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The sum of reciprocals of twin primes converges, whether or not there are infinitely many twin primes. The twin-prime upper bound from a quadratic sieve and reciprocal-sum convergence from a counting bound prove it. This contrasts with the Euler proof that the sum of reciprocals of primes diverges.

Expand the finite Euler product into convergent geometric series. Every positive integer $n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, hence $P(x)\to 0$. If the sum of reciprocal primes were finite, then the inequality $-\log (1-t)\le 2t$ for $0\le t\le 1/2$ would giveuniformly in $x$, contradicting $P(x)\to 0$. Thus the sum of prime reciprocals diverges. Unique factorization justifies the expansion; no estimate for the distribution of primes is required. This is the Euler proof that the sum of reciprocals of primes diverges.

For $x\ge 2$, expand the finite Euler product as a product of geometric series:Every integer $1\le n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, and therefore

Suppose instead that ${\sum }_{p}1/p$ converged. For $0<t\le 1/2$,Since ${\sum }_{p}1/p^2$ converges, this would makeconverge to a finite limit, contradicting the divergence of $P(x)$. Hence the Euler proof that the sum of reciprocals of primes diverges gives