0.999...=1

Theorem: 0.999...=1

Using Geometric Series
We can represent $$0.999...\,$$ as a geometric series with first term $$a=\frac{9}{10}$$, and ratio $$r=\frac{1}{10}$$. Since our ratio is less then 1, then we know that $$\sum_{n=0}^{\infty}\frac{9}{10}\left({\frac{1}{10}}\right)^n$$ must converge to: $$\frac{a}{1-r}=\frac{\frac{9}{10}}{1-\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{9}{10}}=1$$ QED

Using Fractions
$$.333...=(1/3)$$

$$(3)(.333...)=(1/3)(3)$$

$$(.999...)= (3/3)=1$$

Using Multiplication by 10
Let $$c=0.999\dots$$ $$10c=9.999\dots$$ $$10c-c=(9.999\dots)-(.999\dots)$$ $$9c=9\,$$ $$c=1\,$$ $$0.999\dots=1$$