0.999...=1

Theorem

 * $$0.999\ldots=1$$

Proof: Using Geometric Series
We can represent $$0.999 \ldots \,$$ as the sum of an infinite geometric progression with first term $$a=\frac{9}{10}$$, and ratio $$r=\frac{1}{10}$$.

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


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

Proof: Using Fractions
$$ $$ $$ $$

Proof: Using Multiplication by 10
$$ $$ $$ $$ $$

Therefore we have that $$0.999\ldots=1$$.