斐波那契前后两项的比

$\begin{aligned}\dfrac{f_{n+1}}{f_n}&=\dfrac{\dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1}-\dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1}}{\dfrac{1}{\sqrt{5}}\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\dfrac{1}{\sqrt{5}}\left(\dfrac{1-\sqrt{5}}{2}\right)^n}\\&=\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1}}{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}\\&=\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)-\left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1}\left(\dfrac{2}{1+\sqrt{5}}\right)^n}{1-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\left(\dfrac{2}{1+\sqrt{5}}\right)^n}\\&=\left(\dfrac{1+\sqrt{5}}{2}\right)\left(\dfrac{1-\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n+1}}{1-\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n}\right)\end{aligned}$

$\begin{aligned}\lim\limits_{n\to\infty}\dfrac{f_{n+1}}{f_n}&=\lim\limits_{n\to\infty}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)\left(\dfrac{1-\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n+1}}{1-\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n}\right)\right)\\&=\left(\dfrac{1+\sqrt{5}}{2}\right)\left(\dfrac{1-\lim\limits_{n\to\infty}\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n+1}}{1-\lim\limits_{n\to\infty}\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n}\right)\\&=\dfrac{1+\sqrt{5}}{2}\end{aligned}$

可见，斐波那契数列后一项比前一项的比的极限为$\dfrac{1+\sqrt{5}}{2}$。取倒数即可得到前一项和后一项的比的极限为$\dfrac{\sqrt{5}-1}{2}$，这就是黄金分割比，约等于$0.61803398875$。