Consider the sequence {n} defined inductively by x₁ = 1, and Vn € N, Xn+1 = √√4xn +5. Prove that {n} converges, and find its limit. [Hint: See Example 2.5.8.] Example 2.5.8 Consider the sequence {n} defined inductively by ₁ = 1, and Vn € N, Xn+1 = √+2. Prove that {n} converges, and find its limit. Solution: Part 1-Prove that {n} converges. (a) First, we prove that {} is monotone increasing. We use mathematical induction to prove the following: Vn € N, 0≤ n ≤ n+1- (i) ₁ = 1 and ₂ =√1+2= √3. Thus, 0 ≤ ≤ ₂. (ii) Now, assume 0 ≤ x ≤ x+1. Then, 0≤ x + 2 ≤ J+1+2 0<√x+2 < √x+1+2 2+1 0 Ik+2. Therefore, by (i), (ii), and mathematical induction, VnN, 0≤n n+1. That is, (a) is monotone increasing. (b) Now, we prove that {} is bounded above. We use mathematical induction to prove the following: Vn € N, 2, 3. (i) = 1 so 1 ≤ 3. (ii) Now, assume 2 <3. Then +25 √ +2 < √5 Ik+1 <3. Therefore, by (i), (ii), and mathematical induction, Vn N, 2 ≤ 3. That is, (n) is bounded above by 3. (c) Therefore, by (a), (b), and the monotone convergence theorem, {} converges. 72-00 Part 2-find lim . By Part 1, we know that 3L= lima, R. We proceed to find L. Consider the defining equation: Xn+1 = √n+2. By squaring both sides, we have (+1)² = +2. Thus, lim (+1) lim (x+2). = 12-400 Applying the algebra of limits to both sides of this equation, L²=L+2 L²-L-2=0 (L-2)(L+1)=0 L = 2 or L = -1. Now, L -1 (see Theorem 2.3.12 (b) with K = 0). Thus, L= 2.