0.999.../Proof by the geometric series formula

Assumptions

• Definition from series
• The geometric series formula

Proof

Using the series definition of the value of an infinite decimal,

$0.999\ldots =9\left({\tfrac {1}{10}}\right)+9\left({\tfrac {1}{10}}\right)^{2}+9\left({\tfrac {1}{10}}\right)^{3}+\cdots .\,$

This is a geometric series with a common ratio of 1/10. Applying the geometric series formula,

$0.999\ldots =9\left({\tfrac {1}{10}}\right)+9\left({\tfrac {1}{10}}\right)^{2}+9\left({\tfrac {1}{10}}\right)^{3}+\cdots ={\frac {9\left({\tfrac {1}{10}}\right)}{1-{\tfrac {1}{10}}}}=1.\,$

Notes

Recall that,

$\sum _{n=0}^{\infty }ax^{n}=\lim _{t\to \infty }{\frac {a(1-x^{t})}{1-x}}\underbrace {=} _{{\text{Given }}|x|<1}{\frac {a}{1-x}}$