Tetration convergence

[Daniel]

\[(1+a)^n = \sum_{k=0}^n {n \choose k}a^k \]

Consider the following transseries


\[\text{Let } n=\;^{m-1}(1+a) \text{ with } a\in\mathbb{N}\]

\[^m(1+a) =(1+a)^{(^{m-1}(1+a))} = \sum_{k=0}^{^{m-1}(1+a)} {{^{m-1}(1+a)} \choose k}a^k\]

\[{{^{m-1}(1+a)} \choose k} = 0 \text{ for } ^{m-1}(1+a)>k\]

\[\text{Thus } ^m(1+a) \text{ is convergent.}\]

[tommy1729]

why not use the gamma function and generalized binomium theorem ?

Your condition of being 0 fails then.
But we get the usual interpretation of powers.

And things are analytic.

Or maybe this is about rounding to the closest integer ?
And thereby making a shortcut computation ?


regards

tommy1729

[leon]

i like this.

the convergence is such that two variables like several functions and superfunctions should fit the sums