# 0.999...=1/Proof 1

## Theorem

$0.999 \ldots = 1$

## Proof

$0.999 \ldots = \dfrac a {1 - r}$

where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.

Since our ratio is less than $1$, then we know that $\displaystyle \sum_{n \mathop = 0}^\infty \frac 9 {10}\left({\frac 1 {10}}\right)^n$ must converge to:

$\dfrac a {1-r} = \dfrac{\frac 9 {10}} {1 - \frac 1 {10}} = \dfrac {\frac 9 {10}} {\frac 9 {10}} = 1$

$\blacksquare$