<s>[INST] 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
 [/INST] 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.
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