How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$?

Question

How do I prove this inequality with sequential Fibonacci numbers: $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$ (i.e. $\frac{5}{3}>\frac{3}{2}$)? I know these formulas alternately overestimate and underestimate the golden ratio as $n$ gets large.

Answer 1

$$\begin{align*} f_{2n+1}f_{2n-1} - f_{2n}^2 &= \det\pmatrix{f_{2n+1}&f_{2n}\\f_{2n}&f_{2n-1}}\\ &= \det\pmatrix{f_{2n+1}-f_{2n}&f_{2n}\\f_{2n}-f_{2n-1}&f_{2n-1}}&&C_1 \to C_1-C_2\\ &= \det\pmatrix{f_{2n-1}&f_{2n}\\f_{2n-2}&f_{2n-1}}\\ &= \det\pmatrix{f_{2n-1}&f_{2n}-f_{2n-1}\\f_{2n-2}&f_{2n-1}-f_{2n-2}} && C_2 \to C_2-C_1\\ &= \det\pmatrix{f_{2n-1}&f_{2n-2}\\f_{2n-2}&f_{2n-3}}\\ &= \vdots\\ &= \det\pmatrix{f_{3}&f_{2}\\f_{2}&f_{1}}\\ &= \det\pmatrix{2&1\\1&1}\\ &= 1\\ &> 0\\ \frac{f_{2n+1}}{f_{2n}}&> \frac{f_{2n}}{f_{2n-1}} \end{align*}$$

(for $n \ge 1$, so that $f_{2n-1}$ and $f_{2n}$ are both positive)

Answer 2

The characteristic equation $^2−−1=0$ has two roots $\varphi=\frac{1+\sqrt 5}{2}, \psi=\frac{1-\sqrt 5}{2}$. Vieta’s formulas give $\varphi+\psi=1, \varphi \psi = -1$. Then we can rewrite $f_{n+1} = f_n + f_{n-1}$ as $$f_{n+1}−\varphi f_=\psi (f_−\varphi f_{-1}).$$

Note that $\psi<0$. Then $$f_{+1}−\varphi f_=\psi^ (f_1− \varphi f_0 )=\psi^ \begin{cases} >0 & \text{ if } n \text{ is even;} \\ <0 & \text{ if } n \text{ is odd.} \end{cases}$$ Hence $$ \frac{f_{2n+1}}{f_{2n}}>\varphi >\frac{f_{2n}}{f_{2n-1}} $$

Answer 3

The base case of OP's equality is, for $n=1$,

$$\frac21>\frac11.$$

Assume that OP's inequality holds for some $n = k$, $k \ge 1$.

$$\frac{f_{2k+1}}{f_{2k}} > \frac{f_{2k}}{f_{2k-1}}$$

Then for $n=k+1$, using the mediant inequality twice,

$$\begin{array}{rcrclcl} \dfrac{f_{2k+1}}{f_{2k}} &>&&&&& \dfrac{f_{2k}}{f_{2k-1}}\\ \dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+1}+ f_{2k}}{f_{2k}+f_{2k-1}} &> &\dfrac{f_{2k}}{f_{2k-1}}\\ \dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+2}}{f_{2k+1}}\\ \dfrac{f_{2k+1}}{f_{2k}} &>& \dfrac{f_{2k+1}+f_{2k+2}}{f_{2k}+f_{2k+1}} &>& \dfrac{f_{2k+2}}{f_{2k+1}}\\ &&\dfrac{f_{2k+3}}{f_{2k+2}} &>& \dfrac{f_{2k+2}}{f_{2k+1}} \end{array}$$

(The intermediate step also proves the other half of the inequality pair:

$$\frac{f_{2n+2}}{f_{2n+1}} < \frac{f_{2n+1}}{f_{2n}})$$