[Mathematics, infinities, proof by contradiction] Is the following statement valid?

Hello everyone. I am new to this forum and my first language is not English so I may make mistakes concerning the mathematical vocabulary (mathematical concepts as well) : so bear with me.

I had a question related to infinities.

Here is some background(if I make mistakes, or you do not understand how I formulated something, do please reply):

Let a(x) = x/2 and b(x) = 3x+1
Let f be a function: f(x) returns a(x) if x is even and b(x) if x is odd.

Let n be a positive integer.
Let (s_n) denote the series where s_i = f(f(...f(n))) where f is called i times.
Suppose s_i = ((3^x)*n + z)/(2^y) where x, y, and z are positive integers.

Suppose that for all s_i, there exists an s_j such that s_i < s_j.
Consequently, s_i "grows" towards infinity, the larger i is.

After an infinite number of steps:
- if b is called a finite number of times, a is called an infinite number of times (for when a is called b is not and vice versa).
- As a result, if b were not called an infinite number of times when i approaches infinity, s_i would decrease, which contradicts our hypothesis. b is therefore called an infinite number of times.
- because b(n) is even, a is called after it (because there is an infinite number of steps), so a is called at least as often as b: so a is also called an infinite number of times.

Here is where I need your help (if there was no mistake on what I just wrote): is the following statement valid?

Because s_i = ((3^x)*n + z)/(2^y) , if a is called an infinite number of times, 2^y is arbitrarily big, so is 3^x because b is called an infinite number of times.
However, after an infinite number of steps, by hypothesis s_i is arbitrarily big.
We couldn't conclude that s_i is arbitrarily big if a and b were called the same infinity of time, so b is called an "infinity higher" than a.

However, a is called at least as much as b: there is a contradiction so such an n with the given hypothesis cannot exist.

Thank you for reading and for your help!