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 $\displaystyle \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to:

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

$\blacksquare$