Proof that the ratio between the logs of the product and LCM of the Fibonacci numbers converges to $\frac{\pi^2}{6}$ I came across this amazing fact on Twitter.
$$\lim_{n\to\infty} \frac{\log\left(F_1 \cdots F_n\right)}{\log \text{LCM}\left(F_1, \ldots,  F_n\right)} = \frac{\pi^2}{6}$$
where $F_i$ is the $i$th Fibonacci number and LCM = Least Common Multiple. This is such an interesting link between the Fibonacci numbers (which are closely linked to the Golden Ratio $\varphi$) and $\pi$. I'm trying to prove why this is the case.
I know that, for the numerator, it'll become something like
\begin{align}
\log\left(F_1 \cdots F_n\right) &\sim \log\left(\varphi^1 \cdots \varphi^n\right) \\
&\sim \frac{\log \varphi}{2} n^2
\end{align}
Reverse engineering, that tells me that the denominator will turn out to be the same, but with an extra factor of $\frac{6}{\pi^2}$. What I'm wondering is, how to prove that this will be the case? Some relevant tools might be that $$\prod_i \left(1 - \frac{1}{p_i^2}\right) = \frac{6}{\pi^2}$$, where $p_i$ is the $i$th prime. Also, it's not hard to show that the LCM of the first $n$ natural numbers is roughly $e^n$. Finally, the probability that the $n$th Fibonacci is prime is roughly $\sim\frac{\log\varphi}{n}$ I think (though I'm not actually sure if this has been proven).
As an aside, even though it's cool that this is done with the Fibonacci sequence, judging by how the $\varphi$ factor just cleanly cancels on top and bottom, I have a feeling that this fact might be true for other linear integer recurrence relations that have an exploding dominant eigenvalue.
 A: A full proof of this fact can be found in the paper A new Formula for $\pi$ by Yuri V. Matiyasevich  and Richard K. Guy.
A brief summary: Notations I will use:

*

*Let $\mu$ denote the Möbius function,

*let $w_n:=\operatorname{LCM}(F_1,F_2,\dots,F_n)$.

As you noticed, we have (for large $n$)
\begin{equation}\tag 1\label 1
\log(F_1\cdots F_n)\sim \frac{n^2 \ln\tau}{2},
\end{equation}
where $\tau=\frac{1+\sqrt 5}{2}$ denotes the golden ratio.
So proving your statement is equivalent to proving that (for large $m$)
\begin{equation}\tag 2\label 2
\ln w_m \sim 3\ln(\tau)\frac{m^2}{\pi^2}.
\end{equation}
In the paper, \eqref{2} is called Chebyshev's form of the prime number theorem for Fibonacci numbers.
By rather lengthy applications of results on Arithmetic functions and the Möbius inversion formula, one gets
\begin{equation}\tag 3\label 3\begin{split}
\ln w_m &= \sum_{d=1}^m \sum_{i \text{ such that } i | d} \mu\left(\frac di\right)\ln F_i \\
&= B(m)+\sum_{d=1}^m  \sum_{i \text{ such that } i | d} \mu\left(\frac di\right) i \ln\tau,
\end{split}
\end{equation}
where $0\le B(m)< 2m^\frac32$.
One can write (by elementary results on Euler's totient function $\phi$)
\begin{equation}\tag 4\label 4
\sum_{d=1}^m  \sum_{i \text{ such that } i | d} \mu\left(\frac di\right) i \ln\tau
= \ln\tau\cdot\sum_{d=1}^m \phi(d)\sim 3\ln(\tau)\frac{m^2}{\pi^2}
\end{equation}
(the last similarity is a proven result from number Theory.)
So, as a result,
$$\ln w_m \sim 3\ln(\tau)\frac{m^2}{\pi^2} + B(m),$$ where $0\le B(m)< 2m^\frac32$.
From this follows \eqref{2} and thus also your result.
