Let F ( x ) be the Fibonacci function such that F ( 0 ) = F ( 1 ) = 1 and F ( x ) = F ( x − 1 ) + F ( x − 2 ) for x ≥ 2 .
Find
x → ∞ lim F ( x ) F ( x + 1 )
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@Gia Hoàng Phạm , sorry, I have change the words of your question without asking you.
The Greeks thought that in hanging a picture, the golden ratio should be used to determine how far from the floor the nail should be inserted. Ed Gray
It converges to the golden ratio ϕ = 2 1 + 5
The relation between the golden ratio and fibonacci is given by the formula F n = 5 ϕ n − ϕ − n . This is called Binet's Formula.
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But your formula should be F n = ⌊ 5 ϕ n − ϕ − n ⌋
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We note that F ( x ) = F x − 1 , where F n is the n th Fibonacci number defined as F 0 = 0 , F 1 = 1 and F n = F n − 1 + F n + 2 for n ≥ 2 . Then we have:
L = x → ∞ lim F ( x ) F ( x + 1 ) = x → ∞ lim F x − 1 F x = x → ∞ lim φ x − 1 − ( − φ ) − ( x − 1 ) φ x − ( − φ ) − x = x → ∞ lim 1 − ( − φ ) − 2 ( x − 1 ) φ − ( − φ ) − 2 x + 1 = φ = 2 1 + 5 ≈ 1 . 6 1 8 Since F n = 5 φ n − ( − φ ) − n , where φ = 2 1 + 5 is the golden ratio. Divide up and down by φ x − 1 .