I read a couple days ago about the Collatz conjecture. Considering that this problem has been around since the '30s, I'm guessing that my idea written below has a few problems. Nonetheless, I want to know if this sounds like a good approach: trying to analyze how the Collatz sequence grows.
Suppose we have a starting number (obviously, the starting number must itself be large) with a Collatz sequence that does not eventually decay to $1$. It follows that the sequence is infinite. Assume further that there are no infinite cycles away from the cycle which leads to $1$: $4,2,1,4,2,1,...$. Then, for such a starting value, it should be true that its Collatz sequence diverges to $\infty$.
The geometric mean of a terminating Collatz sequence of a number $m$ can be written as $$\mu = m\sqrt[n](\prod_{i=1}^{n-1} a_i)$$, where there are $n$ terms in the Collatz sequence, and the $a_i$s are the terms of the sequence with $a_{n-1} = 1$. Now, suppose $m$ is the starting value of an infinite, diverging Collatz sequence. We can define the $n^{th}$ geometric mean of the diverging sequence to be $$\mu_n = m\sqrt[n](\prod_{i=1}^{n-1} a_i)$$ with $a_{n-1} \neq 1$. If the sequence diverges, it should be true that its $n^{th}$ geometric mean also diverges. In particular, there should exist an $n$ such that $$m\sqrt[n](\prod_{i=1}^{n-1} a_i) > m'$$, where $m'<m$ is a lower bound of the sequence, but also large. Observe that if term $a_k$ is odd, then $a_{k+1} = 3a_k+1 \implies \frac{a_{k+1}}{a_k} = \frac{3a_k+1}{a_k}$. In the limit that $a_k \rightarrow \infty$, $\frac{3a_k+1}{a_k} \rightarrow 3$. For $a_k$ even, $\frac{a_{k+1}}{a_k} = \frac{1}{2}$.
Now, for the $n^{th}$ Collatz sequence, we will have a ratio $p$ of the $n$ terms which are odd, and the remainder even. The number of odd terms, excluding the starting value $m$ is approximately $(n-1)p$ because $n$ is large, and the number of even terms is approximately $(n-1)(1-p)$. So, in the limit that $n$ is large, the $n^{th}$ geometric mean can be approximated by $$\mu_n \approx m\sqrt[n]((3)^{(n-1)p}(\frac{1}{2})^{(n-1)(1-p)}) > m'$$ because we will multiply by approximately $3$ $(n-1)p$ times, and we will multiply by $\frac{1}{2}$ $(n-1)(1-p)$ times.
Now, it is obvious that for any Collatz sequence, the proportion of terms which are odd is less than $\frac{1}{2}$. This is because, given odd term $a_k = 2n+1$, $a_{k+1} = 6n+4$, which is even, meaning we cannot have two consecutive odd terms. If we do some algebra to simplify the above expression for $\mu_n$, we find that $$\frac{1}{2} > p > \frac{(\frac{m'}{m})^n}{(n-1)\log(6)} + \frac{\log(2)}{\log(6)}$$.
Because we can take $n>>m$ and $n>>m'$, then this expression can be approximated by $$\frac{1}{2} > p > \frac{\log(2)}{\log(6)}$$
So, for divergence $p > 0.387$ approximately. So, for a diverging Collatz sequence, a minimum of about $38\%$ of its terms must be odd.
The geometric mean is $$\sqrt[n]{m \prod_{i=1}^{n-1} a_i},$$ but your approximation seems to be using $$\sqrt[n]{m \prod_{i=1}^{n-1} \frac{a_i}{a_{i-1}}} = \sqrt[n]{a_{n-1}},$$ which does not need to diverge.
It doesn't seem like taking the geometric mean is helping you. As you note, the geometric mean will diverge if and only if the sequence itself diverges. You have observed that, on the assumption the sequence diverges, $$a_n \approx m (3)^{np} \left(\frac{1}{2}\right)^{n(1-p)} = m \left(\frac{1}{2} \cdot 6^p\right)^n.$$ This does in fact diverge so long as $$p > \frac{\log 2}{\log 6} \approx 0.387.$$