Let's prove that there exists infinitely many primes.
Suppose there exists only n primes.
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = (1/1+1/2+1/4+…)(1/1+1/3+1/9+…)(1/1+1/5+1/25+…)…(1/1+1/(p_n)+1/(p_n)^2+…)
where p_j denotes the jth prime.
By the unique factorization theorem, when we expand this, we get every reciprocal of a natural number once. So,
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Σ[k=1,∞]1/k.
Also, {1/(p_j)^i} is a geometric sequence with first term 1 and common ratio 0<1/(p_j)<1. So,
Π[j=1,n](Σ[i=0,∞]1/(p_j)^i) = Π[j=1,n]1/(1-1/(p_j)) = Π[j=1,n](p_j)/((p_j)-1).
We obtain
Σ[k=1,∞]1/k = Π[j=1,n](p_j)/((p_j)-1).
The L.H.S. is the harmonic series known to diverge, while the R.H.S. is a finite number.
Contradiction.
Therefore, there exists infinitely many primes.

Let's prove that the sum of the reciprocals of the primes diverges.
Π[j=1,∞](p_j)/((p_j)-1) = ∞.
For any x, 1+x ≦ e^x, so,
Π[j=1,a](p_j)/((p_j)-1) = Π[j=1,a]1+1/((p_j)-1) ≦ Π[j=1,a]e^(1/((p_j)-1)) = e^(Σ[j=1,a]1/((p_j)-1)).
Thus
e^(Σ[j=1,∞]1/((p_j)-1)) = ∞
and
Σ[j=1,∞]1/((p_j)-1) = ∞
follows.
For j≧2, (p_j)-1 ≧ p_(j-1), and 1/((p_j)-1) ≦ 1/(p_(j-1)), so,
Σ[j=1,b]1/((p_j)-1) = 1/(2-1) + Σ[j=2,b]1/((p_j)-1) ≦ 1 + Σ[j=2,b]1/(p_(j-1)) = 1 + Σ[j=1,b-1]1/(p_j).
Thus
Σ[j=1,∞]1/(p_j) = ∞
WWWWW.