Reducing erratic Collatz sequence to converging sequences

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This approach refers to a half-of-a-century-old paper: Corrado Böhm, Giovanna Sontacchi, On the existence of cycles of given length in integer sequences like $x_{n+1}=x_n/2$ if $x_n$ even, and $x_{n+1}=3x_n+1$ otherwise, Atti Accad. Naz. Lincei, Rend. Cl. Sci. Fis. Mat. Natur. 8(64), 1978.

Outlook

Regardless of whether the Collatz conjecture is true or not, any positive integer n can be represented via a term $k_i$ of its “shortcut” Collatz sequence $(k_i)$, $i\geq0$:

$n=\frac{2^i\ k_i-b_i}{3^{c_i}}$,

where the integer $c_i$ is the number of odd terms preceding $k_i$, $c_0=0$, and the non-negative integer $b_i$ is sequentially calculated as following:

$b_0=0$; $b_{i>0}=$ [tex]\begin{cases} b_{i-1} \: if \: k_{i-1}≡0 \: (mod \: 2); \\ 3b_{i-1}+2^{i-1} \: if \: k_{i-1}≡1 \:(mod \: 2). \end{cases}[/tex]

As it follows from above, $b_{i>0}=0$ if $c_i=0$, otherwise $b_{i>0}$ has a $c_i$-special 3-smooth representation:

$b_{i>0}=\sum_{j=0}^{c_i-1}{3^{c_i-1-j}\times2^{a_j}},\ \ \ a_j\in N,\ \ \ 0\le a_0<\ a_1<,\ \ldots,\ <a_{c_i-1}<i . $

According to above, $k_{a_j}$ is the $(j+1)$-th odd term of the Collatz sequence, $c_{a_j}=j$, $a_j$ is the number of terms preceding $k_{a_j}$, and $a_{j+1}-a_j-1\geq0$ is the number of even terms between odd terms $k_{a_j}$ and $k_{a_{j+1}}$.

Now, let’s consider the sequence $\left(\frac{3^j}{b_{a_j}}\right)_{j\geq1}$.
Lemma 1. For any Collatz sequence starting with $n\in N^+$, the sequence $\left(\frac{3^j}{b_{a_j}}\right)_{j\geq1}$ is decreasing.
Lemma 2. For any Collatz sequence starting with $n\in N^+$, [tex]\lim_{j \to \infty }{\frac{3^j}{b_{a_j}}=0}[/tex]
At this point, for the Collatz sequence starting with $n\in N^+$, let’s consider the sequence $\left(U_j\right)_{j\geq1}$,$\ \ U_j\in Q$; every term of this sequence is defined as $U_j=\frac{3^jn-2^{a_j}}{b_{a_j}}$.

Lemma 3. For any Collatz sequence starting with $n\in N^+$, $U_{j+1}-U_j<0$ if $k_{a_j}>1$.

Corollary 1. For any Collatz sequence starting with $n\in N^+$, $U_j>1$ if $a_{j+1}-a_j=1$.

Now, let’s look at the diverging Collatz sequence and the Collatz sequence that collapses to the non-trivial cycle. In both cases, all the terms of the Collatz sequence are bigger than 1, and there are infinitely many terms $k_{a_j}$, such that $a_{j+1}-a_j=1$. According to Lemma 3, the sequence $\left(U_j\right)_{j\geq1}$ is decreasing as $k_{a_j}>1$. Then, considering Corollary 1, $U_j>1$ for any $j\geq1$, and thus, [tex]\lim_{j \to \infty }{U_j \ge 1}[/tex] in both cases.

At the same time, according to Lemma 2, [tex]\lim_{j \to \infty }{\frac{3^j}{b_{a_j}}=0}[/tex], so, considering the definition of $U_j$, [tex]\lim_{j \to \infty }{U_j \le 0}[/tex], which is not simultaneously possible with [tex]\lim_{j \to \infty }{U_j \ge 1}[/tex]. This means that both the Collatz sequence that diverges and the Collatz sequence that collapses to the non-trivial Collatz cycle are impossible. The only remaining possibility is that any Collatz sequence starting with $n\in N^+$ collapses to the trivial Collatz cycle (1, 2). This proves the Collatz Conjecture.

More details are available in a 5-page preprint: https://osf.io/preprints/osf/bc4fe