Change of base formula for Tetration

[jaydfox]

Really, I thought it obvious it converges. As n goes to infinity, the epsilon goes to 0 (and quite rapidly at that), making the formula exact in the limiting case.

For publishing this argumentation would not suffice

Well, most of real and complex analysis would fall apart if limiting cases were not sufficient to provide proofs!

To make my point quite plain, I can demonstrate the superlogarithmic constant exists quite easily with bases e and 2.

\(
\begin{eqnarray}
{}^{n} e & < & {}^{(n+2)} 2,\text{ for all } n > -2 \\
{}^{n} e & > & {}^{(n+1)} 2,\text{ for all } n > 3 \\
\end{eqnarray}
\)
And there you have it, the superlogarithmic constant for conversion from base 2 to base e must be between 1 and 2. Once 2^^(n+2) gets ahead of e^^n by significantly more than a factor of log_2(e), it's all over. The tetration of e never has a chance to catch up. The tetration of 2 was given a head start, and it will always stay way ahead of e.

Such a simple demonstration only suffices to show the existence of the superlogarithmic constant. For any two bases, you can define an integer interval within which the constant must lie. But determining its value with any finer precision requires an exact solution for one of the bases.