How to show that any computable function is bounded above by Busy Beaver Function? Despite the endless hours of googling, pretty much all of them used an alternative definition of Busy Beaver function. The definition BB(k) I'm interested in is the maximum number of 1's that can be written by k-state before it halts, but the other definition they used was the maximum number of steps, and they're not equivalent. 

And one of the techniques they used is by reduction from halting problem, that if there was any computable function f(x) bounding above the function BB(k), you can solve halting problem in a finite amount of time using the alternative definition, but this argument doesn't work for the definition of BB(k) I am interested in. How do we show that BB(k) is asymptotically larger than any computable function f(x)?
I think a sketch of the proof could be as follows.

First prove BB(k) is uncomputable. Assume it is computable, then it means there is some machine with some number of states m that can compute this value. Show that there exist a larger machine with some number of states M that calls this function on BB(M) and outputs BB(M)+1 1s, and then halts. Therefore there is a contradiction since it outputed more 1s than predicted by BB(M). It's easy to see M = m + C for some constant.

Now regarding the growth.

Assume G(k+1) >= BB(k+1) - BB(k) is computable for k >= K0.

Now, if the above is true then there must exist a machine with m states that computes G(k).

And there must also exist a machine, with a larger number of states M, that guesses BB(K0) and then computes UBB(M) = sum(K0..M)G(k) + BB(k0) and output UBB(M)+1 1s.

So this machine outputs more 1s than our assumed computable upper bound.

Therefore G(k) can not be computed for any K0.