Proof of Bound for Growth of Divergent Trajectory in $3x+1$ Problem In this paper, Lagarias makes the following claim in section 2.7 (Do divergent trajectories exist?).
Context
$$T(x) = \left\{ \begin{array}{rl} \dfrac{3x + 1}{2}, & 2 \nmid x \\ \dfrac{x}{2}, & 2 \mid x \end{array} \right.$$
$$\begin{align*} \tag{2.30} \lim_{k \to \infty} |T^{(k)}(n_0)| = \infty \end{align*}$$
Claim
If a divergent trajectory $\{T^{(k)}(n_0) : 0 \leq k < \infty\}$ exists, it cannot be equidistributed $\pmod{2}$. Indeed if one defines
$N^*(k) = |\{j : j \leq k \mathrm{\ and\ } T^{(j)}(n_0) \equiv 1 \pmod{2}\}|$,
then it can be proved that the condition (2.30) implies that
$$\begin{align*} \tag{2.31} \liminf_{k \to \infty} \dfrac{N^*(k)}{k} \geq (\log_2 3)^{-1} \approx .63097 \end{align*}$$
Question
How can this statement be proved?
Difficulty
It seems like the author may be ignoring the $+1$ term under the assumption that the factors will dominate. (I've seen this assumption made often for heuristic arguments for the truth of the Collatz conjecture.) I don't see how such an assumption can be justified.
Given any length $n$ sequence of $n - k$ zeros and $k$ ones, we can find an $x \in \mathbb{N}$ such that
$$T^n(x) = \dfrac{3^k x + m}{2^n}$$
where
$$3^k - 2^k \leq m \leq 2^{n-k}(3^k - 2^k)$$
Now, suppose, for example, $n = 2k$. Then, we have the bound
$$T^n(x) \leq \dfrac{3^k x + 2^{n-k}(3^k - 2^k)}{2^n} = \left(\dfrac{3}{4}\right)^k x + \left(\dfrac{3}{2}\right)^k - 1$$
and the exponential "$+1$" term dominates for large $n$. Now, of course, $m$ won't always be as large as possible, but even if we look at "random" $m$, that only introduces a constant factor in front of the exponential.
Additional Questions
Is the proof of this statement difficult? Is that why the author doesn't include it? Is there a paper containing a proof?

The "+1" term is just included in the bound,
$$3^{N^*(k)}\leq\frac{T^{(k)}(n_0)}{n_0}2^{k}$$ 
see construction here: A possible way to prove non-cyclicity of eventual counterexamples of the Collatz conjecture?
If you start with this:
$$T^{(k)}(n_0) = \frac{3^{N^*(k)}}{2^{m_1+m_2+...+m_i}}\cdot n_0+\frac{3^{N^*(k)-1}}{2^{m_1+m_2+...+m_i}}+\frac{3^{N^*(k)-2}}{2^{m_1+m_2+...+m_{i-1}}}+...+\frac{3^0}{2^{m_1}}$$
see construction here: $(n_0) \to (n_i)$ and $(n_0+k\cdot2^j)\to(n_i+k\cdot3^i)$, same Glide?
This can be simplified to
$$T^{(k)}(n_0) = \frac{3^{N^*(k)}}{2^k}\cdot n_0+\alpha N^*(k)$$
where $\alpha < \frac{1}{3}$ as shown in this paper: https://arxiv.org/abs/1710.01525v1
$$2^k = 3^{N^*(k)}\frac{n_0}{T^{(k)}(n_0)-\alpha N^*(k)}$$
$$k = N^*(k)\log_23+\log_2(\frac{n_0}{T^{(k)}(n_0)-\alpha N^*(k)})$$
Now as long as $T^{(k)}(n_0)-\alpha N^*(k)>=n_0$
$$k \leq N^*(k)\log_2(3)$$
$$\frac{N^*(k)}{k} \geq \frac{1}{\log_23}$$
I didn't check Theorem F, but i guess this is linked to his remark "cannot go too slowly"
