Is this a valid statement that would imply the Collatz Conjecture? Let $f$ denote the Collatz transformation: $f(x) = \left\{
        \begin{array}{ll}
            {x\over 2} & \quad x\equiv 0 \mod 2 \\
            3x+1 & \quad x \equiv 1\mod 2
        \end{array}
    \right.$
Let $f^n(x)$ denote $n$ successive applications of $f$ to $x$.
Call a number $x\in\Bbb N$ convergent iff there exists $n\in \Bbb N$ such that $f^n(x) = 1$.

Proposition. If, for any odd number $n$, $f^k(n) < n$ for some $k\in\Bbb N$, then all numbers are convergent (Collatz Conjecture).

We proceed by strong induction. The base case is trivial - 1 is convergent. Suppose $1,2,\dots,n-1$ are convergent. If $n$ is even, then $1\leq {n\over 2} < n$ is convergent by the inductive hypothesis and we are done. If $n$ is odd, by assumption we can find $k$ such that $f^k(n) < n$. But then $f^k(n)$ converges by the inductive hypothesis. The result follows.
If this is valid (which somehow I don't think it is), it gives an interesting outlook on the conjecture as asserting the non-existence of odd numbers that somehow "spiral out" infinitely under iteration of $f$. Is this proof correct, and if so, is it part of a typical description of the Collatz conjecture?
 A: 
Is this proof correct, and if so, is it part of a typical description of the Collatz conjecture?

The answers to these questions are yes, and yes. :)
Indeed, there are at least a couple of standard approaches that include this argument. For example, we can see that, if $n=4k+1$ for some positive integer $k$, then $n$ cannot be the smallest non-convergent odd number, because it leads to the trajectory: $$f(n)=12k+4,\\ f^2(n)=6k+2,\\ f^3(n)=3k+1<n.$$
Thus, if we seek the smallest odd number not converging to $1$, we need only consider odds of the form $4k+3$. Every number of the form $4k+3$ can either be written as $8k+3$ or as $8k+7$; these can then be written in only a few ways relative to $16, 32$, etc. At each step, we can follow the trajectory and see whether $n$ gets smaller in a few steps. Some cases, we cannot eliminate: $$8k+7\to 24k+22\to 12k+11\to 36k+34\to\\ 18k+17\to 54k+52 \to 27k+26\to  \,\,??,$$ and now we're stuck, because we don't know whether $27k+26$ is even or odd without knowing more about $k$. On the other hand: $$16k+3\to 48k+10\to 24k+5\to 72k+16\to\\ 36k+8\to 18k+4\to 9k+2 < n,$$
So any number of the form $16k+3$ is guaranteed to reduce, and we needn't consider $32k+3$ or $32k+19$.
This line of reasoning has been followed, and although we can narrow down the list of candidates for the smallest odd non-convergent, we cannot reduce it to nothing.

Taking a different approach, one can define a stopping-time function $h(n)=\min\{k:f^k(n)<n\}$, with $h(n)=\infty$ if all iterates remain larger than or equal to $n$. The Collatz conjecture is then equivalent to the claim that $h(n)<\infty$ for all $n>1$. This function has also been studied, but its properties are not nice in a way that yields to any obvious analysis, which becomes clear if you start playing with it.
