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Sieve an interval of length $z$ by the prime two and the primes congruent to three modulo four, with local densities $g(p)=1/p$. Take $D=z^{1/2}$ and $L=\sqrt{D}$. The truncated Euler-product lower bound, the Mertens first theorem and the reciprocal-prime sum in residue class one modulo four give $J\gg \sqrt{\log z}$. The interval counts have bounded remainders, whose sum is $O(D{\log }^{2}D)$ by the summatory bound for three to the prime omega. Hence the number surviving is $O(z/\sqrt{\log z})$, uniformly in the location of the interval.

The prime sum ${\sum }_{p\le y}(\log p)/p=\log y+O(1)$ is the first Mertens theorem. It controls the mean logarithmic size in a truncated Euler-product lower bound.

Let $A(x)$ count primes $p\le x$ for which $p+2$ is also prime. We prove $A(x)\ll x/{\log }^{2}x$ using a quadratic Selberg sieve, then apply Abel summation.

For $F(n)=n(n+2)$, each odd prime has exactly two roots modulo that prime. The Chinese remainder theorem gives the polynomial root density in a sievefor odd squarefree integers $d$. Put $R=x^{1/8}$ and choose the optimizing real Selberg sieve weights ${\lambda }_d$, supported on these $d\le R$, with ${\lambda }_1=1$. If $p,p+2$ are prime and $p>R$, their weight $({\sum }_{d\mid F(p)}{\lambda }_d{)}^2$ is one. It is nonnegative for every other integer. Expanding the square therefore givesThe Selberg sieve diagonalization makes the main quadratic form $1/J(R)$, whereThe optimal Selberg weights have modulus at most one. Since ${\rho }_F([d,e])\le [d,e]\le R^2$, the remainder is at most $O(R^4)$.

For completeness, the lower bound on $J$ is where the second logarithm enters. Restrict to primes $3\le p\le y=R^{1/8}$ and put $k(p)=2/(p-2)$. The full product $W={\prod }_{3\le p\le y}(1+k(p))$ is $\asymp {\log }^{2}y$, since its logarithm is $2{\sum }_{3\le p\le y}1/p+O(1)$ and the Mertens second theorem applies. Give the associated squarefree divisor $d$ probability $k(d)/W$. Prime inclusion probabilities are $2/p$, so the Mertens first theorem givesfor large $R$. The Markov inequality shows that at least half the weight has $d\le R$. This proves the truncated Euler-product lower bound $J(R)\gg W\gg {\log }^{2}R$. Consequently the twin-prime upper bound from a quadratic sieve isFinally, Abel summation givesThe tail is dominated by ${\int }_3^{\infty }dt/(t{\log }^{2}t)<\infty$. Thus reciprocal-sum convergence from a counting bound proves convergence of the first reciprocal sum. The second is smaller termwise. We have proved Brun's theorem:Counting each twin prime just once changes at most a finite repetition and gives the same conclusion.

Take sieve level $D=x^{1/3}$ and coefficient cutoff $L=\sqrt{D}=x^{1/6}$. Let $\mathcal{P}$ be the finite set of odd primes at most $L$. Give $n=p+2$ one unit of weight for each prime $p\le x$. For an odd squarefree integer $d$ on these primes, the sieve distribution isThe residue $-2$ is coprime to every such $d$, and $g(p)=1/(p-1)\in (0,1)$ for all permitted primes. Values of $g$ outside the permitted set can be chosen arbitrarily in $(0,1)$ at primes and extended multiplicatively on squarefree integers.

For the Selberg sieve normalizing sum, $k(p)=g(p)/(1-g(p))=1/(p-2)$ andHere is a direct proof of this lower bound. Restrict the Euler product to $3\le p\le y=L^{1/4}$. The Mertens first theorem gives ${\sum }_{3\le p\le y}\log p/(p-1)=\log y+O(1)$, which is at most $\frac{1}{2}\log L$ for large $L$. The Mertens second theorem givesThe truncated Euler-product lower bound, with $\theta =1/2$, now gives $J\ge W/2\gg \log L$.

