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.

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 $$\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{2}{10}=0.2\text{.}$$
