Average power of 2 in all even natural numbers Consider all even natural numbers.
Every 4th number has a power of 4 (or $2^2$)
Every 8th number has a power of 8 (or $2^3$)
Every 16th number has a power of 16 (or $2^4$)
What is the average number of power of 2 in any random even number?
Since every 2nd number has a power of 2,and every 4th number is a power of 4 and so on
$\left(2 \times \dfrac{1}{2}\right) + \left(4 \times \dfrac{1}{4}\right) +  \left(8 \times \dfrac{1}{8}\right) + \ldots$
This goes all the way to Infinity and hence the answer is $\infty$. Is this correct?
 A: A multiple of $4$ only has two powers of $2$ in its factorization and a multiple of $8$ only has three.  Your sum should therefore be 
$$\left(1 \times \dfrac{1}{2}\right) + \left(2 \times \dfrac{1}{4}\right) +  \left(3 \times \dfrac{1}{8}\right) + \ldots=2$$
because half the evens have exactly one factor of two, one quarter have exactly two factors of two and so on.  Although it you can't pick a random natural number, this makes sense as the limit of the average number of factors of two in the even numbers up to $n$ as $n \to \infty$.
A: I think you are double counting the multiples of $4$ since half of them have been counted as multiples of $2$. That multiple counting continues.
I think the sum (corrected as I suggest) for the power of $2$ dividing a "random integer", is 
$$
\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots
$$
which sums to $1$. That makes sense for the average power of $2$.
The result for a random even integer would be $2$. 
I put "random integer" in quotes since defining a 
"random integer" is a little subtle. 
