>>779
>>792
If it is defined 
\(lim_{x \rightarrow +\infty} \frac{1}{x}=0^+ \)
Than 
\( 0.\bar{9} = \frac{9}{10}\Sigma_{n=0}^{+\infty [minus \, 1 ?]} \frac{1}{10}^n ≥ \frac{9}{10}\Sigma_{n=0}^{\log_{10} (+\infty) -1} \frac{1}{10}^n = \frac{9(\frac{1}{10}^{\log_{10}(+\infty)}-1)}{1-10}=1 \) but \( \Sigma_{n=\log_{10} (+\infty)}^{+\infty} \frac{1}{10}^n =0 \) so \( 0.\bar{9}=1 \) 
Note that \(  1^{\log_{10}(+\infty)}=1  \), because \( 10>1  \) so \( \log_{10}{n}<n \, \,\,\, \forall n \in \mathbb{R^+} \)