The Golden Ratio, i.e.

$\varphi = \frac{1+\sqrt{5}}{2}$

and Fibonacci sequence, i.e.

$F_n=F_{n-1}+F_{n-2}$ with the initial conditions $F_0=0$ and $F_1=1$

are clearly connected, but not perfectly so, and I'm seeking to understand this. I've been trying to read up a bit to understand the similarities and differences between these 2.

A Live Science article, for instance, says:

Around 1200, mathematician Leonardo Fibonacci discovered the unique properties of the Fibonacci sequence. This sequence ties directly into the Golden ratio because if you take any two successive Fibonacci numbers, their ratio is very close to the Golden ratio. As the numbers get higher, the ratio becomes even closer to 1.618. For example, the ratio of 3 to 5 is 1.666. But the ratio of 13 to 21 is 1.625. Getting even higher, the ratio of 144 to 233 is 1.618. These numbers are all successive numbers in the Fibonacci sequence.

Or, as another source put it:

The quotient of any Fibonacci number and it's predecessor approaches Phi, represented as ϕ (1.618), the Golden ratio.

Based on these descriptions, it sounds like the the ratio of consecutive Fibonacci numbers and the Golden Ratio converge asymptotically but are not identical (especially with the initial ratios). I would like to understand this a little bit better.

Asides from Live Science and Google, I did some preliminary research on Mathematics SE and Cross Validated. The closest question I could find was an unanswered one which primarily focused on the relationship between Arctangents and the Fibonacci sequence.

  • $\begingroup$ youtube.com/watch?v=dTWKKvlZB08 might help. $\endgroup$
    – user451844
    Sep 1, 2017 at 15:43
  • $\begingroup$ @RoddyMacPhee Thanks, I will watch it right now. $\endgroup$
    – Hack-R
    Sep 1, 2017 at 15:43
  • $\begingroup$ The ratio of two successive Fibonacci numbers is not equal to $\phi$, but it gets closer and closer to $\phi$ when $n\rightarrow \infty$. That's what "the ratio converges asymptotically to $\phi$" means. In order to help you "understand this a little bit better", maybe we would need to know your mathematical background a bit better. Ever worked on sequences ? $\endgroup$
    – Evargalo
    Sep 1, 2017 at 15:45
  • 1
    $\begingroup$ @Evargalo You're correct. For the record that was me who said that they converge asymptotically (not a source), so I'll take it as a good sign that I got that right ;) What I'm wanting to understand is why they converge asymptotically. Good question about my background, I wasn't sure if it was appropriate to put it in the question - I'm a Machine Learning and stats guy, but not a pure math guy. I'm asking this question in part due to the use of the Golden Ratio in an algorithm. $\endgroup$
    – Hack-R
    Sep 1, 2017 at 15:47
  • 1
    $\begingroup$ @Hack-R Tie the dog to a tree with a long rope, then have him circle the tree. As the rope wraps around, the dog will eventually converge to the tree, but it's less common to say the dog and the tree converged ;-) $\endgroup$
    – dxiv
    Sep 1, 2017 at 20:30

2 Answers 2


If we set $\varphi = \frac{1+\sqrt{5}}{2}$ and $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$, then the Binet formula for Fibonacci numbers says:

$F_n=\frac{\varphi^n-(\bar\varphi)^{ n}}{\sqrt{5}}$.

Because $\mid \bar\varphi \mid < 1$, the term $(\bar\varphi)^n$ decays fairly quickly toward $0$. This points to why successive ratios of Fibonacci numbers get close to $\varphi$.

Note: Here is a reference for the Binet formula: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html

  • 1
    $\begingroup$ In particular, $F_n$ is the nearest integer to $\varphi^n/\sqrt 5$ for $n\in\mathbb N$. $\endgroup$ Sep 1, 2017 at 16:05

I've seen a few proofs of this. Here is one I like.

We start with the definition of Fibonacci number as the sum of the previous two terms: $F(n+1) = F(n) + F(n-1)$. We can then express the ratio of successive Fibonacci terms as follows \begin{equation} \begin{aligned} \frac{F(n+1)}{F(n)} & = \frac{F(n) + F(n-1)}{F(n)} \\ & = 1 + \frac{F(n-1)}{F(n)} \end{aligned} \label{eq:fibonacci_limit_varphi_proof1_1} \end{equation} We now define a quantity $x$ as \begin{equation} x = \lim_{n \to \infty}\ \frac{F(n+1)}{F(n)} \end{equation} The inverse of this quantity is \begin{equation} \frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n)}{F(n+1)} \label{eq:fibonacci_limit_varphi_proof1_2} \end{equation} In plain english, this expression says ``take the limit as $n$ approaches infinity of a Fibonacci term and the next term. We can simply make a change in notation and get the exact same limit: \begin{equation} \frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)} \label{eq:fibonacci_limit_varphi_proof1_3} \end{equation} That is, if you spell out what the above limit is seeking, we'll see that \begin{equation} \frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n)}{F(n+1)} = \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)} \label{eq:fibonacci_limit_varphi_proof1_4} \end{equation} Returning to the first equation and applying the limit as $n \rightarrow \infty$ to both sides, we get \begin{equation} \begin{aligned} \lim_{n \to \infty}\ \frac{F(n+1)}{F(n)} & = \lim_{n \to \infty}\ \left[1 + \frac{F(n-1)}{F(n)} \right] \\ {} & = 1+ \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)} \\ \end{aligned} \end{equation} The left-hand side is what we defined as $x$ and the limit on the right-hand side is what we discussed to be $\displaystyle \frac{1}{x}$. So we re-write the above as \begin{equation} \begin{aligned} x & = 1 + \frac{1}{x} \\ x^2 - x - 1 & = 0 \end{aligned} \end{equation} Solving this quadratic gives us \begin{equation} \begin{aligned} \varphi & = {1 \pm \sqrt{5} \over 2} \nonumber \\ & = 1.6180339887.... , -0.618033887... \end{aligned} \end{equation} The positive root of this quadratic equation is $\varphi$. So \begin{equation} x = \lim_{n \to \infty}\ \frac{F(n+1)}{F(n)} = \varphi \end{equation}


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