A Fibonacci conjecture: $\frac{n-1}{n}<\log_{F_{n+1}}{F_n}<\frac{n}{n+1}$.

Question

Given the Fibonacci sequence $F_n$, that is $$ F_0=F_1=1,F_{n+2}=F_{n+1}+F_n. $$ Then we have

the sequence $\{\log_{F_{n+1}}{F_n}\}$ is increasing.

for any $n\geqslant2$, $$ \frac{n-1}{n}<\log_{F_{n+1}}{F_n}<\frac{n}{n+1}. $$

In fact, I have tried to prove $(1)$ in a simple way, but failed. (See this previous question.) Because when $n$ is odd, the inequality $\frac{\ln F_n}{\ln F_{n+1}}<\frac{\ln F_{n+1}}{\ln F_{n+2}}\ $ is easy to get by using AM-GM inequality and Cassini's identity; when $n$ is even, things will be very different.

As for $(2)$, I have no idea how to prove it, except for using Binet's formula.

(2) might be more difficult to prove because it implies (1). — Martin R, May 16 '21 at 07:01
@Martin R I have check it again unless I make the index wrong. — Mr.He, May 16 '21 at 07:26
And why to duplicate (1) which you asked recently in A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ ? — Sil, May 16 '21 at 07:44
I would suggest to re-write the conjecture in terms of the standard definition ($F_0 = 0$, $F_1 = 1$) in order to avoid confusion when people try to verify it with computer algebra systems, or try to apply Binet's formula. – I have deleted my previous comments which were based on that confusion. — Martin R, May 16 '21 at 07:51
@ Sil I still expect a nice proof for (1) , and it is a corollary of (2). — Mr.He, May 16 '21 at 07:59
The second property can also be written without logs as $(\frac{F_{n+1}}{F_n})^{n-1}<F_n<(\frac{F_{n+1}}{F_n})^n$, or perhaps as a pair of inequalities $F_{n+1}^{n-1}<F_n^n$ and $F_n^{n+1}<F_{n+1}^n$. — Sil, May 16 '21 at 09:00

score 4 · Answer 1 · edited Mar 28 '24 at 21:01

Let $\phi=\frac{1+\sqrt{5}}{2}$ and $\eta=1/\phi^2<1$. With your definition of $F_n$: \begin{equation} F_n = \frac{1+\sqrt{5}}{2\sqrt{5}} \phi^n ( 1 + O(\eta^{n})) =: c \ \phi^n (1+O(\eta^n)) \ \ \end{equation} with $c\approx 0.72...$. Then, $$ \frac{\log(F_n)}{\log(F_{n+1})} = \frac{n+ k + O(\eta^n)}{n+1 +k+O(\eta^n)} =: \frac{n+k_n}{1+n+k_n}$$ with $k=\log(c)/\log(\phi) \approx -0.67227...$. Here, $k_n -k =O(n \ \eta^n) $ so asymptotically you have the wanted bounds (2). You may obtain rigorous bounds for the error term, though it is cumbersome, after which it suffices to verify for some finite number of initial terms.

The above analysis is not particularly related to the Fibonacci sequence but works for any recursive definition of a sequence $x_n$ that leads to an asymptotic behavior of the form $x_n = c \phi^n(1+O(\eta^n))$ with $\eta<1$ and with the corresponding $k=\log(c)/\log(\phi)$.

A Fibonacci conjecture: $\frac{n-1}{n}<\log_{F_{n+1}}{F_n}<\frac{n}{n+1}$.

1 Answers1