Erdős' proof that the sum of the reciprocals of the primes diverges is far more modern than the "classical" (Euler's?) proof that the sum of the reciprocals of the primes diverges. It is almost completely elementary, and refreshingly different.

Suppose

∑_{p prime} 1/p < ∞

converges. Then for some P,

∑_{p prime, p> P}1/p < 1/2

by the very definition of

series convergence. Call primes p>P "large primes", and the other primes 2≤p≤P "small primes". By noting that any prime p divides approximately 1/p of the numbers

(we can make this precise by using the notion of density, but I shall refrain from doing so as it is unnecessary here), we shall show that "not enough" numbers are divisible by the large primes.

Fix some N>P (we will later set a precise bound on N). Let N_{1} be the number of numbers 1≤n≤N which have all prime divisors "small primes", and let N_{2}=N-N_{1} be the number of numbers with at least one divisor which is a "large prime". We always have

N_{2} ≤ ∑_{p prime, p>P} ⌊N/p⌋ ≤
∑_{p prime, p>P} N/p <
N/2,

since every number with a large divisor is divisible by at least one p>P, and each such p divides ⌊N/p⌋ numbers between 1 and N.

So if we can show N_{1}≤N/2 for some N, we'll have a definite contradiction: N=N_{1}+N_{2} cannot be the sum of 2 numbers, each smaller than N/2.

How many numbers are divisible only by "small primes"?
Suppose there are exactly k prime numbers p_{1}=2 ≤ p_{2}=3 ≤ ... ≤ p_{k}=P up to P. If x≤N is some number with only these k primes as its prime divisors, write x=y⋅z^{2} with y squarefree and z≤√N some integer. Since y is squarefree, every prime p either divides it exactly once (p|y but p^{2}~~|~~y) or not at all (p~~|~~y). Only only the k primes p_{1},...,p_{k} can divide x, so only these k primes can divide y -- there are no more than 2^{k} possible choices for y. And there are no more than √N possible choices for z, as z≤√N.

We've shown that

N_{1} ≤ 2^{k}⋅√N.

Now pick N>2

^{2k+2}. Then we have our contradiction:

N = N_{1} + N_{2} < 2^{k}⋅√N + N/2 <
√N/2⋅√N + N/2 = N.

So no such P=p

_{k} can exist -- the

series ∑

_{p prime} 1/p cannot converge.