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.

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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.