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