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There is a common argument as to whether ${\displaystyle 0.999\ldots =1}$ or not. It does.

Proof

${\displaystyle 0.999\ldots }$	${\displaystyle =9\times 0.111\ldots }$
	${\displaystyle =9\times \left({\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)}$
	${\displaystyle =9\times \left(-1+{\frac {1}{1}}+{\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)}$
	${\displaystyle =9\times \left(-1+\sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}\right)}$
	${\displaystyle =9\times \left(-1+{\frac {1}{1-{\frac {1}{10}}}}\right)}$
	${\displaystyle =9\times \left(-1+{\frac {10}{9}}\right)}$
	${\displaystyle =1\,}$

Explaination

The key step to understand here is that ${\displaystyle \sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}={\frac {1}{1-{\frac {1}{10}}}}}$. This rests on a geometric series converging if the common ratio is between -1 and 1 exclusive.

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