the partial sums (convergence is all we really need) Combining all these inequalities, we see that Dividing through by and taking the natural logarithm of both sides gives as desired. Q.E.D. Using (see the Basel problem), the above constant can be improved to ; in fact it turns out that where is the Meissel–Mertens constant (somewhat analogous to the much more famous Euler–Mascheroni constant). Proof from Dusart's inequality From Dusart's inequality, we get Then by the integral test for convergence. This shows that the series on the left diverges. Geometric and harmonic-series proof Suppose for contradiction the sum converged. Then, there exists such that . Call this sum . Now consider the convergent geometric series . This geometric series contains the sum of reciprocals of all numbers whose prime factorization contain only primes in the set . Consider the subseries . This is a subseries because is not divisible by any . However, by the Limit comparison test, this subseries diverges by comparing it to the harmonic series. Indeed, . Thus, we have found a divergent subseries of the original convergent series, and since all terms are positive, this gives the contradiction. We may conclude diverges. Q.E.D. Partial sums While the partial sums of the reciprocals of the primes eventually exceed any integer value, they never equal an integer. One proof is by induction: The first partial sum is , which has the form . If the th partial sum (for ) has the form , then the st sum is as the st prime is odd; since this sum also has an form, this partial sum cannot be an integer (because 2 divides the denominator but not the numerator), and the induction continues. Another proof rewrites the expression for the sum of the first reciprocals of primes (or indeed the sum of the reciprocals of any set of primes) in terms of the least common denominator, which is the product of all these primes. Then each of these primes divides all but one of the numerator terms and hence does not divide the numerator itself; but each prime does divide the denominator. Thus the expression is irreducible and is non-integer. See also Euclid's theorem that there are infinitely many primes Small set (combinatorics) Brun's theorem, on the convergent sum of reciprocals of the twin primes List of sums of reciprocals References Sources External links Mathematical series Articles containing proofs