# How do you use a geometric series to prove that 0.999…=1?

$0.999 \ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$
$= \frac{9}{10} + \frac{9}{10} \left(\frac{1}{10}\right) + \frac{9}{10} {\left(\frac{1}{10}\right)}^{2} + \cdots$
$= {\sum}_{n = 0}^{\infty} \frac{9}{10} {\left(\frac{1}{10}\right)}^{n}$,
$a = \frac{9}{10}$ and $r = \frac{1}{10}$.
$\frac{a}{1 - r} = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1$