The factors in each of the finite geometric means are fractions chosen like this: The numerator of each individual fraction is an odd number in between 2 consecutive powers of 2, and the denominator is the lower power of 2 that the numerator is in between.
There are as many fractions in each mean as there are odd numbers in between each power of 2, and the mean is different depending on which power of 2 you choose to be your upper and lower bound, but as you make your powers of 2 larger and larger, the mean converges.
In straight math, this means every factor in each geometric mean takes the form: $$ a\over b$$
where:
$$n \in \mathbb{N}$$
$$b=2^n$$
$$b \lt a \lt 2b$$
and $a$ is an odd number.
The original conditions inherently prove that the limit of means to infinity has to converge to something because, in each fraction, $$1 \lt \frac{a}{b} \lt 2$$ since: $$\frac{b \lt a \lt 2b}{b}$$ $$1 \lt \frac{a}{b} \lt 2$$
It doesn't matter how many of these fractions you get a geometric mean of, because the geometric mean of any infinite number of 2's, or $\sqrt[x]{2}^x$ when $x$ can be as large as you want, is always 2, and the product of all $x$ fractions in a mean has to be smaller than $2^x$, since (as proved earlier) each fraction has to be smaller than 2. The same applies for 1, except each mean must be greater than 1.
Now to find each mean in this sequence of means, you use as many fractions as there are odd numbers in between $b$ and $2b$. This means the number of terms, and also the number you take a root of for the geometric mean, is $b/2$, since there are always $b/2$ odd numbers in between $b$ and $2b$ (if $b$ is as described above).
You can prove there are always $b/2$ odd numbers in between $b$ and $2b$, since the odd numbers (lets call them $t$) take the form:
$$k \in \mathbb{N}$$
$$t=b+1+2(k-1) \lt 2b$$
The largest number of $k$ that satisfies this at any given $b$ is the number of odd numbers in between $b$ and $2b$. Let's call the largest $k$, $k_b$. What $t$ is if you plug in $k_b$ where $k$ was will have to be equal to $2b-1$, since the largest odd number in the bound has to be right under the higher power of 2. For this new variable $k_b$, we see that:
$$b+1+2(k_b-1)=2b-1$$ $$2(k_b-1)=b-2$$ $$k_b-1=\frac{b}{2}-1$$ $$k_b=\frac{b}{2}$$
This proves the number of odd numbers between any of these given bounds must be $b/2$, and that the number of fractions in any of the means must be $b/2$.
This also leads to the fact that each mean has to be to the root of $b/2$.
The first term in this series (which can be thought of as when n=1) of geometric means would just be: $$\sqrt[1]{(\frac {3} {2})}$$ since there is only 1 odd number in between b and 2b, (2 and 4 respectively), meaning there is only one fraction --3/2-- and the geometric mean of 1 number is just itself.
The second term would be: $$ \sqrt[2]{\frac{5}{4}*\frac{7}{4}}$$ since 5 and 7 are the only odd numbers in between 4 and 8 (again b and 2b)
Third term is:
$$\sqrt[4] {\frac{9}{8} * \frac{11}{8} * \frac{13}{8} * \frac{15}{8}}$$
Now, you can also write this as:
$$\sqrt[4]{\frac {9 * 11 * 13 * 15}{8 * 8 * 8 * 8}}$$
This is where it gets interesting.
Let's make a function in terms of $n$, our old definition for $b$. It'll be $G(n)$, where each output of the function is the mean for each $n$.
For reference to the kind of fraction root we made earlier, we split the function into: $$ \sqrt [c_g] {\frac {a_g}{b_g}}$$ since, every single mean here can take that form, as we showed before.
How would you go about defining $a_g$, $b_g$, and $c_g$ in terms of $n$, and consequentially defining $O(n)$ in terms of $n$?
Well, the most obvious part of this is $c_g$, which is always $b/2$ because the number of fractions is always b/2. Since $b=2^n$: $$c_g=b/2=(2^n)/2 = 2^{n-1}$$
The denominator of this new big fraction, or $b_g$, would have to be $b^\frac{b}{2}$, since there are always $b/2$ terms, meaning $b$ is multiplied by itself $b/2$ times. Since we know $b=2^n$ and $b/2=2^{n-1}$, then: $$b_g=b^{\frac{b}{2}}=(2^n)^{2^{n-1}}$$
This lines up with us having to multiply 2 by itself once (when $n=1$), 4 by itself twice (when $n=2$), 8 by itself 4 times (when $n=3$) and so on and so forth for any number $n$.
Now what about defining $a_g$? This seems trickier, since we need to multiply a specific set of odd numbers within a bound. If you sat there and thought about it, you might remember that double factorials are a similar function, multiplying consecutive numbers of the same parity (odd/even). You might even notice that dividing a double factorial by a smaller double factorial with the same parity would essentially shave off the lower part.
In the case of $n=1$'s fraction this would look like:
$$\frac{3!!}{1!!}$$ $$3$$
In the case of $n=2$ and $n=3$, their respective fractions are:
$$\frac{7!!}{3!!}$$ $$\frac{7*5*3}{3}$$ $${7*5}$$ and $$\frac{15!!}{7!!}$$ $$\frac{15*13*11*9*7*5*3}{7*5*3}$$ $${15*13*11*9}$$
As you can see, this gives us the exact pattern we need. Writing the insides of the double factorials in terms of n is much simpler now.
