Why is the fraction of a Fibonacci pair always the best estimation of the golden ratio in its range?

Question

It seems, that fib(n)/fib(n-1) is always the best approximation for the golden ratio you can get, when the numerator is smaller than fib(n) and denominator can be chosen arbitrarily.

fib(n) is the n'th Fibonacci number.

I tested this algorithmically, and also the statement seems intuitively right, but I lack the skills to prove it.

Is there any prove to this?

Edit: (because it was marked as duplicate) I know, that the limit of these Fibonacci pairs is converging to the golden ratio, and I also know the proof to it. But why are these pairs the best convergents?

Edit Nr2: There are "para-Fibonacci"-sequences like 1,3,4,7,11,18 that have the same convergent property, but converge slower than the original series. So the only fact, that it converges, says nothing about the speed of convergence.

These are the convergents to the continued fraction of the golden ratio. — Angina Seng, Dec 24 '17 at 11:58
Possible duplicate of Limit of the ratio of consecutive Fibonacci numbers — , Dec 24 '17 at 11:59
This is absolutely clear. But why are these the best convergent? — Art1st, Dec 24 '17 at 12:00
Those are general facts about continued fractions. I recommend Khinchin's short book "Continued Fractions" for a reference for all such results. — Wojowu, Dec 24 '17 at 12:05
@Wojowu There is also a good account in Hardy and Wright "An Introduction to the Theory of Numbers" — Mark Bennet, Dec 24 '17 at 12:34

score 1 · Accepted Answer · answered Dec 24 '17 at 14:50

The Fibonacci ratios are the continued fraction convergents to $(1+\sqrt{5})/2$. For any irrational number $\alpha$, each of its continued fraction convergents $p/q$ is the best rational approximation to $\alpha$ among all fractions (in reduced form) with denominator at most $q$. But for a positive integer $d$ that is not a denominator of a continued fraction convergent to $\alpha$, the best rational approximation to $\alpha$ with denominator at most $d$ need not be among the continued fraction convergents to $\alpha$ with denominator at most $d$. This point is often misunderstood.

Here is an example. The continued fraction convergents to $\sqrt{3}$ with denominator at most $10$ are 1, 2, 5/3, and 7/4, but the best rational approximation to $\sqrt{3}$ with denominator at most $10$ is not any of those fractions. It is 12/7.

It turns out that the best rational approximations to an irrational $\alpha$ up to an arbitrary bound on the denominator are found among the convergents and intermediate convergents to the continued fraction of $\alpha$. Find the definition of intermediate convergents on the Wikipedia page for continued fractions. The intermediate convergents to $\sqrt{3}$ include 12/7.

To build intermediate convergents to a standard continued fraction $[a_1,a_2,a_3,\ldots]$, where $a_i$ is a positive integer for $i>1$, you use positive integers less than $a_i$. The special thing about $(1+\sqrt{5})/2$ is that its continued fraction is $[1,1,1,1,\ldots]$, with each $a_i$ equal to $1$, so there are no intermediate convergents. Therefore the best rational approximations to $(1+\sqrt{5})/2$ are always its continued fraction convergents, which are the Fibonacci ratios.

Why is the fraction of a Fibonacci pair always the best estimation of the golden ratio in its range?

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