides:

\[
L^2 = 2 + L
\]

This can be rewritten as:

\[
L^2 - L - 2 = 0
\]

Now, we solve this quadratic equation using the quadratic formula:

\[
L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}
\]

This gives two possible solutions:

\[
L = \frac{4}{2} = 2 \quad \text{or} \quad L = \frac{-2}{2} = -1
\]

Since \( X_n \) is defined as the square root of a positive quantity, \( X_n \geq 0 \), we discard the negative solution \( L = -1 \). Therefore, the limit is:

\[
L = 2
\]

### Step 3: Show that the sequence is monotonic and bounded.

To show convergence, we also need to show that the sequence is monotonic (either increasing or decreasing) and bounded.

- **Boundedness:** We already know from the recursive definition that \( X_n \geq 0 \) for all \( n \). We will show that \( X_n \leq 2 \) for all \( n \).
  
  Assume \( X_0 \leq 2 \). Then:
  
  \[
  X_1 = \sqrt{2 + X_0} \leq \sqrt{2 + 2} = \sqrt{4} = 2
  \]
  
  By induction, if \( X_{n-1} \leq 2 \), then \( X_n = \sqrt{2 + X_{n-1}} \leq 2 \). Hence, the sequence is bounded above by 2.

- **Monotonicity:** Assume \( X_n \leq X_{n+1} \) (the sequence is increasing). Then:

  \[
  X_{n+1} = \sqrt{2 + X_n} \geq X_n
  \]

  Squaring both sides, we get:
  
  \[
  2 + X_n \geq X_n^2
  \]

  Rearranging:

  \[
  X_n^2 - X_n - 2 \leq 0
  \]

  This can be factored as:

  \[
  (X_n - 2)(X_n + 1) \leq 0
  \]

  Since \( X_n \geq 0 \), this inequality holds, implying \( X_n \) is increasing.

Thus, \( \{X_n\} \) is monotonic and bounded, so by the Monotone Convergence Theorem, the sequence converges. Since we already found that the limit must be \( L = 2 \), we conclude that:

\[
\lim_{n \to \infty} X_n = 2
\]

### Conclusion

The sequence \( X_n = \sqrt{2 + X_{n-1}} \) is convergent, and its limit is 2.

---

Do you want details or have any questions? Here are some related questions to explore:

1. What is the Monotone Convergence Theorem, and why is it important?
2. How does boundedness affect the convergence of a sequence?
3. Can we apply the same method to sequences defined by other recursive formulas?
4. What would happen if the initial value \( X_0 \) was different? Would the limit change?
5. Is there a general approach to finding the limit of recursively defined sequences?
6. Can the sequence \( X_n \) have more than one limit under different conditions?
7. How does the choice of the initial value \( X_0 \) affect the convergence speed?
8. How can you determine if a sequence is monotonic?

**Tip:** When dealing with recursive sequences, try to identify whether the sequence is monotonic and bounded. These properties are key to proving convergence.

Explore the convergence of the recursive sequence Xn = sqrt(2 + Xn-1). Understand how to prove convergence using the Monotone Convergence Theorem and find the limit of the sequence. Detailed step-by-step solution provided.

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