.999...

Explanation 0

$\begin{align} \frac{1}{3} &= 0.333\dots \\ 1 = \frac{1}{3} \cdot 3 &= 0.999\dots \\ \end{align}$

$\sum_{k=1}^\infty r^k=\frac{r}{1-r}$

$0.999\dots = 9 \times \sum_{k=1}^\infty .1^k=9 \times \frac{.1}{1-.1} = 1$

$\begin{align} a &= 0.999\dots \\ 10a &= 9.999\dots \\ 10a - a &= 9.999\dots - 0.999\dots = 9 \\ 9a &= 9 \\ a &= 1 \end{align}$