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+--- Thread: Tetration convergence (/showthread.php?tid=1742)
Tetration convergence - Daniel - 04/24/2023
\[(1+a)^n = \sum_{k=0}^n {n \choose k}a^k \]
Consider the following transseries
\[\text{Let } n=\;^{m-1}(1+a) \text{ with } a\in\mathbb{N}\]
\[^m(1+a) =(1+a)^{(^{m-1}(1+a))} = \sum_{k=0}^{^{m-1}(1+a)} {{^{m-1}(1+a)} \choose k}a^k\]
\[{{^{m-1}(1+a)} \choose k} = 0 \text{ for } ^{m-1}(1+a)>k\]
\[\text{Thus } ^m(1+a) \text{ is convergent.}\]
RE: Tetration convergence - tommy1729 - 04/25/2023
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