The solutions of the quadratic equation $x^2-x-1=0$ are $x=(1\pm \sqrt{5})/2$ and the Golden Ratio $\varphi$, which is positive, is $(1+\sqrt{5})/2$. Notice that $L=\varphi -1$ and $L\varphi =(\sqrt{5}-1)(\sqrt{5}+1)/4=1$ so $L=\frac{1}{\varphi }$.

We next prove that

${\displaystyle X_n=\frac{F_n}{F_{n+1}}.}$

This is true for $n=1$ and $n=2$:

${\displaystyle X_1=\frac{F_1}{F_2}=\frac{1}{1}=1,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{X}_2=\frac{F_2}{F_3}=\frac{1}{2}.}$

If the theorem is true for $X_k$ then

${\displaystyle X_{k+1}=\frac{1}{1+X_k}=\frac{1}{1+\frac{\backslash strut{F}_k}{F_{k+1}}}=\frac{F_{k+1}}{F_{k+1}+F_k}=\frac{F_{k+1}}{F_{k+2}},}$

and hence it is rue for $n=k+1$ so by the axiom of induction it is true for all $n$.

Hence the ratio of successive terms of the Fibonacci sequence

${\displaystyle \frac{F_{n+1}}{F_n}=\frac{1}{X_n}}$

and as $X_n\to \frac{1}{\varphi }$ as $n\to \infty$ this ratio tends to the Golden Ratio as $n\to \infty$.