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

  1. the sequence $\{\log_{F_{n+1}}{F_n}\}$ is increasing.
  2. 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.

  • 1
    $\begingroup$ (2) might be more difficult to prove because it implies (1). $\endgroup$
    – Martin R
    May 16, 2021 at 7:01
  • $\begingroup$ @Martin R Yes, you are right. $\endgroup$
    – Mr.He
    May 16, 2021 at 7:11
  • $\begingroup$ Binet's formula should work perfectly for this. $\endgroup$ May 16, 2021 at 7:11
  • 1
    $\begingroup$ And why to duplicate (1) which you asked recently in A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ ? $\endgroup$
    – Sil
    May 16, 2021 at 7:44
  • 1
    $\begingroup$ 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. $\endgroup$
    – Martin R
    May 16, 2021 at 7:51

1 Answer 1


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 $k=\log(c)/\log(\phi)$.


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