Use the minimizing Selberg sieve weights from Question 3. Their absolute values are at most one, so the Selberg least-common-multiple weights satisfy $|{\lambda }_d|\le 3^{\omega (d)}$. Parts (a) and (b) control the remainder:The first factor uses the Bombieri–Vinogradov theorem; $D=x^{1/3}$ is below $x^{1/2}/{\log }^{B}x$ for all sufficiently large $x$. The second uses the permitted ${\sum }_{q\le D}{9}^{\omega (q)}/\varphi (q)\ll {\log }^{9}D$. The PDF has the exponent nine, which the TeX transcribes incorrectly. Choose $A=14$ to obtain $O(x/{\log }^{3}x)$.

Every twin prime pair with $p+2>L$ survives the finite sieve, while the pairs with $p+2\le L$ number at most $L$. The Selberg upper-bound sieve thus givesTherefore the twin-prime count is $O(x/{\log }^{2}x)$. The finite cutoff on the forbidden primes is essential: sieving by all odd primes would also discard the large prime values $p+2$ that we are trying to count.

In the level-$D$ convention used here, the Selberg upper-bound sieve states that a nonnegative sequence with sieve distribution $A_d=Xg(d)+r(d)$ satisfiesIt follows by minimizing the quadratic form with the Selberg sieve weights and using the Selberg least-common-multiple weights bound $|{\lambda }_d|\le 3^{\omega (d)}$.

Sieve the interval $x<n\le x+z$ by two and by the primes congruent to three modulo four, up to $L=\sqrt{D}$. The interval divisor counts giveThus $X=z$, $g(p)=1/p$, and $r(d)=O(1)$ uniformly in the location of the interval. Take $D=z^{1/2}$ and $L=z^{1/4}$.

To bound $J$, let $y=L^{1/4}$ and form the Euler product over the forbidden primes at most $y$. The given reciprocal-prime sum in residue class one modulo four, subtracted from the Mertens second theorem, yieldsMoreover, the Mertens first theorem bounds the logarithmic mean by ${\sum }_{p\le y}(\log p)/p=\log y+O(1)\le \frac{1}{2}\log L$ for large $L$. The truncated Euler-product lower bound gives $J\ge W/2\gg \sqrt{\log z}$.

The full-range summatory bound for three to the prime omega bounds the error by $O(D{\log }^{2}D)=O(\sqrt{z}{\log }^{2}z)$, which is $O(z/\sqrt{\log z})$. Every integer whose prime factors are all congruent to one modulo four survives this finite sieve. Therefore the half-dimensional interval sieve givesThe constants are independent of $x$. Enlarging them covers the bounded range of $z$ before the asymptotic product estimates apply, so the result holds throughout $1000\le z\le x$.

For $F(n)=n(n+2)$, every odd prime has two forbidden residue classes. Use Selberg sieve weights supported on odd squarefree integers $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J={\sum }_{d\le R}{\prod }_{p\mid d}2/(p-2)$. The truncated Euler-product lower bound applied to primes up to $R^{1/8}$ gives $J\gg {\log }^{2}R$: the mean logarithmic divisor size is $2{\sum }_{3\le p\le R^{1/8}}(\log p)/p<\frac{1}{2}\log R$, by the Mertens first theorem, and the full product is $\asymp {\log }^{2}R$, by the Mertens second theorem. The optimal Selberg weights have modulus at most one, so the remainder is at most $O({\sum }_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.

Apply the Selberg upper-bound sieve to the numbers $p+2$ with $p\le x$ prime. For odd squarefree integers $d$, the local count is $\pi (x;d,-2)$ and the density is $1/\varphi (d)$. With sieve level $D=x^{1/3}$, the normalizing sum is $\gg \log x$ by the truncated Euler-product lower bound. The Selberg least-common-multiple weights have absolute values at most $3^{\omega (d)}$. The weighted arithmetic-progression error bound and Bombieri–Vinogradov theorem make their total error $O(x/{\log }^{3}x)$. Consequently the number of twin prime pairs with smaller member at most $x$ is $O(x/{\log }^{2}x)$.