This is a simple attempt as a proof of the Collatz Conjecture, as defined in Wikipedia. I must tell you that I'm not a math expert and I did it just for fun. Surelly it have serious flaws I just curious to know what could they be.
Introduction
The conjecture holds true if all natural numbers will make the succession to converge to 1 after a finite number of applications of the recursive function $f$, computing $n/2$ if $n$ is even, and $3n+1$ if $n$ is odd.
A useful observation
One useful observation is that if it holds true for $n \geq 2$ an even number, then must hold true for $f(n)=n/2$, since it is the next number in the succession. Analogously if it holds true for $n \geq 3$ an odd number, then it must hold true for $f(n)=3n+1$ too, since it is the next in the chain. In both cases, also the conjecture remains true for $2n$ the previous number in the chain, that is if $a_i=n$, then $f(a_{i-1})=f(2n)=a_i=n$, since $2n$ will always be even, and is in the same chain that $n$.
Proof
The Collatz conjecture holds for all natural numbers, and that the succession is finite (the convergence to reach 1 requires a finite number of steps).
I will proceed by induction on $a_0=n$, for natural $n \geq 1$.
Base case $a_0=1$: The succession goes: $a_0=1; a_1=3.1+1=4; a_2=2; a_3=1$. Also the succession converged in 3 steps.
Inductive hypothesis: The succession is finite an always converge to 1.
The inductive step is divided in two sub cases: the $n$ is even or the $n$ is odd.
Inductive case: $n \geq 2$ is an even number: This means that exist a natural number $q < n$ such that $a_0=n=2q$. Then the next number in the succession is $a_1=f(a_0)=f(2q)=2q/2=q$. As $q < n$ then by inductive hypothesis lets assume that for $q$ the succession converged in $s$ steps, then for $n$ must converge is $s+1$ steps.
Inductive case: $n \geq 3$ is an odd number: This means that the initial number in the succession is $a_0=n$, an odd number. We can construct a virtual succession by using the number $a_{-1}=2n$, which is an even number, and it is in the same chain on the succession, since $f(a_{-1})=a_0$. As shown previously, the conjecture apply for even numbers, so it holds true for the number $2n$, and therefore for $n$ since it is one step further in the succession.
