Define (xn)n≥1 by x1 := 1 and xn+1 := (3xn + 3)/4 for n ≥ 1. Prove thatlimn→∞ xn = 3.(Hint: First show that xn ≤ 3 for all n ≥ 1.)
Question
Define (xn)n≥1 by x1 := 1 and xn+1 := (3xn + 3)/4 for n ≥ 1. Prove thatlimn→∞ xn = 3.(Hint: First show that xn ≤ 3 for all n ≥ 1.)
Solution
Sure, here is the step-by-step proof:
Step 1: Prove that xn ≤ 3 for all n ≥ 1.
We will use induction to prove this.
Base case: For n=1, x1 = 1 which is less than or equal to 3.
Inductive step: Assume that xn ≤ 3 for some n ≥ 1. We need to show that xn+1 ≤ 3.
xn+1 = (3xn + 3)/4 ≤ (3*3 + 3)/4 = 3.
So, the inductive step is proven. Hence, xn ≤ 3 for all n ≥ 1.
Step 2: Prove that the sequence (xn) is increasing.
Again, we will use induction to prove this.
Base case: x2 - x1 = (3*1 + 3)/4 - 1 = 1/2 ≥ 0. So, the sequence is increasing for n=1.
Inductive step: Assume that xn+1 - xn ≥ 0 for some n ≥ 1. We need to show that xn+2 - xn+1 ≥ 0.
xn+2 - xn+1 = (3xn+1 + 3)/4 - (3xn + 3)/4 = 3/4 * (xn+1 - xn) ≥ 0.
So, the inductive step is proven. Hence, the sequence (xn) is increasing.
Step 3: Prove that limn→∞ xn = 3.
Since the sequence (xn) is increasing and bounded above by 3, by the Monotone Convergence Theorem, it must converge to a limit L.
We have xn+1 = (3xn + 3)/4. Taking limits as n→∞ on both sides, we get L = (3L + 3)/4.
Solving this equation for L, we get L = 3. Hence, limn→∞ xn = 3.
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