0.999.../Proof by the geometric series formula

ProofEdit

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

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

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

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

NotesEdit

Recall that,

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