The top half has to be the highest odd number in the bound, or $2b-1$. Plugging in $2^n$ for $b$ gives you:
$$2(2^n)-1$$ $$2^{n+1}-1$$
The bottom half is the odd number right under the lower bound, or $b-1$, which is just:
$$2^n-1$$
Combining all of this, you get:
$$a_g=\frac{(2^{n+1}-1)!!}{(2^n-1)!!}$$
And we did it! The whole explicit equation that gives you the $n$'th mean can be written as:
$$G(n)=\sqrt[2^{n-1}]{\frac{\frac{(2^{n+1}-1)!!}{(2^n-1)!!}}{(2^n)^{2^n-1}}}$$
And, the limit as $n$ goes to infinity would be the answer to our question.
$$\lim_{n\to \infty}\sqrt[2^{n-1}]{\frac{\frac{(2^{n+1}-1)!!}{(2^n-1)!!}}{(2^n)^{2^n-1}}}$$
Now, when plugging this into Desmos, I found that it couldn't handle calculating $n$ for values past $\approx25$ , it was too much for it to calculate, which makes sense considering the double factorial of an exponent must grow very quickly as $n$ increases. This means the formula converges very quickly. I found it converged to around 1.47151.
When I asked this question on YouTube, somebody replied that the answer is 4/e, which is 1.47151... so that's most likely correct. The only thing now is that I have no idea how to prove it, which is what interests me.
Another thing worth noting is that it seems like if you just make one constantly growing geometric mean with each of the terms, it still converges to the same number. I don't know for sure though.
$$\sqrt[1]{\frac{3}{2}}$$ $$\sqrt[2]{\frac{3}{2}*\frac{5}{4}}$$ $$\sqrt[3]{\frac{3}{2}*\frac{5}{4}*\frac{7}{4}}$$ $$\sqrt[4]{\frac{3}{2}*\frac{5}{4}*\frac{7}{4}*\frac{9}{8}}$$
Finding an explicit formula for this version seemed easier yet harder in some ways. If we let $n$ be an input again, for a new function H(n) (such that $n \in \mathbb{N}$), the root is just $n$, and the top half is just $2n+1!!$, which is much simpler. What I don't know would be how to define the bottom half, which would go:
For $n=1$, just $2$.
For $n=2$, $2*4$.
$n=3$, $2*4*4$
$n=4$, $2*4*4*8$
$n=5$, $2*4*4*8*8$
$n=6$, $2*4*4*8*8*8$
$n=7$, $2*4*4*8*8*8*8$
I don't know which function could satisfy that, but maybe that version is easier to solve, or a more essential version of the same idea. Either way I'd love it if somebody could prove that it equals 4/e. Thanks for reading.
Update:
I have noticed that one way that might prove that this is 4/e, is if you can put the limit in terms of:
$$\lim_{x\to \infty}\frac{4}{(1+\frac{1}{x})^x}$$
Here, $x$ is some manipulation of the original function that diverges to infinity, but the rest of the manipulation would take the form shown here.
A solution follows, but first is an attempt to simplify the discussion:
We have $$\begin{align} G(n) &:= \sqrt[\LARGE{|S_n|}]{\frac{(2^n+1)(2^n+3)(2^n+5)\cdots(2^{n+1}-1)}{(2^n)^{|S_n|}}} \\[8pt] &=2^{-n}\;\;\sqrt[\LARGE{|S_n|}]{\;(2^n+1)(2^n+3)(2^n+5)\cdots(2^{n+1}-1)\;} \end{align}$$
Note that $|S_n|=\frac12\left(2^{n+1}-2^n\right)=\frac12\cdot 2^{n}(2-1)=2^{n-1}$. Also, recall that the double-factorial of an odd integer $m$ is defined as $$m!! \;:=\; 1\cdot 3\cdot 5\cdot\,\cdots\,\cdot m$$ Therefore, we can write concisely $$ G(n) = 2^{-n}\left(\frac{(2^{n+1}-1)!!}{(2^n-1)!!}\right)^{\large{1/2^{n-1}}} $$ A respondent on YouTube (a link to where this was asked on YouTube would be nice) suggested that, as $n$ grows, $G(n)$ approaches $4/e = 1.47151\ldots$, which agrees with some numerical estimates.
(At the point OP writes "Another thing worth noting ...", I get a little lost, so I'll stop here.)
Solution:
To determine the limit, express the double-factorials in terms of single-factorials (see Wikipedia) via $$(2k-1)!! = \frac{(2k-1)!}{2^{k-1}(k-1)!}$$ and then invoke Stirling's approximation for large $n$: $$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
If I have my exponent arithmetic (and TeX formatting) correct, we get the following: $$\begin{align} \frac{(2^{n+1}-1)!!}{(2^n-1)!!} &= \frac{\quad\dfrac{ (2^{n+1}-1)!}{2^{2^n-1}(2^n-1)!}\quad}{\quad\dfrac{(2^n - 1)!}{2^{2^{n-1}-1}(2^{n-1}-1)!}\quad} \\[8pt] &= \frac1{2^{2^{n-1}}}\cdot \frac{ (2^{n+1}-1)!\,(2^{n-1}-1)!}{((2^n-1)!)^2} \\[8pt] \text{(for large $n$)}\;&\approx \frac1{2^{2^{n-1}}}\cdot\frac{ \sqrt{ 2 \pi \cdot 2^{n+1} } \left( \dfrac{2^{n+1}}{e} \right)^{2^{n+1}} \sqrt{ 2 \pi \cdot 2^{n-1} } \left( \dfrac{2^{n-1}}{e} \right)^{2^{n-1}}}{ (2 \pi \cdot 2^n )\left( \dfrac{2^n}{e} \right)^{2^{n+1}}} \\[8pt] &=\left(\frac{2^{n+2}}{e}\right)^{2^{n-1}} \end{align}$$ Therefore, for large $n$, $$G(n) \approx 2^{-n} \cdot \left(\frac{2^{n+2}}{e}\right)=\frac{4}{e}$$ confirming the conjecture. $\square$