Given: - \( a_1 = 1 \) - The recursive sequence: \( a_{n+1} = \sqrt{2 + a_n} \) We need to **prove that the sequence \( \{a_n\} \) converges**. *Hint: Show that the sequence is increasing and bounded above.* --- ### **Step 1: Show that the sequence is increasing** We want to show \( a_{n+1} > a_n \) for all \( n \). #### **Base Case:** - \( a_1 = 1 \) - \( a_2 = \sqrt{2 + a_1} \) - \( a_2 = \sqrt{2 + 1} \) - \( a_2 = \sqrt{3} \) - Since \( \sqrt{3} \approx 1.732 > 1 \), \( a_2 > a_1 \) #### **Inductive Step:** Assume \( a_n > a_{n-1} \) for some \( n \geq 1 \). Consider \( a_{n+1} = \sqrt{2 + a_n} \) and \( a_n = \sqrt{2 + a_{n-1}} \) Since the function \( f(x) = \sqrt{2 + x} \) is **increasing** (the derivative \( f'(x) = \frac{1}{2\sqrt{2 + x}} > 0 \)), if \( a_n > a_{n-1} \), then \( a_{n+1} = f(a_n) > f(a_{n-1}) = a_n \). **Conclusion:** By induction, the sequence is increasing. --- ### **Step 2: Show that the sequence is bounded above** We want to find an upper bound \( M \) such that \( a_n < M \) for all \( n \). Suppose \( a_n < 2 \) for all \( n \). **Base Case:** - \( a_1 = 1 < 2 \) **Inductive Step:** Assume \( a_n < 2 \). - \( a_{n+1} = \sqrt{2 + a_n} < \sqrt{2 + 2} \) - \( a_{n+1} < \sqrt{4} \) - \( a_{n+1} < 2 \) So by induction, \( a_n < 2 \) for all \( n \). --- ### **Step 3: Conclude convergence** The sequence \( \{a_n\} \) is **increasing** and **bounded above**. By the **Monotone Convergence Theorem**, an increasing and bounded sequence converges. --- ### **Step 4: Find the limit (optional)** Let the limit be \( L \): - \( L = \sqrt{2 + L} \) - Square both sides: \( L^2 = 2 + L \) - Rearranged: \( L^2 - L - 2 = 0 \) - Factor: \( (L - 2)(L + 1) = 0 \) - Solutions: \( L = 2 \) or \( L = -1 \) Since the terms are all positive and less than 2, the limit is \( \boxed{2} \). --- ### **Final Answer** **The sequence \( \{a_n\} \) is increasing and bounded above, so it converges. The limit of the sequence is \( 2 \).**