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.
∑p prime 1/p < ∞
s. Then for some P,
∑p prime, p> P1/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 N1 be the number of numbers 1≤n≤N which have all prime divisors "small primes", and let N2=N-N1 be the number of numbers with at least one divisor which is a "large prime". We always have
N2 ≤ ∑p prime, p>P ⌊N/p⌋ ≤
∑p prime, p>P N/p <
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 N1≤N/2 for some N, we'll have a definite contradiction: N=N1+N2 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 p1=2 ≤ p2=3 ≤ ... ≤ pk=P up to P. If x≤N is some number with only these k primes as its prime divisors, write x=y⋅z2 with y squarefree and z≤√N some integer. Since y is squarefree, every prime p either divides it exactly once (p|y but p2
|y) or not at all (p |y). Only only the k primes p1,...,pk can divide x, so only these k primes can divide y -- there are no more than 2k possible choices for y. And there are no more than √N possible choices for z, as z≤√N.
We've shown that
N1 ≤ 2k⋅√N.
Now pick N>22k+2
. Then we have our contradiction:
N = N1 + N2 < 2k⋅√N + N/2 <
√N/2⋅√N + N/2 = N.
So no such P=pk
can exist -- the series
1/p cannot converge.