What fraction of all $\mathbb{N}$ are powers of 2? (I go into more detail bellow, but a "fraction of all $\mathbb{N}$" is like saying "half of all numbers are even.")
When thinking about this, I had some trouble trying to figure out what is the best way to approach this question. I narrowed it down to three possible answers, and I am not sure which one would be considered correct:
A1) There is no such thing as a fraction of all $\mathbb{N}$ consisting of only the powers of 2, it does not exist, or it cannot exist.
A2) There is a fraction of all $\mathbb{N}$ that contain all of the powers of 2.
A3) There is no such thing for all $\mathbb{N}$, but this can be answered if the given set was a sequence ending at some number $c$.

Are any one of these three answers the right answer? Or is there a
  different explanation that best answers this question?

For more context, here is how I came across this question and what I worked out on this problem:
I came up with this question while studying the Collatz Conjecture. (As a quick summary/refresher, if a number $x$ is odd, multiply by 3 and add 1, but if even divide $x$ by two, and go on forever until you reach 1.)
I looked at a modified Collatz Rule $3x+5$ ($3x+5$ replaces the odd rule), and not only are there other loops, but it seems to be that the different loops have different "sizes". Afterwards, I wondered if the sequence of the powers of 2 can be tied to some ratio to all $\mathbb{N}$, and whether or not that gives any context to my previous question or the Collatz Conjecture in general.
When I think of "a fraction of all $\mathbb{N}$", I am thinking of something like "half of all $\mathbb{N}$ are even" or for $3x+5$, "It seems to be all multiples of 5 go to the 20-10-5 loop, and exactly 1/5 of all $\mathbb{N}$ are divisible by 5, therefore the 20-10-5 loop for the Collatz rule $3x+5$ contains 1/5 of all $\mathbb{N}$, assuming the Collatz Conjecture is true."
I came to the "1/5" conclusion like this:
If {$\mathbb{N}$} = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
and the multiples of 5 are = 5, 10, 15, ...
the multiples of 5 are found every 5th term for {$\mathbb{N}$}.
...So I did exactly that for the powers of 2... The only problem is there is a power of 2 for every $2^n$th term, which basically iterates the question.
I know the sequence of the powers of 2 converges to zero, which in my mind is not very helpful, but A1) may be the answer....
...except that I found this "formula" that redefines the even numbers, which made me more confused and less convinced all of the powers of 2 are a bunch of added up zeros to get 1.
While studying the Collatz Conjecture, I came across the sequence A007814 a lot:

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, ...

And according to this sequence, I could pull off the same trick I did with the number of 5's for {$\mathbb{N}$}, only instead count the number of powers of 2:
0, 1, 2, 3, 4, 5, 6, 7, ...

which is really (1 * $2^0$), (1 * $2^2$), (1 * $2^3$), (1 * $2^4$), ... and so on.
Therefore, I came up with $a*2^N$ to represent this sequence, where $a$ is any positive odd number and $N$ is the starting point of the sequence {N} = 0, 1, 2, 3, 4, 5, 6, ...
Therefore, assuming all $a$ are iterated at once, $a*2^N$ where $N$ starts at zero = {$\mathbb{N}$}, half of {$\mathbb{N}$} when $N$ starts at 1, and so on. Since this new sequence also converges to zero, I suspected A2) to be the most plausible answer.
 A: Talking about asymptotic/natural density, you can play with it. Perhaps you'll find what you seek.
e.g. in the Condensed Collatz tree (nodes containing only odd numbers) build with the condensed Collatz function $\frac{3n+1}{2^k}$, you see that 
the children of $6x+1$ are $8x+1; 32x+5; 128x+21; 512x+85....$ (with the well known series $\frac{4^i-1}{3}: 1, 5, 21, 85, ...$), each representing a fraction of naturals (e.g. $8x+1$ -> $\frac{1}{8}$ of naturals).
The children of $6x+5$ are $4x+3;16x+13;64x+53;256x+213....$ (with the well known series $\frac{10\cdot4^i-1}{3}: 3, 13, 53, 213, ...$), and multiple of 3 have no children.
Now you can find the densities/repartitions for the children with these formulas:
$\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}...= \sum\limits_{k\geq 2}\frac{2}{4^k} = 2\sum\limits_{k\geq 2}\frac{1}{4^k}=2((\sum\limits_{k\geq 1}\frac{1}{4^k})-\frac{1}{4})=2(\frac{1}{3}-\frac{1}{4})=\frac{1}{6}$
$\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}...= \sum\limits_{k\geq 1}\frac{1}{4^k}=\frac{1}{3}$ (all fractions of naturals)
For instance, you can see that for the $6x+1$ branchs, the first child account for a density of $\frac{1}{8}$ and all the other children bellow, which seems infinitely more numbered, account for only $\frac{1}{24}$ ($\frac{1}{6}-\frac{1}{8}$), which is counter-intuitive.
Note, a branch is constructed on all power of 2 multiplied by it's odd root $6x+1$, and half of them leads to a(n odd) child. (put in another way: $\frac{3(child)+1}{2^k}=root$ -> $3(child)+1=root\cdot2^k$ for any k of the same parity)
A: There is a notion of natural density which seems like what you are asking for. For each $n$, there are approximately $\log(n)$ powers of 2 which are less than $n$, so that the powers of two have natural density $0$.
A: This is an old question, but I am seeing it for the first time. For what it's worth, this is how I immediately thought of it. Think in binary, for a power of $2$ you need a lead bit of $1$ will no other bits or all other bits being $0$. Consider only binary strings that start with a leading $1$ in the most significant bit. Within that set, for one bit starting with lead bit $1$ (which is basically just $1$), one of one, i.e $1/1$ will be a power of $2$. For a string of two bits starting with lead bit $1$, one of two, i.e $1/2$ will be a power of $2$. For a string of $k$ bits starting with lead bit $1$, one of $2^{k-1}$, i.e. $\frac 1{2^{k-1}}$ will be a power of $2$.
Summing over all binary strings starting with a lead bit $1$ up to $n$ bits, you have $\frac n{2^n - 1}$ which is the fraction of powers of $2$ among binary strings of length $n$ starting with $1$.
By one definition of the naturals (the one I prefer), zero is excluded, so you've basically covered the first $2^n-1$ naturals by doing that. If you also consider $0$ a natural, just add $1$ to the denominator to get $\frac n{2^n}$ as the required answer.
If you thought about it in terms of the first $m$ natural numbers such that $m$ is one less than a power of $2$, then the answer can be transformed into $\frac{\log_2 m}{m}$, which is an exact expression (because you're choosing the denominator with care). In the general case, truncating at an arbitrary natural number, you will get the same expression as an approximation, exactly as stated in TomGrubb's answer that references natural density.
In any case, the limit as $n \to \infty$ or $m \to \infty$ is still zero, as expected, exactly as stated in TomGrubb's post.
