0.999...=1

Theorem: 0.9999...=1

Using Geometric Series
We can represent $$0.9999...\,$$ 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
$$.33...=(1/3)$$

$$(3)(.33...)=(1/3)(3)$$

$$(.99...)= (3/3)=1$$

Using Multiplication by 10
Let $$c=0.99\dots$$ $$10c=9.9\dots$$ $$10c-c=(9.9\dots)-(.99\dots)$$ $$9c=9\,$$ $$c=1\,$$ $$0.99\dots=1$$

Using Density of Real Numbers
Two numbers are distinct iff a third number can be place between them. Since no number can be fit between .99... and 1 they must be equal.