Average power of 2 in all even natural numbers

Question

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?

Answer 1

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 $n \to \infty$.

Answer 2

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.