How high is the percentage of primes in $\mathbb{N}$?
($\mathbb{N} := \lbrace { 1, 2, 3, \ldots \rbrace }$ ; a prime is only divisible by itself and 1 in $\mathbb{N}$)
The percentage has to be lower than 50% as all even numbers (except for 2) aren't primes.
So the percentage has to be lower than $1 – (\frac{1}{2} + (\frac{1}{3} – \frac{1}{6})) = 1 – (\frac{1}{2} + \frac{1}{6}) = 1 – (\frac{1}{2} + \frac{1}{2 \cdot 3}) = \frac{1}{3}$
I guess it will be something like that:
$$\frac{\text{primes in } \mathbb{N}}{\text{numbers in } \mathbb{N}} = 1 – \sum_{i=\text{first Prime}}^\text{primes} \frac{1}{\prod_{j=\text{first prime}}^{i\text{-th prime}} j}$$
But calculating this (exact) value goes definitely beyond my math skills. Can somebody help me?
Best Answer
People are quoting the Prime Number Theorem here, but that is pretty serious overkill. Even Chebyshev's theorem is a level above, in my opinion.
In my undergraduate number theory course, I prove that the prime numbers have density zero in the positive integers (including a careful statement of what this means). The proof is given as Theorem 6 in these notes (Wayback Machine). In fact it builds on the OP's observation that the density must be at most $\frac{1}{2}$ because half of all numbers are divisible by $2$. Similarly the density must be at most $\frac{1}{3}$ because every prime number $p > 5$ is relatively prime to $6$, and $\frac{\varphi(6)}{6} = \frac{1}{3}$. One can get a complete proof by showing that for each $\epsilon > 0$, there exists a positive integer $d$ such that $\frac{\varphi(d)}{d} < \epsilon$. This was proved earlier in the course: it is Proposition 6 in this set of notes (Wayback Machine). The proof doesn't use any analytic fact deeper than the divergence of the harmonic series.