A common convention is to denote the number of primes below N as π(N). The fact that there are infinitely many primes is equivalent to π(N) → ∞ when N → ∞.
q = 2·3·5·7·...·p + 1.
for n > 1.
Lemma
pn < 22n
Proof
The proof is by induction. Suppose the claim holds for 1 ≤ n < N. Then, following Euclid,
pn+1 < p1p2·...·pN + 1 < 22+4+...+2N+1 < 22N+1.
Suppose now that n> 4 and
een-1 < x ≤ een.
Then, en-1 > 2n and een-1 > 22n, so that
π(x) ≥ π(een-1) ≥ π(22n) ≥ n.
Since ln(ln(x)) ≤ n, we deduce that
π(x) ≥ ln(ln(x)).
Now, this is a matter of direct verification that the same inequality is also true for x = 2 and n = 3. Therefore, it holds for n ≥ 2. In any event, it implies that
π(N) → ∞, for N → ∞.
We may actually do better. Elsewhere we defined N(x) as the number of terms not exceeding x and not divisible by any prime above pk, for a fixed k. We found that
N(x) ≤ 2k√x.
Take now k = π(x), so that pk+1 > x and N(x) = x. Then,
x = N(x) ≤ 2π(x)√x,
implying √x ≤ 2π(x). Thus we have
π(x) ≥ ln(x)/ln(4).
Reference
- G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford Science Publications, 1996, p. 12, p.17
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