Let's solve ** 1** step by as requested. ## 1 Sequence Defined Rec Given:\( a_ = \sqrt{2} \) \( a_{n+1} = \sqrt{2 + a_n} \) for \( n \geq 1 \) --- ### (a) Show that \( a_n < 2 \) for all \( n \) #### **Step 1: Base Case** - For \( n = 1 \), \( a_1 = \sqrt{2} < 2 \). #### **Step 2: Inductive Step** - Assume \( a_n < 2 \). - Then \( a_{n+1} = \sqrt{2 + a_n} < \sqrt{2 + 2} = \sqrt{4} = 2 \). - So, by induction, \( a_n < 2 \) for all \( n \). **Explanation (Step 1 and 2):** We check the first term and use induction to show the inequality holds for all terms. --- ### (b) Show that the sequence \( (a_n) \) is increasing #### **Step 1: Show \( a_{n+1} > a_n \)** - Need to show: \( \sqrt{2 + a_n} > a_n \). - Square both sides (valid since all terms are positive): \[ 2 + a_n > a_n^2 \] \[ a_n^2 - a_n - 2 < \] \[ (a_n - 2)(a_n + 1) < \] - Since \( a_n < 2 \) and \( a_n > \), \( a_n - 2 < \), \( a_n + 1 > \), so the product is negative. **Explanation:** By showing the difference is negative, we prove \( a_{n+1} > a_n \); thus, the sequence is increasing. --- ### (c) The sequence is bounded and increasing. Find the limit \( L \). #### **Step 1: Let \( L = \lim_{n \to \infty} a_n \).** - Then \( L = \sqrt{2 + L} \). #### **Step 2: Solve for \( L \)** - Square both sides: \[ L^2 = 2 + L \] \[ L^2 - L - 2 = \] \[ (L - 2)(L + 1) = \] So, \( L = 2 \) or \( L = -1 \). - Since all \( a_n > \), the limit is \( L = 2 \). --- ## **Final Answer** - **a)** \( a_n < 2 \) for all \( n \) (by induction). - **b)** \( (a_n) \) is increasing (as shown by algebraic manipulation). - **c)** The sequence converges to \( \boxed{2} \).

# Solution to the Given Problem ## Given Recurrence Relation The sequence is defined as follows: - \( a_1 = \sqrt{2} \) - \( a_{n+1} = \sqrt{2 + a_n} \) for \( n \geq 1 \) We will solve the problem in three parts: proving the sequence is bounded, showing it is increasing, and finding its limit. --- ## Part (a): Show that \( a_n < 2 \) for all \( n \) ### Step 1: Base Case - For \( n = 1 \): \[ a_1 = \sqrt{2} < 2 \] ### Step 2: Inductive Step - Assume \( a_n < 2 \) for some \( n \). - Then, \[ a_{n+1} = \sqrt{2 + a_n} < \sqrt{2 + 2} = \sqrt{4} = 2 \] - By induction, \( a_n < 2 \) for all \( n \). **Explanation:** We verified the base case and used mathematical induction to confirm the inequality holds for all terms in the sequence. --- ## Part (b): Show that the sequence \( (a_n) \) is increasing ### Step 1: Show \( a_{n+1} > a_n \) - We need to prove: \[ \sqrt{2 + a_n} > a_n \] - Squaring both sides (valid since all terms are positive): \[ 2 + a_n > a_n^2 \] - Rearranging gives: \[ a_n^2 - a_n - 2 < 0 \] - Factoring the quadratic: \[ (a_n - 2)(a_n + 1) < 0 \] - Since \( a_n < 2 \) and \( a_n > -1 \), the product is negative, thus: \[ a_{n+1} > a_n \] **Explanation:** The inequality shows that each term is greater than the previous one, proving the sequence is increasing. --- ## Part (c): Find the limit \( L \) ### Step 1: Let \( L = \lim_{n \to \infty} a_n \) - Then, substituting the limit into the recurrence relation: \[ L = \sqrt{2 + L} \] ### Step 2: Solve for \( L \) - Square both sides: \[ L^2 = 2 + L \] - Rearranging gives: \[ L^2 - L - 2 = 0 \] - Factoring: \[ (L - 2)(L + 1) = 0 \] - The solutions are \( L = 2 \) or \( L = -1 \). Since \( a_n > 0 \) for all \( n \), we have: \[ L = 2 \] **Explanation:** The limit was determined by solving the recurrence relation, confirming the limit is 2, as \( a_n \) remains positive. --- ## Final Answer - **(a)** \( a_n < 2 \) for all \( n \) (by induction). - **(b)** The sequence \( (a_n) \) is increasing (as shown by algebraic manipulation). - **(c)** The limit of the sequence is \( \boxed{2} \).