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Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$ I find that $$ \log_21<\log_32<\log_53<\log_85<\cdots, $$ i.e. we have $$ \log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1} $$ If $n=2k$, $(1)$ is easy to get, but if $n=2k+1$, I have no idea to prove it.

What's more, if we let $F_1=a,F_2=b$ be two positive integers, then $(1)$ still exstis when $n$ is greater than some integer $m$.

I do not konw whether my conjecture is true.

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  • $\begingroup$ @Sil I just use the Cassini's identity, so only get a half of proof. $\endgroup$
    – Mr.He
    May 15, 2021 at 9:53
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    $\begingroup$ Can you add your “easy proof” for even $n$ to the question? $\endgroup$
    – Martin R
    May 15, 2021 at 10:27
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    $\begingroup$ @Martin R Just use Cassini's identity and the Base inequality, you can get $\frac{\ln F_n}{\ln F_{n+1}}-\frac{\ln F_{n+1}}{\ln F_{n+2}}<0$. $\endgroup$
    – Mr.He
    May 15, 2021 at 10:37
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    $\begingroup$ The same question has been asked and self-answered here: math.stackexchange.com/q/4009705/42969. I haven't checked if the solution is complete. $\endgroup$
    – Martin R
    May 15, 2021 at 10:42
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    $\begingroup$ @Sil Good point, I seemed to have had a bad confusion myself there! Deleted the comment, it was actually pretty bad. The same mistake was made in the comment below. $\endgroup$ May 15, 2021 at 10:54

4 Answers 4

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For even $n$ you can use:

$$F_nF_{n+2}<F_{n+1}^2$$

$$\implies \log(F_nF_{n+2})<\log(F_{n+1}^2)$$

$$\implies \log(F_n)+\log(F_{n+2})<2\log(F_{n+1})$$

$$\implies \frac{\log(F_n)+\log(F_{n+2})}{2}<\log(F_{n+1})$$

By the AM-GM:

$$\sqrt{\log(F_n)\log(F_{n+2})}<\frac{\log(F_n)+\log(F_{n+2})}{2}< \log(F_{n+1})$$

as required.

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  • $\begingroup$ $F_nF_{n+2}< F_{n+1}^2$ holds only for even $n$. $\endgroup$
    – Martin R
    May 15, 2021 at 15:37
  • $\begingroup$ @MartinR; good point! $\endgroup$
    – JMP
    May 15, 2021 at 15:41
  • $\begingroup$ Yep. If $n$ is even, it is easy to get. But it seems to be hard when $n$ is odd. $\endgroup$
    – Mr.He
    May 15, 2021 at 16:51
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In general, starting with any two numbers, you get $$F_n=A\phi^n+B(-\phi)^{-n}$$ where $\phi=(1+\sqrt5)/2$ is the golden ratio. The second term soon becomes very small. So
$$\ln F_n\ln F_n-\ln F_{n-1}\ln F_{n+1}\approx \\ [n\ln\phi+\ln A]^2-[(n-1)\ln\phi+\ln A][(n+1)\ln\phi+\ln A]\\ \approx (\ln\phi)^2$$

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Conjecture: $ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$

This is equivalent to: $$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$ Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ here is the natural logarithm).

One can use the Cassini identity ($F_{n+1}^2 = F_nF_{n+2} \pm 1$) to prove your conjecture, but first let's establish an inequality equivalent to the above one, $$\log(F_n)\log(F_{n+2})< (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$

Therefore it suffices to prove that $\log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2}$

Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$ $$(\log(F_{n+2}/F_{n}))^2 > (\log 2)^2 > \dfrac{2 \log F_n}{F_n} + \dfrac{2\log 3F_n}{F_n} - \dfrac{1}{F_n^2}$$ $$\geq \dfrac{2 \log F_n + 2\log (2F_n + F_{n-1})}{F_n}- \dfrac{1}{F_n^2} = \dfrac{2 \log F_n + 2\log (F_{n+2})}{F_n}- \dfrac{1}{F_n^2}$$ $$\implies (\log F_{n+2} - \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2}$$ for a sufficiently large $n$($n \geq 8$ for the Fibonacci sequence starting with $1$ and $2$). Now add $4\log F_n \log F_{n+2}$ to both sides: $$(\log F_{n+2} + \log F_n)^2 > \dfrac{2 \log F_nF_{n+2}}{F_n} - \dfrac{1}{F_n^2} + 4\log F_n \log F_{n+2} $$ $$\implies (\log F_{n+2} + \log F_n)^2 - \dfrac{2 \log F_nF_{n+2}}{F_n} + \dfrac{1}{F_n^2} > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n} + \log F_{n+2} - \dfrac{1}{F_n})^2 > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n} + \log (F_{n+2} - \dfrac{1}{F_n}))^2 = (\log(F_n\cdot (F_{n+2} - \dfrac{1}{F_n})))^2 = \log(F_{n+1}^2)^2 > 4\log F_n \log F_{n+2} $$

Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$ $$(\log F_{n+2} - \log F_n)^2 > 0 $$ $$\implies (\log F_{n+2} +\log F_n)^2 > 4\log F_n \log F_{n+2}$$ $$\implies (\log F_{n+1}^2)^2 = (\log F_{n+2}F_n + 1)^2 > (\log F_{n+2}F_n)^2 > 4\log F_n \log F_{n+2}$$

Verifying the inequality for $n < 8$ manually completes the proof using Cassini's identity.

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Conjecture:$ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$

This is equivalent to: $$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$ Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ here is the natural logarithm).

One can use the Cassini identity ($F_{n+1}^2 = F_nF_{n+2} \pm 1$) to prove your conjecture, but first, $$\log(F_n)\log(F_{n+2})< (\log(F_{n+1}))^2 = (\frac{1}2\log(F_{n+1}^2))^2 = \frac{1}{4}(\log(F_{n+1})^2)^2$$

Case 1: when $F_{n+1}^2 = F_nF_{n+2} - 1$ $$4\log(F_n)\log(F_{n+2})< (\log(F_{n+1})^2)^2 = \log(F_nF_{n+2} - 1)^2 < (\log F_n + \log F_{n+2})^2.$$

Case 2: when $F_{n+1}^2 = F_nF_{n+2} + 1$

$$ 4\log(F_n) \log(F_{n+2})< (\log(F_{n+1})^2)^2 = (\log(F_n F_{n+2} + 1))^2 < (\log(F_n F_{n+2}) + 1)^2 = (\log F_n + \log F_{n+2})^2 + 2\log(F_n F_{n+2}) + 1.$$

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    $\begingroup$ I have a confusion now, we need to prove that $\log (F_n)\log (F_{n+2})<(\log (F_{n+1})^2$, and you just use it? $\endgroup$
    – Mr.He
    May 15, 2021 at 14:37
  • $\begingroup$ @Mr.He Okay, I will add the proof to that. $\endgroup$ May 15, 2021 at 14:39
  • $\begingroup$ I mean that you want to prove $ \log (F_n)\log (F_{n+2})<(\log(F^2_{n+1}))^2$, but in your statement, you have used it. So I think there is some Logical problem in the proof. $\endgroup$
    – Mr.He
    May 15, 2021 at 16:41
  • $\begingroup$ @Mr.He you are right. I have uploaded another proof using Cassini's identitiy. $\endgroup$ May 16, 2021 at 7:42

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