04/24/2023, 12:05 PM
\[(1+a)^n = \sum_{k=0}^n {n \choose k}a^k \]
Consider the following transseries
\[\text{Let } n=\;^{m-1}(1+a) \text{ with } a\in\mathbb{N}\]
\[^m(1+a) =(1+a)^{(^{m-1}(1+a))} = \sum_{k=0}^{^{m-1}(1+a)} {{^{m-1}(1+a)} \choose k}a^k\]
\[{{^{m-1}(1+a)} \choose k} = 0 \text{ for } ^{m-1}(1+a)>k\]
\[\text{Thus } ^m(1+a) \text{ is convergent.}\]
Consider the following transseries
\[\text{Let } n=\;^{m-1}(1+a) \text{ with } a\in\mathbb{N}\]
\[^m(1+a) =(1+a)^{(^{m-1}(1+a))} = \sum_{k=0}^{^{m-1}(1+a)} {{^{m-1}(1+a)} \choose k}a^k\]
\[{{^{m-1}(1+a)} \choose k} = 0 \text{ for } ^{m-1}(1+a)>k\]
\[\text{Thus } ^m(1+a) \text{ is convergent.}\]
Daniel