Recall that the Fibonacci sequence (zn)n≥1 is defined byz1 = z2 = 1 and zn+2 = zn+1 + zn for all n ≥ 1.Define a sequence (xn)n≥1 byxn = znzn+1.3. Prove that the sequence (xn)n≥1 is (δ-)contractive for some δ ∈ (0, 1)

To prove that the sequence (xn)n≥1 is (δ-)contractive for some δ ∈ (0, 1), we need to show that there exists a δ such that for all n, |xn+1 - xn| ≤ δ |xn - xn-1|.

First, let's express xn and xn+1 in terms of the Fibonacci sequence:

xn = zn/zn+1 and xn+1 = zn+1/zn+2.

Now, let's compute |xn+1 - xn|:

|xn+1 - xn| = |zn+1/zn+2 - zn/zn+1| = |(zn+1^2 - znzn+2) / (zn+1zn+2)|.

We know that for the Fibonacci sequence, zn+1^2 - znzn+2 = -(-1)^n. So, |xn+1 - xn| = |-(-1)^n / (zn+1zn+2)| = 1/(zn+1*zn+2).

Next, let's compute |xn - xn-1|:

|xn - xn-1| = |zn/zn+1 - zn-1/zn| = |(zn^2 - zn-1zn+1) / (znzn+1)|.

We know that for the Fibonacci sequence, zn^2 - zn-1zn+1 = (-1)^(n-1). So, |xn - xn-1| = |-(-1)^(n-1) / (znzn+1)| = 1/(zn*zn+1).

Now, we can see that |xn+1 - xn| / |xn - xn-1| = (znzn+1) / (zn+1zn+2).

We know that zn+1/zn -> φ (the golden ratio) as n -> ∞. So, (znzn+1) / (zn+1zn+2) -> 1/φ.

Therefore, we can choose δ = 1/φ, which is in (0, 1), and the sequence (xn)n≥1 is (δ-)contractive.