0. An Infinity of Primes 3 We now use this same approach to prove a much stronger result: Theorem 0.2. The summation of the reciprocals of the primes di- verges. Proof. Again, assume not. So (0.6) p 1 p = C for some constant C. (We shall use ∑ p to indicate the sum over all primes p.) Label the primes p1,p2,.... As equation (0.6) is a convergent sum of positive terms, at some point it reaches C − 1 2 . That is, there exists r such that (0.7) ir 1 pi 1 2 . Call the primes p1,...,pr small and the other primes large. Call an integer s rare if all of its prime factors are small otherwise, call s ordinary. Again consider the s, 1 ≤ s ≤ n. The rare integers have a factorization (0.1) and so, as with Theorem 0.1, their number is at most most 2r(log2 n)r, polylog to the cognoscenti. What about the ordinary s? For each ordinary s there is some (perhaps several) large prime p dividing it. For a given prime p, the number of elements in 1 ≤ s ≤ n divisible by it is n p , which is at most n p . Thus, the total number of ordinary s is at most ∑ n p , where p now ranges over the large primes. From (0.7), this is less than n 2 . Of the n values of s, less than n 2 are ordinary, so at least n 2 are rare. Thus (0.8) 2r(log2 n)r ≥ n 2 . As with (0.5), for n suﬃciently large (0.5) must fail! We have achieved our reductio ad absurdum: the assumption that the sum of the recip- rocals of the primes is finite must be false, and Theorem 0.2 is true. Remark. There was no need to cut large and small primes precisely via (0.7). The same argument works if the sum of the reciprocals of the large primes is less than one.

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