Here’s a quick demonstration of a connection between the Fibonacci sequence and geometric sequences.

The famous Fibonacci sequence starts out 1, 1, 2, 3, 5, 8, 13 … The first two terms are both 1, then each subsequent terms is the sum of the two preceding terms.

A **generalized Fibonacci sequence** can start with any two numbers and then apply the rule that subsequent terms are defined as the sum of their two predecessors. For example, if we start with 3 and 4, we get the sequence 3, 4, 7, 11, 18, 29, …

Let φ be the **golden ratio**, the positive solution to the equation 1 + x = x^{2}. Let φ’ be the conjugate golden ratio, the negative solution to the same quadratic equation. Then φ = (1 + √ 5)/2 and φ’ = (1 – √ 5)/2.

Now let’s look at a generalized Fibonacci sequence starting with 1 and φ. Then our terms are 1, φ, 1 + φ, 1 + 2φ, 2 + 3φ, 3 + 5φ, … Let’s see whether we can simplify this sequence.

Now 1 + φ = φ^{2} because of the quadratic equation φ satisfies. That tells us the third term equals φ^{2}. So our series starts out 1, φ, φ^{2}. This is looking like a geometric sequence. Could the fourth term be φ^{3}? In fact, it is. Since the fourth term is the sum of the second and third terms, it equals φ +φ^{2} = φ(1 + φ) = φ(φ^{2}) = φ^{3}. You can continue this line of reasoning to prove that the generalized Fibonacci sequence starting with 1 and φ is in fact the **geometric sequence **1, φ, φ^{2}, φ^{3}, …

Now start a generalized Fibonacci sequence with φ’. Because φ’ is also a solution to 1 + x = x^{2}, it follows that the sequence 1, φ’, 1 + φ’, 1 + 2φ’, 2 + 3φ’, … equals the geometric sequence 1, φ’, (φ’)^{2}, (φ’)^{3}, …

**Related posts**:

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Fibonacci numbers at work (advanced)

Another, and perhaps more direct, relationship between the original Fibonacci sequence F_1, F_2, … and the golden ratio is that the limit of F_{n+1}/F_n as n goes to infinity is φ. This is quite easy to prove given the observation you have already made about the golden ratio.

Another beautiful property of the golden ratio that you can arrive at easily is that it has, in some sense, the simplest continued fraction expansion: [1; 1, 1, ...].

@Mark the reason why the ratio of successive members of the Fibonacci series tends towards the golden ratio is because the sequence can be expressed as a linear combination of the above geometric sequences of phi and phi’. Specifically, there is an explicit formula for elements in the Fibonacci series, being: F(n) = phi^n/sqrt(5) – (-phi)^(-n)/sqrt(5) (this is also detailed in the “Fibonacci numbers at work” post), note that (-phi)^(-n) = (phi’)^n, so this translates to (phi^n – phi’^n)/sqrt(5)

Thus, the ratio of F(n+1) to F(n) is equal to (phi^(n+1) – phi'(^n+1))/(phi^n – phi’^n). Since abs(phi’) < 1, phi’^n will trend towards 0 as n gets larger, and this ratio will trend toward phi^(n+1)/phi^n, which is just phi.

@Wedge, that’s one way to prove it but the approach I had in mind doesn’t need to invoke an extra proposition about Fibonacci sequence (though that is yet another nice relationship between them and the golden ratio).

Notice that R_{n+1} := F_{n+1} / F_n = (F_n + F_{n-1}) / F_n = 1 + F_{n-1}/F_n = 1 + 1/R_n. The limit R of the sequence R_n is a fixed point of this recurrence. That is, R = 1 + 1/R and so R^2 = R + 1 which is exactly the equation defining φ.