0.1999... and 0.2000 are the same? From Sipser's Introduction to the Theory of Computation:

A slight problem arises because certain numbers, such as 0.1999... and 0.2000..., are equal even though their decimal representations are different.

Can someone explain? Because that's news to me.
 A: $$0.1999999...=\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+...$$
$$=\frac{1}{10}+\frac{9}{100}(1+\frac{1}{10}+\frac{1}{100}...)=\frac{1}{10}+\frac{9}{100}\frac{1}{(1-\frac{1}{10})}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}=0.2$$
A: We can write
$$\begin{align}
0.1999 \cdots &= 0.1+0.09+0.099+ 0.999\cdots \\
&= \dfrac{1}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000} + \cdots\\
&= \dfrac{1}{10}+\dfrac{9}{10^2}+\dfrac{9}{10^3}+\cdots \\
&= \dfrac{1}{10}+9\left(\dfrac{1}{10^2}+\dfrac{1}{10^3}+\cdots\right) \\
&= \dfrac{1}{10}+\dfrac{9}{10}\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots\right) \\
&= \dfrac{1}{10}+\dfrac{9}{10}\left[\dfrac{1/10}{1-1/10} \right]
\end{align}$$
since 
$$\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots = \sum_{k=1}^{\infty}\left(\dfrac{1}{10}\right)^k = \dfrac{1/10}{1-1/10}$$
as given here (see equation (9)). The denominator "cancels out" the $9/10$ since $1 - 1/10 = 9/10$, and so we end up with $$\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{2}{10}=0.2\text{.}$$
A: The point is that same numbers may have different expansions in decimal (or other) notation. We can use two different tricks to show why $0.1999 \dots = 0.2000 \dots $:
1) Let $x=0.1999 \dots$, thus $10x=1.999 \dots$ and then $10x-x=1.8$. Solving this simple equation you get $x=0.2=0.1999 \dots$ by previous definition.
2) Write down the expansion of the number (decimal): $0.1999 \dots= (1)10^{-1}+(9) 10^{-2}+(9) 10^{-3}+(9)10^{-4}+ \dots +(9 )10^{-n} + \dots = \frac{1}{10} + 9 \sum_{n=2}^{+\infty} 10^{-n} =  \frac{1}{10} + 9 \left( \frac{1}{1-\frac{1}{10}} - 1 - \frac{1}{10} \right) = \frac{2}{10} = 0.2 $
