Lets start with a negative number: -7
$$ -7\times3+1=-20\\ -20\div2=-10\\ -10\div2=-5\\ -5\times3+1=-14\\ -14\div2=-7 $$
we can sum up these steps as the following functions:
$$ -7\times3+1=-20\\ {-7\times3+1\over2^2}=-5\\ {-7\times3^2+3-2^2\over2^2}=-14\\ {-7\times3^2+3-2^2\over2^3}=-7\\ $$ $$ f_1(x)={x\times3^2+3-2^2\over2^3}\\ $$
$$ {-7\times3^4+3^3+3^2\times2^2+3\times2^3+2^5\over2^6}=-7\\ $$ $$ f_2(x)={x\times3^4+3^3+3^2\times2^2+3\times2^3+2^5\over2^6}\\ $$
Where $f_1(x)$ is the function that sum up the steps: $-7\to-20\to-10\to-5\to-14\to-7$ and $f_2(x)$ is the function that sum up the steps: $-7\to-20\to-10\to-5\to-14\to-7\to-20\to-10\to-5\to-14\to-7$
And:
$$ {\lim_{x \to \infty} f_1(x)\over\lim_{x \to \infty} f_2(x) }= {-5\times3\over-5\times3+1}\times{-7\times3\over-7\times3+1}= 8/9 $$
Remarks: -5 and -7 are the odd number in the sequence: $-7\to-20\to-10\to-5\to-14\to-7$
In the same way, if we let $x$=-25
$$ {-25\times3^7+3^6+3^5\times2+3^4\times2^2+3^3\times2^4+3^2\times2^5+3\times2^6+2^{10}\over2^{11}}=-25\\ $$
$$ {-25\times3^{14}+3^{13}+3^{12}\times2+3^{11}\times2^2+3^{10}x2^4+3^9\times2^5+3^8\times2^6+3^7\times2^{10}+3^6\times2^{11}+3^5\times2^{12}+3^4\times2^{13}+3^3\times2^{15}+3^2\times2^{16}+3\times2^{17}+2^{21}\over2^{22}}=-25\\ $$ $$ f_3(x)={x\times3^7+3^6+3^5\times2+3^4\times2^2+3^3\times2^4+3^2\times2^5+3\times2^6+2^{10}\over2^{11}}\\ $$
$$ f_4(x)={x\times3^{14}+3^{13}+3^{12}\times2+3^{11}\times2^2+3^{10}x2^4+3^9\times2^5+3^8\times2^6+3^7\times2^{10}+3^6\times2^{11}+3^5\times2^{12}+3^4\times2^{13}+3^3\times2^{15}+3^2\times2^{16}+3\times2^{17}+2^{21}\over2^{22}}\\ $$
Then
$$ {\lim_{x \to \infty} f_3(x)\over\lim_{x \to \infty} f_4(x) }= {-17\times3\over-17\times3+1}\times {-25\times3\over-25\times3+1}\times {-37\times3\over-37\times3+1}\times {-55\times3\over-55\times3+1}\times {-41\times3\over-41\times3+1}\times {-61\times3\over-61\times3+1}\times {-91\times3\over-91\times3+1}= 2187/2048 $$
And as Collatz conecture restricts to positive numbers:
$$ {1\times3+1\over2^2}=1\\ $$ $$ f_5(x)={x\times3+1\over2^2}\\ $$ $$ {1\times3^2+3+2^2\over2^4}=1\\ $$ $$ f_6(x)={x\times3^2+3+2^2\over2^4}\\ $$
And:
$$ {\lim_{x \to \infty} f_5(x)\over\lim_{x \to \infty} f_6(x) }= {1\times3\over1\times3+1}= 3/4 $$
Can we say that in the function $f(n) = \begin{cases} n/2, & \text{if $n$ is even} \\ 3n+1, & \text{if $n$ is odd} \end{cases}$ a loop occurs if:
$$ {\lim_{x \to \infty} f_1(x)\over\lim_{x \to \infty} f_2(x) }= {a_1\times3\over a_1\times3+1}\times{a_2\times3\over a_2\times3+1}\times...{a_n\times3\over a_n\times3+1} $$