Expected number of coin flips for heads The expected number of coin flips to get one heads is $2$. What is wrong with this argument? 

There is a $1/2$ chance of getting $H$, $1/4$ chance of getting $TH$,
  $1/8$ chance of getting $TTH$, etc. so the expected value of flips is 
$$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...\approx \ ?$$

Edit: incorrect value

you need to consider the event $A_k$ ={Head shows up in the $k$ flip}
$$P(A_k) = P(T(k-1 \text{times} ) H) = 2^{-k}$$
The expected value of $X = \sum_k k \chi_{A_k}$ which thakes the value $k$ when $A_k$ occurs is
\begin{align}\Bbb{E}[X] = \sum_k k2^{-k} &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots & = 1\\
& \qquad + \frac{1}{4} + \frac{1}{8} + \ldots &= \frac{1}{2} \\
& \qquad \qquad  +\frac{1}{8} + \ldots &=\frac{1}{4}\\
&\qquad \qquad \qquad\vdots&\vdots\end{align}
So $\Bbb{E}[X] = 1 + \frac{1}{2} + \frac{1}{4} + \ldots = 2$ 

Addendum: If the format is a little unclear, try this:
$\begin{align}
\sum_{k=1}^\infty \frac k{2^k} 
& = \frac 1 2 + \frac 2 4 + \frac 3 8 +\cdots
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & (\frac 1 4 & + \frac 1 8 & +\cdots)+ \\ & & (\frac 1 8 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & \frac 1 2(\frac 1 2 & + \frac 1 4 & +\cdots)+ \\ & & \frac 1 4(\frac 1 2 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = 1 + \frac 1 2 + \frac 1 4 + \cdots
\\[2ex] & = 2
\\[1ex]\Box
\end{align}$
