Golden Ratio as a Fibonacci Limit

Key idea. The ratios satisfy $r_n=1+1/r_{n-1}$ . This map contracts distances to its fixed point, so the ratios converge. The limit therefore solves $L=1+1/L$ .

For $n\ge1$ put $r_n=F_{n+1}/F_n$ , so $r_1=1$ . Dividing $F_{n+1}=F_n+F_{n-1}$ by $F_n$ gives

r_n = 1+\frac{1}{r_{n-1}}\qquad(n\ge2).

Let $\varphi=\frac{1+\sqrt5}{2}$ be the positive root of $x^2=x+1$ , so that $\varphi=1+1/\varphi$ . Subtracting,

r_n-\varphi=\frac{1}{r_{n-1}}-\frac{1}{\varphi}=-\frac{r_{n-1}-\varphi}{\varphi\, r_{n-1}}.

The Fibonacci numbers satisfy $F_n=F_{n-1}+F_{n-2}\ge F_{n-1}$ , so $r_{n-1}\ge1$ . Hence

|r_n-\varphi|\le\frac{|r_{n-1}-\varphi|}{\varphi}.

Iterating this bound,

|r_n-\varphi|\le\varphi^{-(n-1)}|r_1-\varphi|\to0,

since $\varphi>1$ . Thus $r_n\to\varphi$ .

This also justifies the heuristic of the editor’s note. Of the two roots $\frac{1\pm\sqrt5}{2}$ of $L^2=L+1$ , the limit must be the positive one, because every $r_n\ge1$ .

\boxed{\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\frac{1+\sqrt5}{2}\approx1.618}

Alternative. The characteristic equation of the recurrence is $x^2=x+1$ , with roots $\varphi$ and $\psi=\frac{1-\sqrt5}{2}$ . Fitting $F_0=0$ and $F_1=1$ gives Binet’s formula

F_n=\frac{\varphi^n-\psi^n}{\sqrt5}.

With $\rho=\psi/\varphi$ , which satisfies $|\rho|<1$ ,

\frac{F_{n+1}}{F_n}=\varphi\,\frac{1-\rho^{n+1}}{1-\rho^n}\to\varphi .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 33.8a

Let $F_n$ denote the Fibonacci sequence, with $F_0=0$ , $F_1=1$ , and $F_n=F_{n-1}+F_{n-2}$ for $n\geq 2$ . Find $\displaystyle\lim_{n\to\infty}\frac{F_{n+2}}{F_n}$ .

Work it out, then compare with the answer.

Solution

Divide $F_{n+2}=F_{n+1}+F_n$ by $F_n$ :

\frac{F_{n+2}}{F_n}=1+\frac{F_{n+1}}{F_n}.

By the main result, the right-hand side tends to $1+\varphi$ . Since $\varphi^2=\varphi+1$ , the limit equals $\varphi^2$ .

\boxed{\lim_{n\to\infty}\frac{F_{n+2}}{F_n}=\frac{3+\sqrt5}{2}\approx2.618}