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,

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

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

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

Notes

Recall that,

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