Question

The sequence {Fn} described by F0 = 1, F1 = 1, and Fn+2 = Fn+Fn+1, if n ≥ 0, is called a Fibonacci sequence. Its terms occur naturally in many botanical species, particularly those with petals or scales arranged in the form of a logarithmic spiral. Consider the sequence {xn}, where xn = Fn+1/Fn. Assuming that lim --> infinity, xn=x exists, show that x=(1+underrot5)/2.

Answer 1

Answer:

See proof below

Step-by-step explanation:

Let L be the limit of Xn when n goes to infinity. We will use the recursive formula for the Fibonacci numbers to find L. Bur first, we need to guarantee that our operations with the formula make sense:

Notice that Fn>0 for all n≥0. Then Fn+2 = Fn+1 + Fn > Fn+1. Hence, dividing by Fn+1, Xn+1= Fn+2/Fn+1 > 1 for all n≥0, that is, Xn>1 for all n≥1. Taking the limit on both sides of the inequality, L≥1,, thus 1/L exists and equal the limit of the sequence 1/Xn=Fn/Fn+1 (by laws of limits).

To find L, divide by Fn+1 in Fn+2 = Fn + Fn+1 to get

Fn+2/Fn+1 = Fn/Fn+1 +1.

This equation can be written as Xn+1= 1/Xn +1. Take the limit in both sides to get L=1/L +1. Then L²=1+L, and L²-L-1=0,

Solve for L with the quadratic formula to get:

$L=\frac{1+\sqrt{1+4}}{2},L=\frac{1-\sqrt{1+4}}{2}$

We discard the second solution because it is negative, and we proved above that L>0. Hence $L=\frac{1+\sqrt{5}}{2}$