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
 A: 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:
$\begin{align}
\sum_{k=1}^\infty \frac k{2^k} 
& = \frac 1 2 + \frac 2 4 + \frac 3 8 +\cdots
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & (\frac 1 4 & + \frac 1 8 & +\cdots)+ \\ & & (\frac 1 8 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & \frac 1 2(\frac 1 2 & + \frac 1 4 & +\cdots)+ \\ & & \frac 1 4(\frac 1 2 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = 1 + \frac 1 2 + \frac 1 4 + \cdots
\\[2ex] & = 2
\\[1ex]\Box
\end{align}$
A: Here is an alternate method for computing the sum (not better, just different).
To compute the sum, note that $$\frac{1}{1-x} = 1 + x + x^2 + x^3+...$$
Differentiate to get $$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3+...$$
Whence $$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + 4x^4+...$$
In particular, taking x = $\frac{1}{2}$ we have
$$ 2 = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...$$
as desired.
A: I think I get what you're trying to do, although the proof is much easier to show from a more direct definition.
The probability of having one coin flip result in heads is $1/2$. The probability of it taking two flips is $1/4$. In general the probability of it taking n flips is $2^{-n}$. So we automatically know you'll never have to flip the coin an infinite number of times.
To find the average number of flips required, simply take the weighted mean (expected value) of all the probabilities.
$$\sum_{n=1}^{\infty} 2^{-n} \cdot n$$
You'll get $2$. This is the average number of flips needed to get heads.
A: Just another way to get the expected value, if you want to consider probability distributions and avoid doing any summation. 
As each toss is independent and identically distributed, we can consider the waiting time for the first "H" to be a Geometric random variable with $p = \frac{1}{2}$. Then for our Geometric random variable $X$, we know 
$$E(X) = \frac{1}{p} = 2 $$
