Question

In this problem, we are going to prove a sequence is convergent by several steps. Define a sequence {Fn​}n=0[infinity]​ such that F0​=1,F1​=1 and Fn+1​=Fn​+Fn−1​ for any n≥1. We are going to prove the sequence {Fn​Fn+1​​}n=1[infinity]​ is convergent to 21+5​​ (c) In fact, {Fn​Fn+1​​}n=1[infinity]​={A1​,B1​,A2​,B2​,⋯}. In other words, the sequences {An​}n=1[infinity]​ and {Bn​}n=1[infinity]​ are an alternating decomposition of a sequence {Fn​Fn+1​​}n=1[infinity]​ Now we can conclude the sequence {Fn​Fn+1​​}n=1[infinity]​ is convergent by (b) and the lemma in Q3. Let limn→[infinity]​Fn​Fn+1​​=L. Show that L=21+5​​.

Answer 1

The problem requires showing the sequence {Fn/Fn+1}, derived from the Fibonacci sequence, converges to L = 2/(1+√5). This proof might leverage Binet's formula and the properties of the Fibonacci numbers.

Step-by-step explanation:

The student is asking to show that the limit of the sequence {Fn/Fn+1} as n approaches infinity is L = 2/(1+√5). The sequence {Fn} is defined as Fibonacci sequence, with the initial values F0 = 1, F1 = 1, and the recursive relation Fn+1 = Fn + Fn-1 for n ≥ 1.

The convergence to the limit L is based on a lemma that was presumably proven earlier, which indicates alternating subsequences, {An} and {Bn}, converge to the same limit if certain conditions are met.

To show that L = 2/(1+√5), we utilize properties of the Fibonacci sequence and possibly the Binet's formula, which relates Fibonacci numbers to powers of the golden ratio, φ = (1+√5)/2.